# Improper Integrals Comparison Test

Integrals over unbounded intervals. Improper integrals (Sect. Let f and g. Show graphically that the benchmark you chose satisfies the conditions of the Comparison Test. Comparison tests for convergence. The book gives students the prerequisites and tools to understand the convergence, principal value, and evaluation of the improper/generalized Riemann integral. , find the values of p for which this integral converges, and the values for which it diverges. Calculus of parametric curves. Comparison Tests. Math 185 - Calculus II. This articles needs some examples, better references, a discussion of the relation with other convergence tests, an explanation of the comparison test for improper integrals, and so forth. We want to prove that the integral diverges so if we find a smaller function that we know diverges the area analogy tells us that there would be an infinite amount of area under the smaller function. (right) Comparison of test prediction accuracy when using ourIDKkernel to a numerical estimation of the kernel integral using random features, as a function of the number of features used for estimation. Partial Fractions. Does Z 1 0 1 p 1 x2 dxconverge? Notice that the function p1 1 x2 has a vertical asymptote at x= 1, so this is an improper integral and we will need to consider the. Loading Unsubscribe from Safet Penjic? Lecture 2 - Improper integrals (cont. 1 x2 3 + 3 4. Problem 4 (15 points) For p >0 consider the improper integral Derive the p -test for these integrals, i. Where k is going to be some positive number. The comparison test. However, for x>1 we have x2 >x, so that 1/ex2 > 1/ex. Improper integrals - part 2 - integrals with integrand undefined at an endpoint [video; 21 min. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Select the second example from the drop down menu, showing Use the same guidelines as before, but include the exponential term also: The limit of the ratio seems to converge to 1 (the "undefined" in the table is due to the b terms getting so small that the algorithm thinks it is dividing by 0), which we can verify: The limit comparison test says that in this. aa g x dx f x dx ³³ ff When we cannot evaluate an integral directly, we first try to determine whether it converges or diverges by comparing it to known integrals. Frequently we aren't concerned along with the actual value of these integrals. Theorem 1 (Comparison Test). We are covering these tests for definite integrals now because they serve as a model for similar tests that check for convergence of sequences and series -- topics that we will cover in the next chapter. For example, one possible answer is BF, and another one is L. Partial Fractions 32 1. Improper integrals are definite integrals where one or both of the boundaries is at infinity, or where the integrand has a vertical asymptote in the interval of integration. Page 632 Numbers 74. By deﬁnition, these integrals can only be used to compute areas of bounded regions. Convergence test: Direct comparison test Remark: Convergence tests determine whether an improper integral converges or diverges. This is , ∃ a psoitive number ,independent of ,such that ∫ + < M, 0< < −. If Maple is unable to calculate the limit of the integral, use a comparison test (either by plotting for direct comparison or by limit comparison if applicable). 7 Review of limits at infinity:-2 -1 0 1 2 2 4 6 x y y ex 1 2 3-4-2 0 2 x y Use the above comparison test to determine whether 1 1 x2 5. State the Comparison Test. Solutions; List of Topics that may. That means we need to nd a function smaller than 1+e x. Give a reasonable "best" comparison function that you use in the comparison (by "best, we mean that the comparison function has known integral convergence properties, and is a reasonable upper or lower bound for the integrand we are evaluating). Consider the following These are all improper because the function being integrated is not finite at one of the limits of integration. I Examples: I = Z ∞ 1 dx xp, and I = Z 1 0 dx xp I Convergence test: Direct comparison test. 10) Activity: Choosing a Technique of Integration EWA 5. Since the integral has an infinite discontinuity, it is a Type 2 improper integral. Integral Test: If f(x) is a decreasing positive function from [1,∞) to [0,∞), then the series P f(n) converges if and only if the improper integral R ∞ 0 f(x)dx converges (i. Review Problems from your textbook: Integration plus L'Hospital's Rule page 579 #1-15, 33-37, 73-87 odds. In such cases the following test is useful. To be convincing, we need to find a series with smaller terms whose sum diverges. give me a formal proof of comparison test for improper integrals. The rst of these is the Direct Comparison Test (DCT). 7) I Integrals on inﬁnite. Comparison Tests. The key tools are the Comparison Test and the P-Test. Comparison Test which requires the non negativity of relevant functions, we shall consider instead f(x) and g(x). Int[1 to infinity] (sin^2 x)/x^3 dx. We work through several examples for each case and provide many exercises. Improper. We can summarize this line of argumentation with the following: Theorem. Comparison test: Suppose f and g are continuous with f(x) g(x) 0, for x a. via integration by parts with u = t/2, dv = sin(2t) dt = π/8. Improper Integrals. NEET: 45 Day Crash Course | Day 45 Last Day | Alcohols, Phenols, Ethers | Unacademy NEET | Anoop Sir Unacademy NEET 117 watching Live now. Improper Integral. The concepts used include regular inequalities as well as the. There are 2 types of comparison tests: direct comparison and limit comparison. This type of integral may look normal, but it cannot be evaluated using FTC II, which requires a continuous integrand on $[a,b]$. I thought about splitting the integral and using the direct comparison test, but I'm not sure what Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. One way to determine the convergence of an improper integral is by comparing it to other integrals that we do know the convergence of. Using the Comparison Test from Calculus, we determine whether several improper integrals in the question converge or diverge. Step 2: Click the blue arrow to submit. integral(1 to infinity) ln(x) dx/x^(1. This chapter has explored many integration techniques. Convergence tests for improper integrals. We have two main tests that we can use to establish convergence or divergence -- the Direct Comparison Test and the Limit Comparison Test. Although we state it for Type 1 integrals,. The comparison test Identify an improper integral. Ok so from this test i came to the conclusion that the integral of sinx/lnx converges. Here I give a quick idea of what the direct comparison test for improper integrals and use it show whether an improper integral converges or. Z 1 1 1 (2 x+ 1)3 dx 2. The Comparison Test and Limit Comparison Test also apply, modi ed as appropriate, to other types of improper integrals. A similar statement holds for type 2 integrals. Comparing Improper Integrals. 2 The Integral Test and p-series test File WS 11. integral toolbox. 14 Improper integrals While the rst integral on the right-hand side diverges, the second one converges, as can be proved by the same procedure as above. The direct comparison test is a simple, common-sense rule: If you’ve got a series that’s smaller than a convergent benchmark series, then your series must also converge. There are two we will talk about here. The Limit Comparison Theorem for Improper Integrals Limit Comparison Theorem (Type I): If f and g are continuous, positive functions for all values of x, and lim x!1 f(x) g(x) = k Then: 1. Otherwise the integrals are divergent. Here is the statement: Limit Comparison Test: Suppose f(x) and g(x) are continuous and positive for all x 2[a;1). Here I give a quick idea of what the direct comparison test for improper integrals and use it show whether an improper integral converges or. Z 1 1 1 x+ 3ln(x) dx 6. If you're behind a web filter, please make sure that the domains *. Often we aren't concerned with the actual value of these integrals. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. ∫∞ 1 1 x3+1dx Solution. improper integrals (comparison theorem) 5 $\mu$ test for convergence of improper integral of first kind. ) - Duration: 1:02:59. Smith , Founder & CEO, Direct Knowledge. integral 1/(x^2+3) from 1 to infinity 3. Comparison Test: If 0 < a n ≤ b n and P b n converges, then P a n also converges. \\int \\frac{2+ e^{-x} dx}{x} from 1 to infinity Wolframalpha tells me this integral diverges, now i just need to know what to compare it to. Give a reasonable "best" comparison function that you use in the comparison (by "best, we mean that the comparison function has known integral convergence properties, and is a reasonable upper or lower bound for the integrand we are evaluating). Here’s an example. Comparison Test for Improper Integrals In some cases, it is impossible to find the exact value of an improper integral, but it is important to determine whether the integral converges or diverges. Choose "Evaluate the Integral" from the topic selector and click to. Solutions; Summary of the convergence tests from Chapter 11 (ht Libby Runte). Review Problems from your textbook: Integration plus L'Hospital's Rule page 579 #1-15, 33-37, 73-87 odds. if 0 < k < 1, then Z 1 a g(x)dx converges Z 1 a f(x)dx converges 2. ∫∞ 4 e−y y dy Solution. Comparison Test for Integrals Comparison Test for Integrals Theorem If f and g are continuous functions with f(x) g(x) 0 for x a, then (a) If R 1 a f(x)dx is convergent, then R 1 a g(x)dx is convergent. Hence the Comparison test implies. The limit comparison test gives us another strategy for situations like Example 3. The last convergence tool we have is the comparison test. For Example, One Possible Answer Is BF, And Another One Is L. Hint: 0 < E^-x Lesserthanorequalto 1|for X Greaterthanorequalto 1. Question: For Each Of The Improper Integral Below, If The Comparison Test Applies, Enter Either A Or B Followed By One Letter From C To K That Best Applies, And If The Comparison Test Does Not Apply, Enter Only L. In fact, 1 t+t3 < 1 (bigger denominator = smaller fraction), and the p-type integral dt 0t ⌠1 ⌡ ⎮ converges, so by the comparison test, this integral also converges. Improper Integrals (with Examples) May 2, 2020 January 8, 2019 Categories Formal Sciences , Mathematics , Sciences Tags Calculus 2 , Latex By David A. Read lecture notes, page 1 to page 3; An integral with an infinite upper limit of integration to be evaluated. Iff(x) is continuous on ( —x, 00), then f(x) dx + — lim 2. For x > e we surely have Then also and the Comparison test is inconclusive. I Examples: I = Z ∞ 1 dx xp, and I = Z 1 0 dx xp I Convergence test: Direct comparison test. Improper Integral: Comparison Test. If is convergent then is. On the other hand, if L 0, then we must compare f x to a suitable comparison function in order to determine the convergence or divergence of the. The last convergence tool we have is the comparison test. Active 3 years, 5 months ago. Comparison Test and Limit Comparison Test. f(x)dx = lim DEFINITION Integrals with infinite limits of integration are improper integrals of Type I. Z 4 0 1 (x 22) dx Comparison Test for Integrals Theorem If fand gare continuous functions with f(x) g(x) 0 for x a, then (a) If R 1 a f(x)dxis convergent, then R 1 a g(x)dxis convergent. And if your series is larger than a divergent benchmark series, then your series must also diverge. A similar statement holds for type 2 integrals. Improper at x = 0, where the t is much larger than the t3, so this “looks like” the p-type dt 0t ⌠1 ⌡ ⎮ which converges since p < 1. This means the limits of integration include $\infty$ or $-\infty$ or both. Arc Length: Week #5: Feb 10 - 14 The Comparison Tests. Let gðxÞ A 0 for all x A a, and suppose that. LIATE; Trig Integrals. Improper Integrals - Recognizing an improper integral and using a value of an integral to find other values. Hence it is convergent by comparison test. ∫∞ 3 z2 z3−1dz Solution. In this section we use a different technique to prove the divergence of the harmonic series. The improper integral. ∫∞ 1 z−1 z4+2z2dz Solution. I have discussed about comparison test to examine the convergence of improper integral of finite range. The question is simple: Decide whether the following integral converges: There are essentially three methods to answer this question: 1. The Integral Test (14 minutes, SV3 » 41 MB, H. The comparison test. f ( x) = e - x 2 ( ln x) 2. For Example, One Possible Answer Is BF, And Another One Is L. integral e^-x/x^2 from 1 to infinity 2. If f(x) has a tail you must do further tests. integral 1/(x^2+3) from 1 to infinity 3. We have seen that the integral test allows us to determine the convergence or divergence of a series by comparing it to a related improper integral. There is a discontinuity at $$x = 0,$$ so that we must consider two improper integrals: \[{\int\limits_{ - 2}^2 {\frac{{dx}}{{{x^3}}}} } = {\int\limits. 5 Direct Comparison Test. So it seems plausible that results which tell you about the convergence of improper integrals would have. We can handle only one 'problem' per integral. And in order to handle this, the thing that I need to do is to check the integral from 0 up to N, e^(-kx) dx. Thus the integral converges. Trig Substitution. Hence it is convergent by comparison test. Both the Direct and Limit Comparison Tests were given in terms of integrals over an infinite interval. with bounds) integral, including improper, with steps shown. (Note this is a positive number when a is negative, so this answer makes sense. Consider the improper integral ∫ 1 1 1 xp dx: Integrate using the generic parameter p to prove the integral converges for p > 1 and. One cannot apply numerical methods like LEFT or RIGHT sums to approximate the value of such. Page 631 Number 48. Improper integrals. The Organic Chemistry Tutor 409,621 views 20:18. integral 1/(x^2+3) from 1 to infinity 3. Finally we need to discuss a second type of improper integral. NOVA COLLEGE-WIDE COURSE CONTENT SUMMARY. Use the Comparison test. Convergence test: Direct comparison test Remark: Convergence tests determine whether an improper integral converges or diverges. Improper Integrals. {Hint: for implies that. Proof: harmonic series diverges. (b) If R 1 a g(x)dxis divergent, then R 1 a f(x)dxis divergent. The instructions were to show that the integral from 1 to infinity of sinx/sqrt(x) converges. Alternating series test for convergence. Improper Integrals ( Part 5 ) Infinite Limits. The p-Test. The improper integral Z ∞ 1 1. Here, the answer just provides the base function that the problem most closely resembles. 1 dx p 1 if p > 1 1 x p diverges if p < 1 Example 2 On the surface, the graphs of the last three examples seem very much alike and there is nothing to suggest why one of the areas should be infinite and the other two finite. The following test will be helpful. Improper Integrals: Solutions Friday, February 6 Bounding Functions More Comparison Test Pretty much the only function you care to compare things to here is 1=x (or, more generally, 1=xp). (right) Comparison of test prediction accuracy when using ourIDKkernel to a numerical estimation of the kernel integral using random features, as a function of the number of features used for estimation. 138 Improper Integrals M. In fact, 1 t+t3 < 1 (bigger denominator = smaller fraction), and the p-type integral dt 0t ⌠1 ⌡ ⎮ converges, so by the comparison test, this integral also converges. Usually it's more important to know whether an improper integral converges than it is to know what it converges to. You should not extend the inequality to $\int_{9}^{\infty } \frac{1}{x}dx$ because it's divergent and a convergent integral is always less than a divergent integral, so it's of no use. if k = 0, then Z 1 a g(x)dx converges =) Z 1 a f(x)dx converges 3. Improper Integrals ( Part 5 ) Infinite Limits. NOVA COLLEGE-WIDE COURSE CONTENT SUMMARY. Note the analogy with the geometric series if r ¼ e t so that e tx ¼ rx. Example: Determine whether the series X∞ n=1 1 √ 1+4n2 converges or diverges. Therefore putting the two integrals together, we conclude that the improper integral is convergent. improper integral convergent. Improper Integrals Convergence and Divergence, Limits at Infinity & Vertical Asymptotes, Calculus - Duration: 20:18. Improper Integrals-Section 7. Let us evaluate the corresponding improper integral. {comparison} 6. Integrals over bounded intervals of functions that are unbounded near an endpoint. Comparison test for convergence/divergence. Limit test: Let and be two positive function defined on. Use the Comparison Theorem to decide if the following integrals are convergent or divergent. However, we know that continuity is "almost necessary" to integrate in the sense of Riemann, so teachers do not worry too much about the minimal assumptions under which the theory can be taught. Comparison Test which requires the non negativity of relevant functions, we shall consider instead f(x) and g(x). Z 1 1 1 2x2 x dx 3. 