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Math 141: Lecture 19 Convergence of Infinite Series

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Math 141: Lecture 19 Convergence of Infinite Series Math 141: Lecture 19 Convergence of infinite series Bob Hough November 16, 2016 Bob Hough Math 141: Lecture 19 November 16, 2016 1 / 44 Series of positive terms 1 P1 Recall that, given a sequence fangn=1, we say that its series n=1 an 1 converges if the sequence of partial sums fsngn=1 n X sn = ak ; k=1 converges. We begin by giving some criteria for the convergence in the case when the terms an are non-negative. Bob Hough Math 141: Lecture 19 November 16, 2016 2 / 44 Comparison test Theorem (Comparison test) 1 1 Let fangn=1 and fbngn=1 be two sequences such that an ≥ 0 and bn ≥ 0 for all n ≥ 1. Suppose there is a number N and a constant c > 0 such that an ≤ cbn P1 for all n > N. Then convergence of n=1 bn implies convergence of P1 n=1 an. In this case, we say that the sequence bn dominates the sequence an. Bob Hough Math 141: Lecture 19 November 16, 2016 3 / 44 Comparison test Proof of the comparison test. P1 Suppose that n=1 bn converges. Given > 0, apply the Cauchy property of the sequence of partial sums of bn to choose M > N such that m > n > M implies n X b < : n c k=m+1 It follows that n n X X an ≤ cbn < , k=m+1 k=m+1 so the sequence of partial sums of an is Cauchy, hence converges. Bob Hough Math 141: Lecture 19 November 16, 2016 4 / 44 Limit comparison test Theorem (Limit comparison test) 1 1 Let fangn=1 and fbngn=1 be sequences satisfying an > 0 and bn > 0 for all n ≥ 1. Suppose that a lim n = 1: n!1 bn P1 P1 Then n=1 an converges if and only if n=1 bn converges. Proof. The condition implies that there is N such that n > N implies a n ≤ b ≤ 2a : 2 n n P1 It now follows from the comparison test that n=1 an converges if and P1 only if n=1 bn converges. Bob Hough Math 141: Lecture 19 November 16, 2016 5 / 44 Examples Recall that last class we checked 1 1 1 X 1 X 1 X 1 1 = = − = 1 n2 + n n(n + 1) n n + 1 n=1 n=1 n=1 n2+n 1 since the series telescopes. Since limn!1 n2 = limn!1 1 + n = 1, P1 1 it follows that n=1 n2 converges. P1 1 By comparison, it follows that n=1 ns converges for all s ≥ 2. Bob Hough Math 141: Lecture 19 November 16, 2016 6 / 44 Examples P1 1 Recall that the harmonic series satisfies n=1 n = 1. One may check that 1 1 X 1 X 1 = 1; sin = 1 p n n=1 n(n + 10) n=1 by noting that pn(n + 10) r 10 lim = lim 1 + = 1; n!1 n n!1 n and sin 1 1 1 1 lim n = lim n + O = lim 1 + O = 1: n!1 1 n!1 n n3 n!1 n2 n Bob Hough Math 141: Lecture 19 November 16, 2016 7 / 44 Integral test Theorem (Integral test) Let f be a positive decreasing function, defined for all real x ≥ 1. The P1 infinite series n=1 f (n) converges if and only if the improper integral R 1 1 f (x)dx converges. Bob Hough Math 141: Lecture 19 November 16, 2016 8 / 44 Integral test Proof of the integral test. We first check that the improper integral converges if and only if the R n 1 sequence tn = 1 f (x)dx n=1 converges. Recall that convergence of the improper integral is defined by the R x existence of the limit limx!1 1 f (t)dt. Evidently convergence of the integral entails convergence of the sequence as a special case. To check the reverse direction, note that for n ≤ x ≤ n + 1, R x tn ≤ 1 f (t)dt ≤ tn+1, and hence, if jtn − Lj < for all n ≥ N, then Z x f (t)dt − L < 1 for all x ≥ N also. Bob Hough Math 141: Lecture 19 November 16, 2016 9 / 44 Integral test Proof of the integral test. 1 We now show that ftngn=1 converges if and only if the sequence of Pn 1 partial sums fsn = k=1 f (k)gk=1 converges. Given m > n, note that by taking upper and lower step functions for the integral, m−1 m X Z m X f (k) ≥ tm − tn = f (t)dt ≥ f (k) k=n n k=n+1 1 1 and hence ftngn=1 is Cauchy if and only if fsngn=1 is Cauchy, which proves the claim. Bob Hough Math 141: Lecture 19 November 16, 2016 10 / 44 Examples P1 1 We can show that the series n=1 ns , s real, converges if and only if s > 1. To check this, note that ( 1−s Z n dt n −1 1−s s 6= 1 tn = s = : 1 t log n s = 1 This sequence converges if and only if s > 1. 