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Math 4111 October 9, 2020 Lecture Advanced Convergence Tests Math 4111 October 9, 2020 Lecture Steven G. Krantz Steven G. Krantz Math 4111 October 9, 2020 Lecture Advanced Convergence Tests Figure: This is your instructor. Steven G. Krantz Math 4111 October 9, 2020 Lecture Advanced Convergence Tests Now we have the first nontrivial convergence test for series. Theorem (The Root Test) P1 1=j Consider the series j=1 aj and let r = lim sup jaj j : j!1 P1 (a) If r < 1, then the series j=1 aj converges. P1 (b) If r > 1, then the series j=1 aj diverges. Steven G. Krantz Math 4111 October 9, 2020 Lecture Advanced Convergence Tests r+1 Proof: (a). By our hypothesis, for β = 2 < 1 there is an 1=j integer N > 1 such that, for all j > N, it holds that jaj j < β; j namely, jaj j < β : Since 0 < β < 1, by the Geometric Series Test, P1 j P1 j=1 β converges. By the Comparison Test, j=1 aj converges. Steven G. Krantz Math 4111 October 9, 2020 Lecture Advanced Convergence Tests (b). When r > 1, it can be seen that there are infinitely many 1=j 1+r positive integers j such that jaj j > 2 = β > 1, that is, j P1 jaj j > β . Thus (aj ) is unbounded, and j=1 aj diverges. Note that the Root Test is inconclusive if r = 1. Indeed, for P1 P1 2 both n=1 1=n and n=1 1=n , we have r = 1, but only the second series converges. Steven G. Krantz Math 4111 October 9, 2020 Lecture Advanced Convergence Tests Example P1 n2 Determine if the series n=1 2n converges. n2 Solution. Let a = : Then n 2n 2 n1=n ja j1=n = : n 2 As n ! 1, we see that 1=n 1 lim sup janj = < 1: n!1 2 By the Root Test, the series converges. Note that the Root Test is not easy to apply when complicated 5n factors such as n! are involved, as in P1 . n=1 n! Steven G. Krantz Math 4111 October 9, 2020 Lecture Advanced Convergence Tests Theorem (The Ratio Test) 1 X aj+1 Consider a series aj : If lim sup < 1; then the series aj j=1 j!1 1 X aj converges. j=1 aj+1 Proof: Let λ = lim sup < 1: Select a real number µ such j!1 aj that λ < µ < 1. By the definition of lim sup, there is an N so large that if j > N then aj+1 < µ : This may be rewritten as aj jaj+1j < µ · jaj j ; for all j ≥ N: Thus (much as in the proof of the Root Test) we have for k ≥ 0 that k jaN+k j ≤ µ · jaN+k−1j ≤ µ · µ · jaN+k−2j ≤ · · · ≤ µ · jaN j : It is convenient to denote N + k by n; n ≥ N. Thus the last inequality reads n−N janj < µ · jaN j Steven G. Krantz Math 4111 October 9, 2020 Lecture Advanced Convergence Tests or 1=n (n−N)=n 1=n janj < µ · jaN j : Remembering that N has been fixed once and for all, we pass to the lim sup as n ! 1. The result is 1=n lim sup janj ≤ µ : n!1 Since µ < 1, we find that our series satisfies the hypotheses of the Root Test. Hence it converges. Remark. The proof of the Ratio Test shows that if a series passes the Ratio Test then it passes the Root Test (the converse is not true, as you will learn in Exercise 2). Why do we therefore learn the Ratio Test? The answer is that there are circumstances when the Ratio Test is easier to apply than the Root Test. Steven G. Krantz Math 4111 October 9, 2020 Lecture Advanced Convergence Tests aj+1 P1 Note that lim sup > 1 does not imply that j=1 aj diverges. j!1 aj P1 1 For instance, consider j=1 (3+(−1)j )j . However, we have Theorem (The Ratio Test for Divergence) aj+1 P1 If lim infj!1 > 1 ; then the series aj diverges. aj j=1 P1 j! For example, the series j=1 5 j diverges as aj+1 limj!1 = 1 > 1: aj Steven G. Krantz Math 4111 October 9, 2020 Lecture Advanced Convergence Tests Example P1 2 j Show that the series j=1 j! converges. j Proof: We have aj = 2 =j! and j+1 aj+1 2 =(j+1)! 2 = j = : aj 2 =j! j+1 The lim sup of the last expression is 0. By the Ratio Test, the series converges. Example P1 n! Determine if n=1 nn converges. n! Solution. We have an = nn and n+1 n an+1 (n + 1)!=(n + 1) n 1 1 = = n = ! < 1: an n!