9: Feb 1: Feb 2 Week 4 Feb 3 Areas of Regions and Volumes of Solids of Revolution(6. Improper Integrals. This is because Z b · f(t)dt = F(b)−F(·) and F(b) is increasing and F(b) has a limit if and only if F(b) is bounded. In fact, 1 t+t3 < 1 (bigger denominator = smaller fraction), and the p-type integral dt 0t ⌠1 ⌡ ⎮ converges, so by the comparison test, this integral also converges. If is convergent then is. (a) integrate limit 6 to 7 x/x-6 dx Since the integral has an infinite interval of integration, it is a Type 1 improper integral. Improper Integrals. Spending a class hour to introduce students to the concepts in the flipped class lesson is found to be helpful. I thought about splitting the integral and using the direct comparison test, but I'm not sure what Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The Limit Comparison Test. Comparison Test and Limit Comparison Test. Comparison, Limit comparison and Cauchy condensation tests. We already know the second integral is finite, so the first one has to be finite as well. 9: Feb 1: Feb 2 Week 4 Feb 3 Areas of Regions and Volumes of Solids of Revolution(6. And if your series is larger than a divergent benchmark series, then your series must also diverge. Usually it's more important to know whether an improper integral converges than it is to know what it converges to. Part 3: Tests for Convergence and Divergence. 16 Improper Integrals. First, try the comparison. Unit 8 (Chapter 8): Sequences, l'hospital, & improper integrals Limits, Sequences Finding Limits Analytically Comparison Test Notes Extra Practice. Both the Direct and Limit Comparison Tests were given in terms of integrals over an infinite interval. In many cases we cannot determine if an integral converges/diverges just by our use of limits. 6 Show that the improper integral R 1 1 1+x2 dxis convergent. If we look at the other one, and we decide the other one is bursting at the seems, we know it's safe to open ours up. (b) If R 1 a g(x)dxis divergent, then R 1 a f(x)dxis divergent. Select the second example from the drop down menu, showing Use the same guidelines as before, but include the exponential term also: The limit of the ratio seems to converge to 1 (the "undefined" in the table is due to the b terms getting so small that the algorithm thinks it is dividing by 0), which we can verify: The limit comparison test says that in this. Improper Riemann Integrals is the first book to collect classical and modern material on the subject for undergraduate students. You solve this type of improper integral by turning it into a limit problem where c approaches infinity or negative infinity. For Example, One Possible Answer Is BF, And Another One Is L. that the limit is the value of the improper integral. if 0 < k < 1, then Z a 0 g(x)dx converges Z a 0 f(x)dx converges 2. I thought about splitting the integral and using the direct comparison test, but I'm not sure what Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Lecture 4 - Comparison tests for improper integrals Safet Penjic. Discrete calculus-How to integrate discrete functions Numerical ODEs-Using sequences to solve ODEs Numerical integration-Using sequences to solve definite integrals Series-Infinite series as improper discrete integrals Convergence tests 1-Comparison-type tests Convergence tests 2-Geometric series-type tests. UNSOLVED! Use the comparison test to determine if the following integral is convergent or divergent. We will only state these theorems for integrals that are improper at b. Suppose 0 f(x) g(x) for x aand R b a f(x)dxexists for all b>a. MATH1014 Tutorial 8 The comparison test for improper integrals Iff(x) andg(x) arecontinuousandf(x) ≥g(x) ≥0,then a f(x)dxconvergesimplies a g(x)dxconverges. If 0 f(x) g(x) on [a;1), it can be shown that the convergence of R 1 a g(x)dx implies the convergence of R 1 a f(x)dx, and the divergence of R 1 a f(x)dx implies the divergence of R 1 a g(x)dx. Integrals over unbounded intervals. Improper Integrals. Theorem 1 (Comparison Test). We know that 0 1 sin(x) 2 so 0 1 sin(x) x2 2 x2. (a) integrate limit 6 to 7 x/x-6 dx Since the integral has an infinite interval of integration, it is a Type 1 improper integral. I thought about splitting the integral and using the direct comparison test, but I'm not sure what Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Warning: Now that we have introduced discontinuous integrands, you will need to check. BC Calculus Improper Integrals Day 3 Notesheet Name: _____ Sometimes, we cannot find the antiderivative of an integrand of an improper integral. Since G(t) is an increasing function, it follows that a L G(t) L - ε y 0 FIGURE 1 If G(t) is increasing with least upper bound L, then G(t) eventually lies within of L L− < G(y 0. (a) ∫ 0 1 1 2x−1 dx (b) ∫ 1 1 xe x2 dx (c) ∫ 2 0 x−3 2x−3 dx (d) ∫ 1 0 sin d 3. The primary tool in that toolbox is the set of integrals of power functions. ∫∞ 6 w2+1 w3(cos2(w)+1)dw Solution. Frequently we aren't concerned along with the actual value of these integrals. THE INTEGRAL AND COMPARISON TESTS 93 4. (May need to break up integral into several. Express it as a limit and determine whether it converges or diverges; if it converges, find the value. There is a discontinuity at $$x = 0,$$ so that we must consider two improper integrals: \[{\int\limits_{ - 2}^2 {\frac{{dx}}{{{x^3}}}} } = {\int\limits. Trig Substitution. We will only state these theorems for integrals that are improper at b. How do you use basic comparison test to determine whether the given series converges or diverges How do you use the Comparison Test to see if #1/(4n^2-1)# converges, n is going to infinity? See all questions in Direct Comparison Test for Convergence of an Infinite Series. {comparison} 6. This chapter has explored many integration techniques. An improper integral of type 1 is an integral whose interval of integration is infinite. Here is the statement: Direct Comparison Test: Suppose that f(x) and g(x) are continuous functions on the interval [c;1). Integration using Tables and CAS 39 1. Sometimes it is impossible to find the exact value of an improper integral and yet it is important to know whether it is convergent or divergent. Remainder estimation. There are two classes of improper integrals: (1) those in which at least one of the limits of integration is infinite (the interval is not bounded); and (2) those of the type where f (x) has a point of discontinuity (becoming infinite) at x = c, a c b (the function is not bounded). 2020-04-21T09:24:18Z http://oai. The Comparison Test Suppose 0 ≤ f(x) ≤ g(x) for all x ≥ a. This technique is important because it is used to prove the divergence or convergence of many other series. Section 1-9 : Comparison Test for Improper Integrals. Solution 2 (b). Improper Integrals — One Infinite Limit of Integration. is divergent. Improper Integrals In general, To determine if an integral converges or diverges, you can use the Comparison Test described below. Mansi Vaishnani 2. In my assignment I have to evaluate the (improper) integral, by means of the "comparison theorem". Comparison Test for Improper Integrals. o If an > bn for all n and if ba diverges, then an also diverges. Express it as a limit and determine whether it converges or diverges; if it converges, find the value. Comparison Test Suppose [and. Testing Convergence of Improper Integrals. A Comparison Test Sometimes it is not possible to obtain the value of improper integrals. Improper Integrals - Recognizing an improper integral and using a value of an integral to find other values. David Jerison. If 0 f(x) g(x) on [a;1), it can be shown that the convergence of R 1 a g(x)dx implies the convergence of R 1 a f(x)dx, and the divergence of R 1 a f(x)dx implies the divergence of R 1 a g(x)dx. 3 Integral and Comparison Tests The Integral Test: Suppose a function f(x) is continuous, positive, and decreasing on [1;1). The calculator will evaluate the definite (i. Example: Z 1 1 1 p x dx We have the following general result related to the last two examples. Improper integrals are definite integrals where one or both of the boundaries is at infinity, or where the integrand has a vertical asymptote in the interval of integration. An integral is called an improper integral if one of, or both, of the conditions hold: The interval of integration is infinite. Comparison test: Suppose f and g are continuous with f(x) g(x) 0, for x a. Solution 2 (a). {comparison} 4. Evaluate the …. Show that the improper integral. Partial Fractions. And in order to handle this, the thing that I need to do is to check the integral from 0 up to N, e^(-kx) dx. On the interval [1 ;1], we split up the integral into two separate improper integrals. NOVA COLLEGE-WIDE COURSE CONTENT SUMMARY. Note that Z t 1 1 x dx= [lnx]t 1 = lnt!1 as t!1: Hence, R 1 1 1 x dxdiverges. and is clearly finite, so the original integral is finite as well. Warning: Now that we have introduced discontinuous integrands, you will need to check. are all improper because they have limits of integration that involve ∞. {comparison} 8. 2 (Improper Integrals with Inﬁnite Discontinuities) Consider the following three. Convergence test: Direct comparison test Remark: Convergence tests determine whether an improper integral converges or diverges. For each of the improper integrals below, if the comparison test applies, enter either A or B followed by one letter from C to K that best applies, and if the comparison test does not apply, enter only L. Improper Integrals - Recognizing an improper integral and using a value of an integral to find other values. xml 03/07/2013 12:21:24 mchinn [Discussion Draft] [Discussion Draft] March 7, 2013 113th CONGRESS 1st Session Rules Committee Print 113-4 of H. Loading Unsubscribe from Safet Penjic? Lecture 2 - Improper integrals (cont. Example: Determine whether the series X∞ n=1 1 √ 1+4n2 converges or diverges. Hint: 0 < E^-x Lesserthanorequalto 1|for X Greaterthanorequalto 1. with bounds) integral, including improper, with steps shown. In my assignment I have to evaluate the (improper) integral, by means of the "comparison theorem". Practice: Limit comparison test. Both of the limits diverge, so the integral diverges. integral e^-x/x^2 from 1 to infinity 2. Solution to this Calculus Improper Integral problem is given in the video below!. ∫∞ 1 z−1 z4+2z2dz Solution. The p-Test implies that the improper integral is convergent. ∫∞ 4 e−y y dy Solution. Logistic Model problems from Pg414-415. Even though you were instructed not to use the Comparison Test to solve this problem; you can use it to check your result. 4 Day 2 Improper Integrals and the Comparison Test Thursday, January 30, 2020 7:54 AM Notes Page 1. Improper Integral Comparison Test example #9. Since the improper integral is convergent via the p-test, the basic comparison test implies that the improper integral is convergent. This is called a comparison test. (b) Let's guess that this integral is divergent. o If an > bn for all n and if ba diverges, then an also diverges. BC Calculus Improper Integrals Day 3 Notesheet Name: _____ Sometimes, we cannot find the antiderivative of an integrand of an improper integral. Here’s the mumbo jumbo. {comparison} 5. However, we might be able to draw a conclusion about its convergence or divergence if we can compare it to something similar for which we do something. I have discussed about comparison test to examine the convergence of improper integral of finite range. One way to determine the convergence of an improper integral is by comparing it to other integrals that we do know the convergence of. Explain briefly how the test supports your conclusion. For x > e we surely have Then also and the Comparison test is inconclusive. We know that 0 1 sin(x) 2 so 0 1 sin(x) x2 2 x2. Compute integrals of functions with vertical asymptotes. In my assignment I have to evaluate the (improper) integral, by means of the "comparison theorem". We deﬁne Z b a f(t)dt def= lim x→a+ Z b x f(t)dt Similarly, if f(x) is deﬁned on [a,b), Z b a f(x)dx def= lim x→b. Similar tests exist where an Comparison test for integrals with non-negative integrands. Simply look at the interval of integration. Accordingly, some mathematicians developed their own. The integral is a proper integral. Similar comparison tests are true for integrals of the form Rb 1 f(x)dx and for improper integrals with discontinuous integrands. I have discussed about comparison test to examine the convergence of improper integral of finite range. We now develop comparison tests for integrals of the form R1 a f. (Note this is a positive number when a is negative, so this answer makes sense. Topics explored in this course include integrals, Riemann sums, techniques of integration, improper integration, differential equations, and Taylor series. Lecture 25/26 : Integral Test for p-series and The Comparison test In this section, we show how to use the integral test to decide whether a series of the form X1 n=a 1 np (where a 1) converges or diverges by comparing it to an improper integral. It may help determine whether we have absolute convergence, conditional convergence, or neither. If the improper integral is finite. f(x) g(x) = c where cis a postive number. The idea with this test is that if each term of one series is smaller than another, then the sum of that series must be smaller. (2) If R∞ a g(x)dx if divergent then R∞ a f(x)dx is divergent. {Hint: for implies that. ∫ 1 ∞ ln  x x d x ∫ − ∞ 1 x e 2 x d x ∫ 0 2 x 4 − x 2 d x; Use a comparison test to determine if the improper integral ∫ 1 ∞ sin 2  x x 2 d x converges or. {integrate by parts and absolute. Even though one of our bounds is an asymptote, we can use limits to determine the area!. Let's try n^-2: This limit is positive, and n^-2 is a convergent p-series, so the series in question does converge. This comparison of the two types of cloud events suggested that evaporation was the most likely cause of upward droplet fluxes for the smaller droplets (dia13µm) during cloud with variable LWC (σL >0. Definite Integral and The Fundamental Theorem of Calculus. From Calculus. Explain briefly how the test supports your conclusion. Loading Unsubscribe from Safet Penjic? Lecture 2 - Improper integrals (cont. Both 𝑓(𝑥) = 1∕𝑥³ and 𝑔(𝑥) = 1∕𝑥 are non-negative over [1, ∞), but 𝑓(𝑥) is not greater than or equal to 𝑔(𝑥), and thereby we cannot use the comparison test to tell if the improper integral of 𝑓(𝑥) diverges. These improper integrals are called convergent if the corresponding limit exists and divergent if the limit does not exist. but similar versions hold for the other improper integrals. The Comparison Test for Improper Integrals allows us to determine if an improper integral converges or diverges without having to calculate the antiderivative. The Limit Comparison Test Let and be series with positive terms and let If then either both series converge, or they both diverge. The calculator will evaluate the definite (i. 7 Review of limits at infinity:-2 -1 0 1 2 2 4 6 x y y ex 1 2 3-4-2 0 2 x y Use the above comparison test to determine whether 1 1 x2 5. And in order to handle this, the thing that I need to do is to check the integral from 0 up to N, e^(-kx) dx. Use the comparison test to show the integral. ∫∞ 1 z−1 z4+2z2dz Solution. We have for. If these limits exist and are finite then we say that the improper integrals are convergent. 