1 1 For complex s = σ + it, ns = nσ exp (it log n) : Since jexp (it log n)j = 1, the sum 1 X 1 ζ(s) = ns n=1 converges absolutely if σ = <(s) > 1. This function is called the Riemann zeta function. Bob Hough Math 141: Lecture 19 November 16, 2016 11 / 44 Examples The study of the analytic properties of the Riemann zeta function and related functions is responsible for a great deal of the information that we know about the prime numbers. Bob Hough Math 141: Lecture 19 November 16, 2016 12 / 44 Root test Theorem (Root test) P1 Let n=1 an be a series of non-negative terms. Suppose that 1 n an ! R; n ! 1: If R < 1 then the series converges. If R > 1 then the series diverges. Bob Hough Math 141: Lecture 19 November 16, 2016 13 / 44 Root test Proof of the root test. R+1 Let x = 2 so that R < x < 1 if x < 1 and 1 < x < R if R > 1. Suppose R < 1. Then, for all n sufficiently large, n an < x : P1 n The series converges by comparison with the geometric series n=1 x . Suppose instead that R > 1. Then for all n sufficiently large, n an > x so that the series diverges by comparison with the geometric series P1 n n=1 x . Bob Hough Math 141: Lecture 19 November 16, 2016 14 / 44 Example n2 P1 n We apply the root test to the series n=1 n+1 : Calculate n !n 1 n 1 1 a n = = ! : n n + 1 1 e 1 + n Thus the series converges. Bob Hough Math 141: Lecture 19 November 16, 2016 15 / 44 Ratio test Theorem P1 Let n=1 an be a sum of positive terms such that a n+1 ! L; n ! 1: an The series converges if L < 1 and diverges if L > 1. Bob Hough Math 141: Lecture 19 November 16, 2016 16 / 44 Ratio test Proof of the ratio test. If L > 1 then the series does not converge since the terms do not tend to 0. If L < 1, let L + 1 L < x = < 1: 2 Choose N sufficiently large, so that if n ≥ N then an+1 < x. It follows an that, for all n > N, n−1 Y ak+1 n−N aN n an = aN ≤ aN x = N x : ak x k=N aN Since xN is just a constant, the series converges by comparison with the P1 n geometric series n=1 x . Bob Hough Math 141: Lecture 19 November 16, 2016 17 / 44 Examples P1 n! Consider the series n=1 nn . Applying the ratio test, as n ! 1, n n an+1 (n + 1)! n n 1 1 = n+1 · = = n ! : an (n + 1) n! n + 1 (1 + 1=n) e Hence the series converges. Bob Hough Math 141: Lecture 19 November 16, 2016 18 / 44 Edge cases The ratio and root tests are not useful in the case an+1 ! 1 as n ! 1. an P1 1 P1 1 Note that n=1 n = 1 while n=1 n2 converges. The ratios of consecutive terms in these cases are given by n 1 1 n2 2 1 = 1 − + O ; = 1 − + O : n + 1 n n2 (n + 1)2 n n2 The following two tests of Raabe and Gauss sometimes help to decide convergence in cases where the limiting ratio is 1. Bob Hough Math 141: Lecture 19 November 16, 2016 19 / 44 Raabe's test Theorem (Raabe's test) P1 Let n=1 an be a sum of positive terms. If there is an r > 0 and N ≥ 1 such that, for all n ≥ N, a 1 r n+1 ≤ 1 − − an n n P1 then n=1 an converges. If, for all n ≥ N, a 1 n+1 ≥ 1 − an n P1 then n=1 an = 1: Bob Hough Math 141: Lecture 19 November 16, 2016 20 / 44 An inequality for the logarithm In the first part of the proof of Raabe's test we use the inequality log(1 + x) ≤ x; −1 < x: To check this inequality, note equality holds at x = 0. Also, x − log(1 + x) 1 has derivative 1 − 1+x , which changes sign at x = 0. It follows that the function decreases to 0, and increases thereafter.
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