=nn (n+1) (1 + 1=n)n e By the Ratio Test, the series converges. Steven G. Krantz Math 4111 October 9, 2020 Lecture Advanced Convergence Tests The Integral Test We conclude this section with the Integral Test. Proposition (The Integral Test) Let f be a continuous and decreasing function on [0; 1). The 1 X Z 1 series f (j) converges if and only if the integral f (x) dx j=1 1 converges. Z 1 Proof Outline. The improper integral f (x) dx is the area 1 below the curve y = f (x) over the interval [1; 1]. Observe that 1 1 Z 1 X X Z 1 f (x) dx ≥ f (j) and f (j) ≥ f (x) dx: 1 j=2 j=1 1 Steven G. Krantz Math 4111 October 9, 2020 Lecture Advanced Convergence Tests Example 1 X 2 Let p > 1: Show that the series converges. j(ln j)p j=1 1 Proof. Note that x(ln x)p is continuous and decreasing on [2; 1), and Z 1 1−p N 1−p 1 (ln x) (ln 2) p dx = lim = converges. 2 x(ln x) N!+1 1 − p 2 p − 1 Therefore, the series converges. Steven G. Krantz Math 4111 October 9, 2020 Lecture Advanced Convergence Tests In this section we consider convergence tests for series which depend on cancellation among the terms of the series. One of the most profound of these depends on a technique called summation by parts. You may wonder whether this process is at all related to the \integration by parts" procedure that you learned in calculus|it has a similar form. Indeed it will turn out (and we shall see the details of this assertion as the book develops) that summing a series and performing an integration are two aspects of the same limiting process. The summation by parts method is merely our first glimpse of this relationship. Steven G. Krantz Math 4111 October 9, 2020 Lecture Advanced Convergence Tests Proposition (Summation by Parts) 1 1 Let faj gj=0 and fbj gj=0 be two sequences of real or complex numbers. For N = 0; 1; 2;::: set N X AN = aj j=0 (we adopt the convention that A−1 = 0). Then, for any 0 ≤ m ≤ n < 1, it holds that n X aj · bj = [An · bn − Am−1 · bm] j=m n−1 X + Aj · (bj − bj+1) : j=m Steven G. Krantz Math 4111 October 9, 2020 Lecture Advanced Convergence Tests Proof: We write n n X X aj · bj = (Aj − Aj−1) · bj j=m j=m n n X X = Aj · bj − Aj−1 · bj j=m j=m n n−1 X X = Aj · bj − Aj · bj+1 j=m j=m−1 n−1 X = Aj · (bj − bj+1) + An · bn − Am−1 · bm : j=m This is what we wished to prove. Steven G. Krantz Math 4111 October 9, 2020 Lecture Advanced Convergence Tests Now we apply summation by parts to prove a convergence test due to Niels Henrik Abel (1802{1829). Theorem (Abel's Convergence Test) Consider the series 1 X aj · bj : j=0 Suppose that PN 1. The partial sums AN = j=0 aj form a bounded sequence; 2. b0 ≥ b1 ≥ b2 ≥ ::: ; 3. limj!1 bj = 0. Then the original series 1 X aj · bj j=0 converges. Steven G. Krantz Math 4111 October 9, 2020 Lecture Advanced Convergence Tests Proof: Suppose that the partial sums AN are bounded in absolute value by a number K. Pick > 0 and choose an integer N so large that bN < /(2K). For N < m ≤ n < 1 we use the partial summation formula to write n n−1 X X aj · bj = An · bn − Am−1 · bm + Aj · (bj − bj+1) j=m j=m n−1 X ≤ K · jbnj + K · jbmj + K · jbj − bj+1j : j=m Steven G. Krantz Math 4111 October 9, 2020 Lecture Advanced Convergence Tests Now we take advantage of the facts that bj ≥ 0 for all j and that bj ≥ bj+1 for all j to estimate the last expression by 2 n−1 3 X K · 4bn + bm + (bj − bj+1)5 : j=m [Notice that the expressions bj − bj+1; bm, and bn are all nonnegative.] Now the sum collapses and the last line is estimated by K · [bn + bm − bn + bm] = 2 · K · bm : By our choice of N the right side is smaller than . Thus our series satisfies the Cauchy criterion and therefore converges. Steven G. Krantz Math 4111 October 9, 2020 Lecture Advanced Convergence Tests Example (The Alternating Series Test) As a first application of Abel's convergence test, we examine alternating series. Consider a series of the form 1 X j (−1) · bj ; (3:32:1) j=1 with b1 ≥ b2 ≥ b3 ≥ · · · ≥ 0 and bj ! 0 as j ! 1.
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