1 x2 3 + 3 4. Integrals over bounded intervals of functions that are unbounded near an endpoint. Suppose f(x) and g(x) are two positive functions with 0 f(x) g(x). Z 1 1 6 x4 + 1 dx Z 1 2 x2 p x5 1 dx Z 1 3 5 + e x. 1 Consider the improper integral Z 1 1 1 x dx. Comparison test: Suppose f and g are continuous with f(x) g(x) 0, for x a. The rst of these is the Direct Comparison Test (DCT). We know that 0 1 sin(x) 2 so 0 1 sin(x) x2 2 x2. , 18 MB] Finding the area between curves. This type of integral may look normal, but it cannot be evaluated using FTC II, which requires a continuous integrand on $[a,b]$. The p-Test. Type 1: Improper Integrals (Infinite Limits of Integration) 1. Page 632 Numbers 74. The comparison test tells u to look at ur original function 1/(1+x^2) and determine if it is larger than the parent function or less than the parent function. And if your series is larger than a divergent benchmark series, then your series must also diverge. Unfortunately, the harmonic series does not converge, so we must test the series again. Alternating Series. Answer to: Use comparison test to determine whether each of the following improper integral is convergent: (a) \ \int^\infty_1 \frac {2\sqrt. Section 1-9 : Comparison Test for Improper Integrals. AP Calculus BC. Comparison test for convergence/diverg. As with integrals on inﬁnite intervals, limits come to the rescue and allow us to deﬁne a second type of improper integral. 5 in (i) so by the comparison test so does the given series R1 the improper integral only exists as a limit – too many. 1 Definition of improper integrals with a single discontinuity; 2. You should not extend the inequality to $\int_{9}^{\infty } \frac{1}{x}dx$ because it's divergent and a convergent integral is always less than a divergent integral, so it's of no use. Comparison theorems. 3 Integral and Comparison Tests The Integral Test: Suppose a function f(x) is continuous, positive, and decreasing on [1;1). Type II Improper Integral Suppose f(x) is deﬁned on the half–open interval (a,b] and assume the integral R b x f(t)dt exists for each x satisfying a < x ≤ b. Improper Integrals De–nition (Improper Integral) An integral is an improper integral if either the interval of integration is There is another comparison test. 7) I Integrals on inﬁnite. Math Help Forum. An improper integral might have two infinite limits. Since the integral has an infinite discontinuity, it is a Type 2 improper integral. The Organic Chemistry Tutor 412,594 views 20:18. Mansi Vaishnani 2. We now discuss two kinds of improper integrals, and show that they, too, can be interpreted as Lebesgue integrals in a very natural way. ∑ n = 2 N 1 n ln n. Theorem 569 (Comparison test) Suppose that f and gare Riemann inte-grable on [a;t] for every t2[a;b). 10) Activity: Choosing a Technique of Integration EWA 5. This is because Z b · f(t)dt = F(b)−F(·) and F(b) is increasing and F(b) has a limit if and only if F(b) is bounded. Use the comparison test to determine if the series. The p-Test implies that the improper integral is convergent. The Organic Chemistry Tutor 409,621 views 20:18. Volumes 52 2. Use the Comparison Test to determine whether the improper integral converges or diverges. Comparison theorems. The limit comparison test. Give a reasonable "best" comparison function that you use in the comparison (by "best, we mean that the comparison function has known integral convergence properties, and is a reasonable upper or lower bound for the integrand we are evaluating). • If you believe the integral converges, ﬁnd a function g(x) larger than f(x), whose integral also. We have for. Solution: a. Theorems 60 and 61 give criteria for when Geometric and $$p$$-series converge, and Theorem 63 gives a quick test to determine if a series diverges. If is divergent, then is divergent. Exercise 3. Compute integrals of functions with vertical asymptotes. Evaluate the following improper integral or show that it diverges: Z 27 0 x1 3 dx x2 3 9: 6. Serioes of this type are called p-series. Both of the limits diverge, so the integral diverges. Let gðxÞ A 0 for all x A a, and suppose that. (Direct Comparison) If for all then if converges, so does. php oai:RePEc:spr:testjl:v:11:y:2002:i:2:p:303-315 2015-08-26 RePEc:spr:testjl article. Compare to a geometric series. School: Arizona State University Course: MAT 266 Techniques of integration and improper integrals, sections 5. Improper Integrals - Recognizing an improper integral and using a value of an integral to find other values. Midterm 1 from Spring 2014. There are several keys to handling improper integrals. Smith , Founder & CEO, Direct Knowledge. Answer: The first thing we can do to try to answer this question is to graph the function and figure out what the region looks like. LIATE; Trig Integrals. An improper integral is a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration. We can use limits to integrate functions on unbounded domains or functions with unbounded range. Homework Statement Use the Comparison Theorem to determine whether the integral is convergent or divergent. not infinite) amount of area. is convergent and. Z 1 1 6 x4 + 1 dx Z 1 2 x2 p x5 1 dx Z 1 3 5 + e x. Solutions of boundary-value problems in mathematical physics are written as multiple improper integrals with an unbounded function as integrand. For each of the improper integrals below, if the comparison test applies, enter either A or B followed by one letter from C to K that best applies, and if the comparison test does not apply, enter only L. Instructions for Exercises 1-12. Which of the two functions, , has a thinner tail? c. IMPROPER INTEGRALS If a is positive, then lim b!1 eab = 1, so the integral diverges. The integral test helps us determine a series convergence by comparing it to an improper integral, which is something we already know how to find. {comparison} 4. The actual test states the following: If \(f(x) \ge g(x) \ge 0. Like the integral test, the comparison test can be used to show both convergence and divergence. NOVA COLLEGE-WIDE COURSE CONTENT SUMMARY. Use the comparison test to show the integral. I We will of course make use of our knowledge of p-series and geometric series. Z ∞ a f(x)dx converges if Z ∞ a g(x)dx converges. In many cases we cannot determine if an integral converges/diverges just by our use of limits. There are versions that apply to improper integrals with an infinite range, but as they are a bit wordy and a little more difficult to employ, they are omitted from this text. Oh golly, we have a theorem for this! It’s called the “Comparison Test for Improper Integrals,” and it goes like this: Let $f$ and $g$ be. 5 Direct Comparison Test. In the case of the integral test, a single calculation will confirm whichever is the case. integralx/(sqrt(x^6+3)) from 1 to infinity 4. Comparison test for convergence/diverg. For instance, the integrals. Type 1 - Improper Integrals with Infinite Intervals of Integration. • If you believe the integral converges, ﬁnd a function g(x) larger than f(x), whose integral also. The question is simple: Decide whether the following integral converges: There are essentially three methods to answer this question: 1. Math 185 - Calculus II. 3 For p6= 1. Improper Integrals Convergence and Divergence, Limits at Infinity & Vertical Asymptotes, Calculus - Duration: 20:18. The Comparison Test suggests that, to examine the convergence of a given improper integral, we may be able to examine the convergence of a similar integral. Use the Comparison test. Hence the Comparison test implies. converges or diverges. Spending a class hour to introduce students to the concepts in the flipped class lesson is found to be helpful. and is clearly finite, so the original integral is finite as well. I Convergence test: Limit comparison test. For large n (in which case the 1 in the numerator doesn't matter), this series is approximately equal to the divergent p-series 1/n 1/2, so we can use that for the limit comparison test, in which we'll guess that the series is divergent. This type of integral may look normal, but it cannot be evaluated using FTC II, which requires a continuous integrand on $[a,b]$. g(x)=1/xp,wherep is any real number: Z 1 1 1 xp dx is (< 1 if p>1. This is called a comparison test. integral 1/(x^2+3) from 1 to infinity 3. Tutorial Exercises for Section 8. Let’s try to reduce our work by using a comparison test. Logistic Model problems from Pg414-415. {finite limit comparison} 7. Z ∞ a g(x)dx diverges if Z ∞ a f(x)dx diverges. Then we have the following statements about convergence of. New Resources. Theorem: Comparison Test. Theorem 1 (Comparison Test). Typical comparison functions. Read lecture notes, page 1 to page 3; An integral with an infinite upper limit of integration to be evaluated. 10 (Asymptotic comparison test) Suppose the. The integral test bridges the two notions. Comparison Test: If 0 < a n ≤ b n and P b n converges, then P a n also converges. ∫∞ 3 z2 z3−1dz Solution. g(x)=1/xp,wherep is any real number: Z 1 1 1 xp dx is (< 1 if p>1. This skill is important for determining convergence of improper integrals, and it will become important again when we study convergence of series. comparison test. New Resources. Exercise 3. A necessary and sufficient condition for the convergence of the improper integral ∫ at 'a' where f is positive in [a, b]. Z 1 1 1 x2 3x+ 2 dx 1. 7) I Integrals on inﬁnite. -----Edit: No matter how I try to remember, I always get the terms 'indefinte integral' and 'improper integral. #int_1^infty 1/x^5 dx#. Improper Integrals: Part 2 Sometimes it is di cult to nd the exact value of an improper integral, but we can still know if it is convergent or divergent by comparing it with some other improper integral. 1 Comparison Test If f(x) g(x) 0, then the area under gis smaller than the area under f. LIMIT COMPARISON TEST FOR IMPROPER INTEGRALS UM Math 116 February 13, 2018 The basic question about improper integrals in Math 116 is whether or not they converge. For example, one possible answer is BF, and another one is L. Homework Statement f(x) ~ g(x) as x→a, then \\frac{f(x)}{g(x)} = 1 (that is. Use the Comparison Test to determine whether the improper integral converges or diverges. The integral comparison test involves comparing the series you’re investigating to its companion improper integral. (2) If R 1 a f(x)dx= 1then R 1 a g(x)dx= 1. 1 Improper Integrals with Infinite Limits of Integration. f(x) g(x) = c where cis a postive number. The comparison test for integrals may be stated as follows, assuming continuous real-valued functions f and g on with b either or a real number at which f and g each have a vertical asymptote: If the improper integral converges and for , then the improper integral also converges with. But as c goes to one, ln(c) goes to. The idea of this test is that if the limit of a ratio of sequences is 0, then the denominator grew much faster than the numerator. This is the currently selected item. Introduction to improper integrals. Arc Length: Week #5: Feb 10 - 14 The Comparison Tests. Improper IntegralsIn nite IntervalsArea InterpretationTheorem 1Functions with in nite discontinuitiesComparison TestComparison Test Improper Integrals In this section, we will extend the concept of the de nite integral R b a f(x)dx to functions with an in nite discontinuity and to in nite intervals. Click the following button to go to the top of the page, Navigation Menu, and navigate within this course:. One cannot apply numerical methods like LEFT or RIGHT sums to approximate the value of such. (Direct Comparison) If for all then if converges, so does. Improper at x = 0, where the t is much larger than the t3, so this "looks like" the p-type dt 0t ⌠1 ⌡ ⎮ which converges since p < 1. The following functions can often be used as the comparison function g(x) when applying the comparison tests. We studied improper integrals a while back, and we learned that, if f ≤ g on the interval (c. Improper integrals. The Limit Comparison Test Let and be series with positive terms and let If then either both series converge, or they both diverge. However, we might be able to draw a conclusion about its convergence or divergence if we can compare it to something similar for which we do something. Basic Improper Integrals ; Comparison Theorem ; The Gamma Function; 2 Improper Integrals Definition. If Maple is unable to calculate the limit of the integral, use a comparison test (either by plotting for direct comparison or by limit comparison if applicable). 361 Proof of Theorem 10. Show graphically that the benchmark you chose satisfies the conditions of the Comparison Test. Subsection 1. For Example, One Possible Answer Is BF, And Another One Is L. Continues the study o f calculus of algebraic and transcendental functions i ncluding rectangular, polar, and parametric graphing, indefinite and definite integrals, methods of integration, and power series along with applications. Decide whether each integral is convergent or divergent. Improper Integrals Example Example 1 Example 2 Example 3 Example 5 Example 6 Example 7 Comparison Theorem for Improper Integrals Example 9 Example 10 Homework In the textbook, Section 5. [CALC II]Comparison Test for Improper Integrals. Example: Prove that Z ∞ 0 e. (170120107169) Guided By- Prof. And in order to handle this, the thing that I need to do is to check the integral from 0 up to N, e^(-kx) dx. Limit comparison test. Homework Statement Use the Comparison Theorem to determine whether the integral is convergent or divergent. Improper integrals; Chat × After completing this section, students should be able to do the following. Let’s try to reduce our work by using a comparison test. Worked example: limit comparison test. Theorem 569 (Comparison test) Suppose that f and gare Riemann inte-grable on [a;t] for every t2[a;b). The idea behind the comparison tests is to determine whether a series converges or diverges by comparing a given series to an already familiar (Direct Comparison Test) Suppose that $\sum a_n$ and $\sum b_n$ are series with positive terms. The first example is the integral from 0 to infinity of e^(-kx) dx. Thus this is a doubly improper integral. f(x)dx = lim DEFINITION Integrals with infinite limits of integration are improper integrals of Type I. Direct Comparison Test for ( Improper ) Integrals. 1 1 1 2 2b 2 2 Improper Integrals These examples lead us to this theorem. Homework Statement determine the value of the improper integral when using the integral test to show that \\sum k / e^k/5 is convergent. IMPROPER INTEGRALS. Differentiation and integration of power series. 4 Improper Integrals and L'Hôpital's Rule. Type 2: Discontinous Integrands. 5 in (i) so by the comparison test so does the given series R1 the improper integral only exists as a limit – too many. Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums. Instead we might only be interested in whether the integral is convergent or divergent. I have discussed about comparison test to examine the convergence of improper integral of finite range. if k = 0, then Z 1 a g(x)dx converges =) Z 1 a f(x)dx converges 3. However, we know that continuity is "almost necessary" to integrate in the sense of Riemann, so teachers do not worry too much about the minimal assumptions under which the theory can be taught. Worked example: limit comparison test. Improper Riemann Integrals is the first book to collect classical and modern material on the subject for undergraduate students. Section 1-9 : Comparison Test for Improper Integrals. Numerical. f(x) g(x) = c where cis a postive number. [T] A fast computer can sum one million terms per second of the divergent series ∑ n = 2 N 1 n ln n. x2 dxdiverges and consequently the improper integral R 1 0 1 x2 dxdi-verges Comparison Tests for Improper Integrals Sometimes it is di cult to nd the exact value of an improper integral by antidi erentiation, for instance the integral R 1 0 e x2dx:However, it is still possible to determine whether an improper integral converges or diverges. Then we have If is convergent, then is convergent. New Resources. Improper integrals - part 2 - integrals with integrand undefined at an endpoint [video; 21 min. asked Jun 7, 2019 in Mathematics by GipsyKing. o If an, < bn, for all n and if E bn, converges, then an also converges. We could try to compare the given function to some powers that have convergent integrals to infinity, but this is not possible.
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# Sum of reciprocals of Fibonacci numbers convergence

I am trying to prove the convergence/divergence of the series´ $$\sum_{i=1}^\infty \frac{1}{F_n} = 1+1+\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{8}+...$$ ${F_n}$ being the Fibonacci sequence.
The Fibonacci sequence is defined without recursion by: $${F_n}=\frac{\phi^n-(-\phi)^{-n}}{\sqrt{5}} \quad\land\quad\phi=\frac{1+\sqrt{5}}{2}$$ I have tried to prove its convergence with the Root Test and the Ratio Test because of the $n$ exponent but can't manage to do it because of the difference in the fraction.

Can anyone help me? Thank you

• Both the root and ratio tests work. – Lord Shark the Unknown May 23 '17 at 19:42
• I'm sure they do, but I'm having problems in the limit calculation because of the difference. If someone could hint me in the right direction it would be great! Thank you. – Luis Dias May 23 '17 at 19:45
• Use the usual trick of pulling out the dominant term: $F_n=\phi^n a_n$ where $a_n\to1/\sqrt5$. – Lord Shark the Unknown May 23 '17 at 19:46
• Thank you so much! The exercises I was doing didn't require much limit calculation techniques and when I tried to prove something on my own I didn't even remember the old tricks! – Luis Dias May 23 '17 at 19:54
• Maybe you find the answer of this post useful. – Amin235 May 25 '17 at 20:13

Since $$F_{n}=\frac{\phi^{n}-\left(-\phi\right)^{-n}}{\sqrt{5}}$$ we have $$F_{n}\sim\frac{\phi^{n}}{\sqrt{5}}$$ as $n\rightarrow\infty$ so $$\sum_{n\ge1}\frac{1}{F_{n}}\sim\sqrt{5}\sum_{n\ge1}\frac{1}{\phi^{n}}=\frac{\sqrt{5}}{\phi-1}.$$

You may prove by induction that for any $n\geq 5$ we have $F_{n+5}\geq 11 F_n$. That is enough to deduce convergence by comparison with a geometric series and further get that:

$$S = \sum_{n\geq 1}\frac{1}{F_n} =\frac{17}{6}+\sum_{n\geq 5}\frac{1}{F_n} = \frac{17}{6}+\sum_{n=5}^{9}\frac{1}{F_n}+\sum_{n\geq 10}\frac{1}{F_n}\leq \frac{17}{6}+\frac{88913}{185640}+\frac{1}{11}\sum_{n\geq 5}\frac{1}{F_n}$$ such that: $$\frac{10}{11}\sum_{n\geq 5}\frac{1}{F_n}\leq \frac{88913}{185640},\qquad S\leq \frac{17}{6}+\frac{11}{10}\cdot \frac{88913}{185640}=\frac{2079281}{618800}.$$

For $n \ge 3$, we have

$$\frac{1}{F_n} \le \frac{F_{n-1}}{F_nF_{n-2}} = \frac{F_{n} - F_{n-2}}{F_nF_{n-2}} = \frac{1}{F_{n-2}} - \frac{1}{F_n} = \left(\frac{1}{F_{n-2}} + \frac{1}{F_{n-1}}\right) - \left(\frac{1}{F_{n-1}} + \frac{1}{F_n}\right)\\ = \frac{F_{n}}{F_{n-2}F_{n-1}} - \frac{F_{n+1}}{F_{n-1}F_n}$$

The partial sums is monotonic increasing and bounded from above. $$\sum_{n=1}^N \frac{1}{F_n} = 2 + \sum_{n=3}^N \frac{1}{F_n} \le 2 + \frac{F_3}{F_1F_2} - \frac{F_{N+1}}{F_{N-1}F_N} \le 2 + \frac{2}{1\cdot 1} = 4$$ As a result, the series converges.

$$\frac{F_{n+1}}{F_n}=\frac{\phi^{n+1}-(-\phi)^{-n-1}}{\phi^n-(-\phi)^{-n}}\ge\phi\frac{1-\phi^{-2n-2}}{1+\phi^{-2n}}.$$

The expression on the right is an increasing function of $n$ that exceeds $1$ as of $n=2$ (and quickly tends to $\phi$).

Let $F_n$ be the fibonacci sequence

We know $F_{2n+2}=F_{2n+1}+F_{2n}\geq 2F_{2n}$

Similarly $F_{2n+1}=F_{2n}+F_{2n-1}\geq 2F_{2n-1}$

So $1/F_2+1/F_4+1/F_6\dots<1/F_2+1/2F_2+1/4F_2\dots=1/F_2(1/2+1/4+1/8\dots)$

$1/F_1+1/F_3+1/F_5\dots<1/F_1+1/2F_1+1/4F_1\dots=1/F_1(1/2+1/4+1/8\dots)$

Now I think it's clearly evident that why the sum of reciprocals of the Fibonacci sequence is convergent, only the definition of the Fibonacci sequence is enough!

Suppose $2 \ge \frac{F_{n+1}}{F_n} \ge \frac32$ for $n$ and $n+1$.

Then

$\frac{F_{n+2}}{F_{n+1}} =\frac{F_{n+1}}{F_{n+1}}+\frac{F_{n}}{F_{n+1}} =1+\frac{F_{n}}{F_{n+1}} \ge 1 + \frac12 =\frac32$

and

$\frac{F_{n+2}}{F_{n+1}} =\frac{F_{n+1}}{F_{n+1}}+\frac{F_{n}}{F_{n+1}} =1+\frac{F_{n}}{F_{n+1}} \lt 1 + 1 =2$.

I think I have an interesting proof for the summability of the reciprocals of the FN. It is based on the Kummer's test and the fact that $F_{n-1}\geq 2 F_{n}$, $n\geq 2$. First note that $(1/F_n)$ is summable, if and only if, there exists a sequence $(u_n)$ of positive numbers such that $$u_n F_{n+1}-u_{n+1}F_{n}\geq F_{n},$$ for all $n$ sufficiently large. (This is the Kummer's test applied to the sequence $(1/F_n)$). Now, from the definition of FN: $F_{n+1}-F_{n}=F_{n-1}$, for all $n\geq 2$.

That is, taking $u_n = 2$, for all $n\geq 1$, we have that $2 F_{n+1}- 2 F_{n}= 2 F_{n-1}\geq F_{n}$, for all $n\geq 2$. As so, by the Kummer's test, the sequence $(1/F_n)$ is summable.


What is the name of this curve?

When I was a kid I used to draw this shape below but today I came against it as a problem. I don't know the name of this red curve below. It is enough to say the name if it is a known curve. I will search for it's properties.

The shape is constructed with lines from point $(0,n)$ to $(8-(n-1),0)$ and the curves passes through the intersection of each two lines. In contrast to the figure, the red curve never crosses the axes. $n=0,1,...,9$ however the shape does not need to be discrete, hence $n$ can go to $0$. Probably the values on the axes are not important to know the name of the curve.

• Curve stitching, (quadratic) Bézier curve – peterwhy Nov 4 '13 at 18:43
• @peterwhy thanks – newzad Nov 4 '13 at 18:43
• It looks to be a branch of a hyperbola, but I'm not sure. – Cameron Buie Nov 4 '13 at 18:44
• @CameronBuie No, when you take continuous $n$ the curve will ends at $(0,9)$ and $(9,0)$. – peterwhy Nov 4 '13 at 18:49
• Upvote if you made this with string and pins. :) – Kaz Nov 4 '13 at 20:35

The particular curve suggested, tangent to all of the segments from $(0,a)$ to $(9-a,0)$ as $a$ runs from $0$ to $9$, is this one: $$y = x-6\sqrt{x}+9$$

This is a portion of a parabola, with axis on the $x=y$ line and vertex at $(9/4,9/4)$.

To see this more clearly, add also such lines with $a>9$ and $a<0$ ...

For this particular example, the red curve is "stitched" by a family of straight lines of \begin{align*} \frac x{9-k} + \frac yk =& 1\\ y =& k - \frac{kx}{9-k} \end{align*} where $0<k<9$.

When there are two such straight lines with parameters $h$ and $k$ respectively, $h\ne k$, \begin{align*} y =& h - \frac{hx}{9-h}\\ y =& k - \frac{kx}{9-k}\\ \end{align*}

Their intersection can be calculated as \begin{align*} h - \frac{hx}{9-h} =& k - \frac{kx}{9-k}\\ \frac{kx}{9-k} - \frac{hx}{9-h} =& k - h\\ x\cdot\frac{9k-hk-9h+kh}{(9-k)(9-h)} =& k - h\\ x =& \frac{(9-k)(9-h)}9\\ \end{align*}

When we take $h\to k$, the $x$-coordinates becomes $$x = \frac{(9-k)^2}9$$

and the $y$-coordinates is $$y = k - \frac{kx}{9-k} = k - \frac{k(9-k)}{9} = \frac{k^2}9$$

We can get an implicit curve for our range $$\sqrt x + \sqrt y = 3$$

• It's a quadratic Bezier segment, with control points (0, 9), (0, 0), and (9, 0), I believe. That means that you can write it as $t^2 (9, 0) + 2t*(t-1) * (0, 0) + (1-t)^2 *(0, 9)$, which simplifies to $(9 t^2, 9(1-t)^2)$. Just in case you have access to matlab, here's code to make the picture: t = linspace(0, 1, 200); x = 9 * t.^2; y = 9 * (1-t).^2; plot(x, y); hold on; % draw the lines for i = 0:9 plot( [0, 9-i], [i, 0], 'r'); end set(gca, 'DataAspectRatio', [1 1 1]); hold off; figure(gcf); – John Hughes Nov 4 '13 at 18:58
• @John thank you very much, your comment is very useful. – newzad Nov 4 '13 at 19:04
• This is a good (best?) answer because it directly addresses the construction of the curve from line segments. This process is De Casteljau's algorithm. – Patrick Hew Jul 5 '19 at 2:46

It's a (portion of a) parabola.

By the curve-stitching construction, it's what's called an "envelope" of the family of line segments. The Wikipedia article (starting at "For example, let $C_t$ be the lines whose $x$ and $y$ intercepts are $t$ and $1-t$...") walks somewhat quickly through your problem as its example (except with $x$ and $y$ intercepts of $1$ instead of $9$).

As I wrote elsewhere, we are most probably dealing with a superellipse, a geometric shape described by algebraic equations of the form $x^n+y^n=r^n$, perhaps with $n=\frac12$ or $\frac23$ (astroid) , or maybe some other form of hypocycloid. Their bidimensional generalizations are called superformulas, and their three-dimensional counterparts are known as superellipsoids, superquadrics or supereggs. The case $n=4$ is called a squircle. Hope this helps.

This curve is a rotated parabola. As drawn in the picture, you get a parametric curve with: $$x = \frac{(t-7)(t-8)}{9}, y = \frac{(t+1)(t+2)}{9}$$ Applying a $45$° rotation counterclockwise, you get:

$$\left( \begin{array}{ccc} x' \\ y' \end{array} \right) = \left( \begin{array}{ccc} \sin(45) & -\cos(45) \\ \cos(45) & \sin(45) \end{array} \right) \times \left( \begin{array}{ccc} x \\ y \end{array} \right)$$

Plugging the parametrized forms of $x$ and $y$, you can see, by direct check, that:

$$y' = \frac{\sqrt{2}(x')^2}{162}+20\sqrt{2}$$

Though probably not equivalent, it is similar to a hyperbola in the first quadrant. Try graphing points for the equation $y = \dfrac{1}{x}$ to see a basic example. Changing the value in the numerator above $x$ is what dictates how far away the whole curve is from at the origin at its closest point (which is on the line $x=y$). Also, a numerator that is an integer with several divisors will also have several integer solutions for $(x,y)$.

Example: Graph the equation $y=\dfrac{20}{x}$ and see how it has integer solutions at (1,20), (2,10), (4,5), (5,4), (10,2), and (20,1).

• That was my first thought too. But the other answers convinced me that it's a parabola. – TonyK Nov 4 '13 at 19:38


# Difference between calculator and google calc for power [duplicate]

I tried to compute the power of 2^2^2^2 on google calculator and my casio calculator but both are giving different results. same is true for 3^3^3. Please explain me the difference between two expressions.

## marked as duplicate by kingW3, JMoravitz, Claude Leibovici, Henrik, user91500Jun 3 '17 at 9:22

Your calculator is interpreting it as this:

$${{2^2}^2}^2 = ((2^2)^2)^2 = (4^2)^2 = 16^2 = 256$$

Google is interpreting it as this:

$${{2^2}^2}^2 = 2^{(2^{(2^2)})} = 2^{(2^4)} = 2^{16} = 65536$$

Similarly with the threes.

Technically Google is correct because order of operations says to do exponents first. So when we want to evaluate $2^{\color{red}{2}^{\color{blue}{2^2}}}$, order of operations says to evaluate the $\color{red}{{2^{\color{blue}{2^2}}}}$ first, i.e., evaluate the exponent first. Apply this rule again and it tells us we're supposed to evaluate the $\color{blue}{2^2}$ first, which is $4.$ Therefore $\color{red}{{2^{\color{blue}{2^2}}}} = 2^4 = 16$, and so $2^{\color{red}{2}^{\color{blue}{2^2}}} = 2^{16} = 65536$.

• What is correct mathematically? – shiv garg Jun 2 '17 at 18:05
• @shivgarg, technically Google is correct because order of operations says to do exponents first. Also see my edit in my answer. – tilper Jun 2 '17 at 18:08
• "what is correct mathematically?" is also answered in greater detail in the linked question that you should see a link to above, but here it is again in case you missed it. – JMoravitz Jun 2 '17 at 18:09
• In cases like this it never hurts to be explicit about the order you intend. Use parens to instruct Google (or a more modern calculator) how you want it processed. – SDsolar Jun 2 '17 at 18:13
• Actually, I don't think there is a "correct" answer. It's a matter of convention, and the convention is not set in stone. – Robert Israel Jun 2 '17 at 18:14

The difference lies in the applied order of operations.

The Casio calculator does operations strictly left to right unless you break it with parentheses:

2^2^2^2 = 4^2^2 = 16^2 = 256

But Google evaluates the entire expression (correctly) using right-to-left order or operations for the stacked powers:

2^2^2^2 = 2^2^4 = 2^16 = 65536

• What is correct mathematically? – shiv garg Jun 2 '17 at 18:06
• I stated that in my answer. – John Jun 2 '17 at 18:08


# Difference between methods for sampling without replacement

A bag contains 2 red balls, 3 red cubes, 3 blue balls and 2 blue cubes. A person picks 3 objects at random from the bag and does not replace them.

What is the probability that the person picks 2 blue balls?

My initial idea to solve this was using unordered sampling without replacement, although I tend not to think of these problems in this way and instead just use combinations and permutations, but that gave me an answer I feel was wrong:

Our sample space size is $10 \choose 3$ since we are selecting 3 things from a set of 10 things. We want to select 2 blue balls from a possible total of three balls so that gives us $3 \choose 2$ and we then select the third and final object from a set of 8 remaining objects which gives us $8 \choose 1$.

Finally we have $\mathbb{P}$(2 blue balls)$=\frac{{3 \choose 2} {8 \choose 1}}{10 \choose 3}=\frac{1}{5}$.

This value seems too high to me for it to be correct, so I tried the following method:

Sample space size is $\frac{10!}{(10-3)!}=720$ and the event two blue balls is $\frac{3!}{(3-2)!}=6$

Hence $\mathbb{P}$(2 blue balls)$=\frac{6}{720}=\frac{1}{120}$ and this seems more correct to me.

Could someone please tell me which answer is correct, and why both unordered and ordered sampling don't give the same answer?

Also, following on from the first Q, what is the probability that all the objects drawn are blue?

We have 5 blue objects in total.

Probability of picking first object blue is $5 \choose 1$

Probability of picking second object blue is $5 \choose 2$

Probability of picking third object blue is $3 \choose 1$

Then $\mathbb{P}$(all blue)$=\frac{5 \times 4 \times 3}{10 \choose 3}$

Is this correct?

Thanks!!

• " What is the probability that the person picks 2 blue balls?" -> At least two blue balls, or exactly two blue balls? May 30 '18 at 17:23

Your approach to calculating the probability was not wrong, you made a simple mistake. You chose 2 blue balls, from the total of 3 blue balls, i.e 3C2, then you chose 1 final item from 8 remaining items. However, one of those 8 items is also a blue ball, so if you picked it you would have 3 blue balls, not 2. You should pick from the 7 remaining non-blue-ball items. So the probability is

$$P(\text{2 blue balls}) = \frac{\binom{3}{2}\binom{7}{1}}{\binom{10}{3}} = 7/40$$

You can also see this in the way user247327 explained. You have 3/10 probability to pick the first blue ball, then 2/9 probability to pick the second, and 7/8 ways to pick the last. $(3/10)(2/9)(7/8) = 7/120$. Then there are three orderings of the ball, so $P = 7/120 \cdot 3 = 7/40$.

• Ah thank you I realise what I did wrong now. Any idea on if my solution is correct for the second bit? May 30 '18 at 17:15
• From 5 blue things, choose 3 of them - 5C3. From 5 red things choose 0 of them, 5C0. Then divide by the sample space which is 10C3. So you get (5C3*5C0)/(10C3) = 10/120 = 1/12. Thus your solution is incorrect. I believe that the mistake is in trying to pick each object individually. If you want to do it individually you could do probabilities, so you have 5/10 probability to pick the first, then 4/9, then 3/8, and you get (5/10)(4/9)(3/8) = 1/12. May 30 '18 at 17:24

Initially there are 2 red balls, 3 red cubes, 3 blue balls, and 2 blue cubes. So there are 3 blue balls and 2+ 3+ 2= 7 non-blue-ball objects. The probability that the first object drawn is 3/10. Once that has happened, there are 2 blue balls and 7 non-blue-ball objects. The probability the second object drawn is 2/9. Then there are 1 blue ball and 7 non-blue-ball objects. The probability the third object drawn is NOT a blue ball is 7/8. The probability that two blue balls are drawn [b]in that order[/b] is (3/10)(2/9)*(1/8)= 1/120.

In the same way, the probability that the first item drawn is a blue ball is 3/10. Given that the probability the second item drawn is NOT a blue ball is 7/9. Then the probability the third item is a blue ball is 2/8 so the probability of "blue ball, not blue ball" in that order is (3/10)(7/9)(2/8) which also 1/120- we've just changed the order of the numbers in the numerator.

I will leave it to you to show that the probability of "not a blue ball, blue ball, blue ball" in that order is also 1/120. So the probability of two blue balls out of three is 3/120= 1/40.


# §4.37 Inverse Hyperbolic Functions

## §4.37(i) General Definitions

The general values of the inverse hyperbolic functions are defined by

4.37.1
4.37.2
4.37.3,
4.37.4
4.37.5
4.37.6

In (4.37.1) the integration path may not pass through either of the points , and the function assumes its principal value when is real. In (4.37.2) the integration path may not pass through either of the points , and the function assumes its principal value when . Elsewhere on the integration paths in (4.37.1) and (4.37.2) the branches are determined by continuity. In (4.37.3) the integration path may not intersect . Each of the six functions is a multivalued function of . and have branch points at ; the other four functions have branch points at .

## §4.37(ii) Principal Values

The principal values (or principal branches) of the inverse , , and are obtained by introducing cuts in the -plane as indicated in Figure 4.37.1(i)-(iii), and requiring the integration paths in (4.37.1)–(4.37.3) not to cross these cuts. Compare the principal value of the logarithm (§4.2(i)). The principal branches are denoted by , , respectively. Each is two-valued on the corresponding cut(s), and each is real on the part of the real axis that remains after deleting the intersections with the corresponding cuts.

The principal values of the inverse hyperbolic cosecant, hyperbolic secant, and hyperbolic tangent are given by

4.37.7
4.37.8
4.37.9.

These functions are analytic in the cut plane depicted in Figure 4.37.1(iv), (v), (vi), respectively.

Except where indicated otherwise, it is assumed throughout the DLMF that the inverse hyperbolic functions assume their principal values.

Graphs of the principal values for real arguments are given in §4.29. This section also indicates conformal mappings, and surface plots for complex arguments.

4.37.10
4.37.11.
4.37.12.

## §4.37(iv) Logarithmic Forms

Throughout this subsection all quantities assume their principal values.

### ¶ Inverse Hyperbolic Cosine

the upper or lower sign being taken according as ; compare Figure 4.37.1(ii). Also,

It should be noted that the imaginary axis is not a cut; the function defined by (4.37.19) and (4.37.20) is analytic everywhere except on . Compare Figure 4.37.1(ii).

On the part of the cuts from −1 to 1

the upper/lower sign corresponding to the upper/lower side.

On the part of the cut from to −1

the upper/lower sign corresponding to the upper/lower side.

### ¶ Inverse Hyperbolic Tangent

4.37.24;

compare Figure 4.37.1(iii). On the cuts

the upper/lower sign corresponding to the upper/lower sides.

### ¶ Other Inverse Functions

For the corresponding results for , , and , use (4.37.7)–(4.37.9); compare §4.23(iv).

## §4.37(vi) Interrelations

Table 4.30.1 can also be used to find interrelations between inverse hyperbolic functions. For example, .


# Number of triples of divisors who are relatively prime as a triple

Given a number $n \in \mathbb{N}$, define $a(n)=\{(d_1,d_2,d_3): d_i|n,\ \gcd(d_1,d_2,d_3)=1,\ 1 \leq d_1 \leq d_2 \leq d_3\}$. What is $|a(n)|$?

There were similar questions asked about pairs rather than triples here: Number of pairs of nontrivial relatively prime divisors and here: Number of Relatively Prime Factors but the addition of the third component seems to make the arguments used in these two questions obsolete.

An example:

Let $n=p$ for some prime $p$. Then the divisors of $n$ are $\{1,p\}$, and the triples of those divisors whom are relatively prime (order does not matter) as a triple are $\{(1,1,1),(1,1,p),(1,p,p)\}$. Thus we see that for any prime, the answer is $|a(n)|=3$.

Similarly, (its not hard to check) for $n=p^2$, we have $|a(n)|=6$.

In fact, if you let $n=p^k$ for $0 \leq k \in \mathbb{Z}$, it seems that $|a(n)|={{k+2} \choose {2}}$.

Also for $n=pq$ where $p$ and $q$ are distinct primes, $|a(n)|=13$ (again, not hard to check).

This leads me to believe that like in the other answers for the two referenced questions, the answer may be found using some nice combinatorics, but if looked at just right from a Number Theoretic perspective, may be a multiplicative function or composition of multiplicative functions, based on how the answers seem to only depend on the powers of the primes in the prime factorization of the number.

• If you forget about $d_1\leq d_2\leq d_3$, it becomes multiplicative. Then deal with $d_1=d_2$ and so on. – Empy2 Jul 12 '18 at 2:11

As in the pair case, let's first find the number of ordered triples without restrictions on the order of the entries.

Every prime factor $p_i$ with multiplicity $a_i$ occurs in at most two divisors.

There is $1$ way for $p_i$ to occur in zero divisors.

There are $3a_i$ ways for $p_i$ to occur in one divisor.

There are $3a_i^2$ ways for $p_i$ to occur in two divisors.

Thus, in total there are $3a_i^2+3a_i+1$ ways to distribute $p_i^{a_i}$ over the divisors, and hence there are $\prod_i(3a_i^2+3a_i+1)$ ordered triples. From this we want to deduce the number of unordered triples (or, equivalently, the number of ordered triples with the order restrictions on the entries).

There is one triple with three equal entries, namely $(1,1,1)$. There are $\prod_i(2a_i+1)-1$ ordered pairs of distinct coprime divisors, each of which corresponds to three ordered triples with two equal entries but only one unordered triple with two equal entries. Thus the count of unordered triples is

$$\frac{\prod_i(3a_i^2+3a_i+1)-3\left(\prod_i(2a_i+1)-1\right)-1}6+\prod_i(2a_i+1)-1+1=\frac{\prod_i(3a_i^2+3a_i+1)+3\prod_i(2a_i+1)+2}6\;.$$

You can check that your examples all come out right.

In fact, if you let $n=p^k$ for $0 \leq k \in \mathbb{Z}$, it seems that $|a(n)|={{k+2} \choose {2}}$.

If $n=p^k$ we can write the triple as $(p^r, p^s, p^t)$ with $r \le s \le t$. Since the GCD of the triple is $1$, clearly $r = 0$. So you're counting integer values of $s, t$ such that $0 \le s \le t \le k$. If $s \neq t$ there are $\binom{k+1}{2}$ ways of choosing them from $k+1$ options; if $s = t$ there are $\binom{k+1}{1}$ ways; and $\binom{k+1}{2} + \binom{k+1}{1} = \binom{k+2}{2}$.

On the general question, let $q$ be prime and coprime to $n$, and consider $(d_1, d_2, d_3) \in a(nq^k)$. We can project $d_i$ onto $n$ and $q^k$ by taking the GCD, so $\textrm{sort}(\gcd(d_1, n), \gcd(d_2, n), \gcd(d_3, n)) \in a(n)$ (where $\textrm{sort}$ gives the ordered triple) and similarly for $q^k$. This gives a upper bound $$a(nq^k) \le 3!\, a(n)\, a(q^k)$$ where the factor of $3!$ is the number of ways to sort 3 items (wlog the triple from $q^k$, when pairing up its elements with the elements of the triple from $n$). However, we need to account for the fact that the triples are not required to have 3 distinct values, so that some of the orders will create duplicates.

We can split $a(n)$ up into three parts: triples with one distinct value $a_1(n) = \{(1,1,1)\}$; triples with two distinct values $a_2(n)$; and triples with three distinct values $a_3(n)$. Triples with two distinct values break down as $(r, r, s)$ or $(r, s, s)$. When $n=p^k$ we have $(1, 1, p^r)$ or $(1, p^r, p^r)$ with $0 < r \le k$, so $a_2(p^k) = 2k$. $a_3(p^k) = \binom{k}{2}$ by similar arguments to the first section. As a sanity check, $a_1(p^k) + a_2(p^k) + a_3(p^k) = 1 + 2k + \binom{k}{2} = \binom{k+2}{2}$.

Now we have nine cases for pairing an element of $a(n)$ with an element of $a(q^k)$. Five of them are covered by the observation that $(1,1,1)$ paired with an element of $i$ distinct values gives an element of $i$ distinct values. That leaves:

• Two with two: $(d_1, d_1, d_2)$ paired with $(e_1, e_1, e_2)$ gives $(d_1 e_1, d_1 e_1, d_2, e_2)$ (two distinct elements) and $(d_1 e_1, d_1 e_2, d_2 e_1)$ which (since we're building from coprime parts) has three distinct elements.
• Two with three: three permutations each of three distinct elements.
• Three with three: the full six permutations, each of three distinct elements.

So we get the recurrence $$\begin{eqnarray} a_1(nq^k) &=& a_1(n) a_1(q^k) \\ a_2(nq^k) &=& a_1(n) a_2(q^k) + a_2(n) a_1(q^k) + a_2(n) a_2(q^k) \\ a_3(nq^k) &=& a_1(n) a_3(q^k) + a_3(n) a_1(q^k) + a_2(n) a_2(q^k) +\\&& 3a_2(n) a_3(q^k) + 3a_3(n) a_2(q^k) + 6a_3(n) a_3(q^k) \end{eqnarray}$$

and we can substitute in and simplify to

$$\begin{eqnarray} a_1(nq^k) &=& 1 \\ a_2(nq^k) &=& (2k+1)(a_2(n)+1) - 1 \\ a_3(nq^k) &=& \left(3k^2 + 3k + 1\right) a_3(n) + \frac{(3k+1)k}{2} a_2(n) + \frac{k(k-1)}{2} \end{eqnarray}$$

If we define $a_{1+2}(n) = a_1(n) + a_2(n)$, there's a nice simple recurrence $a_{1+2}(nq^k) = (2k+1)a_{1+2}(n)$, but not for $a_3$.


# For each face of a cuboid, the sum of perimeter and area is known. Find the volume.

I came across this problem in a national level examination.

On each face of a cuboid, the sum of its perimeter and its area is written. Among the six numbers written are $$16$$, $$24$$, and $$31$$.

Then the volume lies between ...

My workout...

From the problem,

$$ab+2(a+b)=16$$

$$bc+2(b+c)=24$$

$$ca+2(c+a)=31$$

I am sure some amount of manipulation is required after this step, but I have tried in vain.

Add $4$ to each equation and factorise \begin{eqnarray*} (a+2)(b+2)=20 = 5 \times 4 \\ (b+2)(c+2)=28 = 4 \times 7 \\ (c+2)(a+2)=35 = 7 \times 5 \\ \end{eqnarray*} which gives $\color{red}{a=3,b=2,c=5}$. So the volume is $\color{blue}{30}$.

• We can also show easily that this is the only solution by considering small variations in $a,b,c$. +1 Oct 21 '17 at 14:13
• You can solve directly by multiplying the first two equations and dividing by the third: $$\frac{(a+2)(b+2)\cdot(b+2)(c+2)}{(c+2)(a+2)} = \frac{20\cdot 28}{35} \;\to\; (b+2)^2 = 16 \;\to\; b+2 = \pm 4 \;\to\; b = 2$$ (taking $b=-6$ to be extraneous).
– Blue
Oct 21 '17 at 19:31

I much prefer the cleverness of @Donald's answer (which is probably the approach the exam writers hoped you'd see), but it's perhaps worth noting that you can attack this problem with a little algebraic brute force and a lot of perseverance.

You have three equations in three unknowns. Our goal is to eliminate two of the unknowns, leaving a single equation in, say, $a$. Notice that the first equation fairly readily allows us to express $b$ in terms of $a$; the third equation allow us to to do likewise for $c$:

\begin{align} ab + 2 a + 2 b = 16 \quad\to\quad b(a+2) = 16-2a \quad\to\quad b &= \frac{16-2a}{a+2} = \frac{2(8-a)}{a+2} \tag{1a} \\[6pt] c &= \frac{31-2a}{a+2} \tag{1b} \end{align}

By substitution, the second equation transforms to involve $a$ alone: $$\frac{2(8-a)}{a+2}\cdot\frac{31-2a}{a+2} + 2\left(\frac{16-2a}{a+2}+\frac{31-2a}{a+2}\right) = 24 \tag{2a}$$ $$\frac{(8-a)(31-2a)}{(a+2)^2}+ \frac{47-4a}{a+2} = 12 \tag{2b}$$ $$(8-a)(31-2a)+ (47-4a)(a+2) = 12 (a+2)^2 \tag{2c}$$

(It's around this point that you should start to suspect that there's a better way. Nevertheless, ...) We can expand everything and combine terms to get $$14 a^2 + 56 a - 294 = 0 \quad\to\quad 14 ( a^2 + 4 a - 21)= 0 \quad\to\quad 14 (a+7)(a-3) = 0 \tag{3}$$ Here, we solve to find that $a=3$ (discarding $a=-7$ as extraneous). Then $b=2$ and $c=5$ follow from $(1a)$ and $(1b)$ above. $\square$

This approach lacks ingenuity, but it's essentially mechanical. (You could even opt to use the Quadratic Formula in $(3)$ instead of thinking-through the factorization. Simplifying the final result is something of a chore, but it doesn't require any insight.) So, even if you don't see the clever approach, there's still a way to proceed ... and to succeed.

It was already done (and far better that what i did) but i isolated a b c in the next way:

a=(16-2b)/(b+2)

b= (24-2c)/(c+2)

c=(31-2a)/(a+2)

then by substitutions you will get the solution.


# Conditional Probabilities- picking balls out of a bag

3 red, 3 green, 3 blue, and 3 orange balls are in a box and 6 of these balls are drawn at random from the box. If 2 of the 6 drawn balls are red and 2 of them are green, what is the probability that the other 2 drawn balls are blue and orange?

The answer I am getting is 12/28. I came to this answer by realizing there are 8 balls left. 3 orange, 3 blue, 1 green and 1 red. Two more balls need to be chosen. So the total number of ways these can be picked are $8\choose 2$=28 ways to choose 2 balls from the remaining 8. Then I got that there are 3!+3! ways to pick a orange and a blue ball from the remaining two balls. So i got my answer to be $(3!+3!)/28$=12/18 as the probability of picking a orange and a blue as the other two of the 6 balls drawn. I don't know if this is right. Can anyone confirm my solution. If it is incorrect, could anyone point out where my error lies?

There are $9$ ways ($3\times 3$) of picking an orange and a blue ball. So the answer should be $9/28$.

• the other thing i just realized is that the remaining unknown two balls aren't necessarily drawn after the 2 red and 2 green balls, the order is unspecified. Will this have any affect on the answer? For example, the orange and blue balls could be the first two balls drawn, followed by two red and two green balls. – sappgob Nov 2 '15 at 18:40
• If you draw all six balls without looking at them, and you then look at four of them, and they are two red and two green, then the answer I gave is correct. If that's not what you mean, then you need to make your question more precise. – rogerl Nov 2 '15 at 18:44

Rogerl has given the probability that the remaining 2 balls will be 1 orange and 1 blue when given that four revealed balls are 2 red and 2 green.

$$\frac{\binom{3}{1}\binom{3}{1}\binom{2}{0}}{\binom{8}{2}}= \frac{9}{28}$$

If you want the probability of picking 1 orange and 1 blue given that you have picked exactly 2 red and 2 green:

$$P(O{=}1,B{=}1\mid R{=}2,G{=}2)=\dfrac{\binom{3}{1}\binom{3}{1}}{\binom{6}{2}}=\frac{3}{5}$$

If you want the probability of picking 1 orange and 1 blue given that you have picked at least 2 red and 2 green:

$$P(O{=}1,B{=}1,R{=}2,G{=}2\mid R{\geq}2,G{\geq}2)=\dfrac{\binom{3}{1}\binom{3}{1}\binom{3}{2}\binom{3}{2}}{\binom{6}{2}\binom{3}{2}\binom{3}{2}+2\binom{6}{1}\binom{3}{3}\binom{3}{2}+\binom{3}{3}\binom{3}{3}}=\frac{81}{172}$$

This is why it is important to be clear about your specifications.

Similarly: I toss two coins behind a screen and tell you about one of them being a head.   What is the probability of the other being a tail when what I told you was ... ?:

• The left coin is a head.
• At least one coin is a head.
• Exactly one of the coins is a head.

The probability space is $\{\rm HH, HT, TH, TT\}$

• I actually like this answer better, Graham. – rogerl Nov 4 '15 at 22:49


# A series to prove $\frac{22}{7}-\pi>0$

After T. Piezas answered Is there a series to show $22\pi^4>2143\,$? a natural question is

Is there a series that proves $\frac{22}{7}-\pi>0$?

One such series may be found combining linearly the series that arise from truncating $$\sum_{k=0}^\infty \frac{48}{(4k+3)(4k+5)(4k+7)(4k+9)} = \frac{16}{5}-\pi$$ to two and three terms, namely

$$\sum_{k=2}^\infty \frac{48}{(4 k+3) (4 k+5) (4 k+7) (4 k+9)} = \frac{141616}{45045}-\pi$$ and $$\sum_{k=3}^\infty \frac{48}{(4 k+3) (4 k+5) (4 k+7) (4 k+9)} = \frac{2406464}{765765}-\pi$$ Solving $$a\left(\frac{141616}{45045}-\pi\right)+b\left(\frac{2406464}{765765}-\pi\right)=\frac{22}{7}-\pi$$ for rational $a,b$ and some algebra manipulation yields the result

$$\frac{16}{21} \sum_{k=0}^\infty \frac{1008 k^2+6952 k+12625}{(4 k+11) (4 k+13) (4 k+15) (4 k+17) (4 k+19) (4 k+21)}=\frac{22}{7}-\pi$$

It is interesting to note that the coefficients needed to multiply the two component series are both positive $$a=\frac{113}{7·8·9}$$ $$b=\frac{391}{7·8·9}$$

because the truncation points have been chosen so that

$$\frac{2406464}{765765}<\frac{22}{7}<\frac{141616}{45045}$$

This procedure yields a result that proves the claim with no need for further processing, and it is readily seen to prove $\frac{p}{q}-\pi>0$ for all fractions between $\pi$ and $\frac{16}{5}$.

Now, in the light of this equivalent form of Lehmer's formula $$\pi-3=\sum_{k=1}^\infty \frac{4!}{(4k+1)(4k+2)(4k+4)}$$

Q1 Is there a series that proves $\frac{22}{7}-\pi>0$ with constant numerator?

Q2 Is there a reason why $113$ is both the numerator of the $a$ coefficient and the denominator of the next convergent from above $\frac{355}{113}$?

Edit: A similar series with smaller coefficients may be obtained by applying the method above to \begin{align} \sum_{k=0}^\infty \frac{960}{(4 k+3) (4 k+5) (4 k+7) (4 k+9) (4 k+11) (4 k+13)} &= \frac{992}{315}-\pi \\ &= \frac{3·333-7}{3·106-3}-\pi \\ \end{align} in order to obtain $$\sum_{k=0}^\infty \frac{96 (160 k^2+422 k+405)}{(4 k+3) (4 k+5) (4 k+7) (4 k+9) (4 k+11) (4 k+13) (4 k+15) (4 k+17)} = \frac{22}{7}-\pi$$

Q3 What is the relationship between $\frac{992}{315}$ and the third convergent to $\pi$ $\frac{333}{106}$?

• Do you have the answer for this question ? – user230452 Feb 13 '16 at 4:56
• No, I don't have them. – Jaume Oliver Lafont Feb 13 '16 at 5:09
• @user230452 Now I do. – Jaume Oliver Lafont Feb 15 '16 at 22:52
• If you like this Math.SE question you may also enjoy reading this MO.SE post. – Qmechanic Feb 17 '16 at 14:44

Q1

Evaluating the following series \begin{align} &\sum_{k=0}^\infty \frac{240}{(4k+5)(4k+6)(4k+7)(4k+9)(4k+10)(4k+11)} \\ &= \sum_{k=0}^\infty \left(\frac{1}{4k+5}-\frac{4}{4k+6}+\frac{5}{4k+7}-\frac{5}{4k+9}+\frac{4}{4k+10}-\frac{1}{4k+11}\right) \\ &= \sum_{k=0}^\infty \int_{0}^1\left(x^{4k+4}-4x^{4k+5}+5x^{4k+6}-5x^{4k+8}+4x^{4k+9}-x^{4k+10}\right)dx \\ &= \int_{0}^1 x^4\sum_{k=0}^\infty \left(x^{4k}-4x^{4k+1}+5x^{4k+2}-5x^{4k+4}+4x^{4k+5}-x^{4k+6}\right)dx \\ &= \int_{0}^1 x^4\frac{1-4x+5x^2-5x^4+4x^5-x^6}{1-x^4}dx \\ &= \int_{0}^1 x^4\frac{(1-x^2)(1-x)^4}{(1-x^2)(1+x^2)}dx=\int_{0}^1 \frac{x^4(1-x)^4}{1+x^2}dx=\frac{22}{7}-\pi \\ \end{align} shows its connection with Dalzell's integral.

This may be rewritten as $$\sum_{k=1}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\frac{22}{7}-\pi$$

which appears in the 2009 document by Peter Bala New series for old functions http://oeis.org/A002117/a002117.pdf (formula 5.1) and shows that $\frac{22}{7}-\pi$ can be obtained by taking one term out of the summation in the series $$\sum_{k=0}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\frac{10}{3}-\pi$$

Consecutive truncations yield the inequality

$$\pi...<\frac{141514}{45045}<\frac{10886}{3465}<\frac{22}{7}<\frac{10}{3}$$

Similar fractions, but now converging to $\pi$ from below, may be obtained from the series

$$\sum_{k=0}^\infty \frac{240}{(4 k+3) (4 k+4) (4 k+5) (4 k+7) (4 k+8) (4 k+9)} = \pi-\frac{47}{15}$$

This yields

$$\frac{47}{15}<\frac{1979}{630}<\frac{141511}{45045}<\frac{9622853}{3063060}<...\pi$$

(See a similar inequality for $\log(2)$)

Correspondence between series and integrals

$$\sum_{k=n}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\int_0^1 \frac{x^{4n}(1-x)^4}{1+x^2}dx$$

$$\sum_{k=n}^\infty \frac{240}{(4 k+3) (4 k+4) (4 k+5) (4 k+7) (4 k+8) (4 k+9)}=\int_0^1 \frac{x^{4n+2}(1-x)^4}{1+x^2}dx$$

Equivalent expressions

The general term for these series may be written in compact form using factorials, binomial coefficients or the Beta integral $B$ (see this comment by N. Elkies).

\begin{align} \frac{22}{7}-\pi &= 3840\sum_{k=1}^\infty \frac{(k+2)!(4k)!}{(4k+8)!k!} \\ \\ &= \frac{4}{21} \sum_{k=1}^\infty \frac{\displaystyle{k+2 \choose 2}}{\displaystyle{4k+8\choose 8}} \\ \\ &= \frac{4}{21} \sum_{k=1}^\infty \frac{k+1}{\displaystyle{4k+7\choose 7}} \\ \\ &= \frac{16}{21} \sum_{k=1}^\infty \frac{B(4k+1,8)}{B(k+1,2)} \end{align}

Interpretation of $\frac{22}{7}-\pi$

Similar series and approximations

If we use the Pochhammer symbol to express this series: $$\sum_{k=0}^\infty \frac{7!(k+1)}{(4k+1)_7}=\frac{7}{4}(10-3\pi)\approx 1$$

we can change the numbers to obtain variants such as

$$\sum_{k=0}^\infty \frac{5!(k+1)}{(3k+1)_{5}} = \frac{5}{9}\left(2\sqrt{3}\pi-9\right)\approx 1,$$

$$\sum_{k=0}^\infty \frac{11! (k+1)}{(6 k+1)_{11}} = 231-\frac{4565 \pi}{36 \sqrt{3}}\approx 1$$

and $$\sum_{k=0}^\infty \frac{15!(k+1)}{(8k+1)_{15}}=\frac{15}{8}(1716-7(99\sqrt{2}-62)\pi)\approx 1$$

Given that all three series evaluate to almost 1, the following corresponding approximations are derived

\begin{align} \pi &=\frac{9\sqrt{3}}{5}+\sqrt{3}\int_0^1\frac{x^3(1-x)^2(1+x)}{1+x+x^2}dx\\ &\approx\frac{9\sqrt{3}}{5} \\ \pi &=\frac{1656\sqrt{3}}{913}- \frac{6\sqrt{3}}{83}\int_0^1 \frac{x^6(1-x)^8}{1+x^2+x^4} dx\\ &\approx\frac{1656\sqrt{3}}{913} \\ \pi &=\frac{1838 \left(62 + 99 \sqrt{2}\right)}{118185}-\frac{62+99\sqrt{2}}{15758}\int_0^1 \frac{x^8(1-x)^{12}}{1+x^2+x^4+x^6}dx\\ &\approx \frac{1838 \left(62 + 99 \sqrt{2}\right)}{118185} \end{align}

which give 1, 5 and 8 correct decimals respectively.

The fraction $\frac{1838}{118185}$ is the eighth convergent of $\frac{\pi}{62+99\sqrt{2}}$

Another series and integral for $\frac{22}{7}-\pi$

\begin{align} &\sum_{k=0}^\infty \frac{285120}{(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)(4k+9)(4k+10)(4k+11)(4k+13)(4k+14)} \\ &= \frac{1}{28}\int_{0}^1 \frac{x(1-x)^8(2+7x+2x^2)}{1+x^2}dx=\frac{22}{7}-\pi \\ \end{align}

• So based on several series versions already found for 22/7 - Pi, it appears that similar to infinite number of integral expressions (as it was shown by Thomas Baruchel), there are infinite number of series as well. – Alex Feb 15 '16 at 23:20
• You are quite wrong wishing ALL the terms be positive. Anyway your example is nice. Regards. – Piquito Feb 16 '16 at 11:58
• @Piquito Reading the proof in reverse shows how the positive integrand naturally leads to all terms positive. – Jaume Oliver Lafont Feb 17 '16 at 10:18

Proof that $\frac{22}{7}$ exceeds $\pi$.

$$0<\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx=\frac{22}{7}-\pi$$

Proof-

$$\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx$$

$$=\int_0^1x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^2}dx$$

$$=\frac {x^7}{7}+\frac{2x^6}{3}+x^5-\frac{4x^3}{3}+4x-4\tan^{-1}(x)\vert_0^1$$

Now,by applying $\tan^{-1}1=45^\circ=\frac\pi4$ and substituting it in the integral and solving the integral yields $\frac{22}{7}-\pi$

• Why is that integral greater than zero ? Is is just because integral represents an area ? Moreover, what is the usual strategy to evaluate this integral ? – user230452 Feb 13 '16 at 5:13
• @tatan: since this does not answer either question, would you please write it as a comment? – Jaume Oliver Lafont Feb 13 '16 at 5:18
• @tatan your answer is about the integral proof for $\frac{22}{7}-\pi>0$, while this question addresses proofs using series with positive terms only. – Jaume Oliver Lafont Feb 13 '16 at 5:34
• @user230452 It's $>0$ because the integrand is always $>0$ in the domain of integration. – YoTengoUnLCD Feb 13 '16 at 5:40
• I have no prove of it, but $\displaystyle \sum_{k=0}^{+\infty}\dfrac{41760+576k}{(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)(4k+13)(4k+15)(4k+17)}=\dfrac{22}{7}-\pi$ – FDP Feb 14 '16 at 19:26

Let $\sum_{k=0}^\infty a_n$ any series converging to $\pi$ and choose any series converging to $\frac{22}{7}$, for instance $\sum_{k=0}^\infty \left(\frac{15}{22}\right)^n$

No problem to show that $$\sum_{k=0}^\infty \left(\left(\frac{15}{22}\right)^n -a_n\right)=\frac{22}{7}-\pi\gt 0$$

• Does this series have all terms positive? – Jaume Oliver Lafont Feb 15 '16 at 23:42
• From a certain range, yes. – Piquito Feb 15 '16 at 23:48
• How can we prove $\frac{22}{7}-\pi>0$ from that series if not all terms are positive? – Jaume Oliver Lafont Feb 15 '16 at 23:54
• Theorem.-Let $a=\sum a_n$ and $b=\sum b_n$ two convergent series. Then , for all pair of constants $\alpha$, $\beta$ the series $\sum(\alpha a_n+\beta b_n)$ converges to $\alpha a+ \beta b$ (Mathematical Analysis, T. M. Apostol). Can you exhibit an example of what you say? – Piquito Feb 16 '16 at 0:46
• This proves $$\sum_{k=0}^\infty \left((\frac{15}{22})^n -a_n\right)=\frac{22}{7}-\pi$$ but not $$\frac{22}{7}-\pi>0$$ – Jaume Oliver Lafont Feb 16 '16 at 0:51