MATH 301 INTRO TO ANALYSIS FALL 2016

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1 MATH 301 INTRO TO ANALYSIS FALL 016 Homework 04 Professional Problem Consider the recursive sequence defined by x 1 = 3 and +1 = 1 4 for n 1. (a) Prove that ( ) converges. (Hint: show that ( ) is decreasing and bounded below. For decreasing, use induction: if x k+1 x k 0 then x k+ x k+1 0.) (b) Explain that lim +1 exists and is equal to lim. (c) Use (a) and (b) and compute lim. (Abbott,.4.1) Written Problems There are 5 available points. The highest score you can receive is 35 points, 15 of which replaces the canceled Quiz 04 (in/out). 1. (a) (4 points) What is wrong with the following argument? Consider the recursive sequence y 1 = 1; y n+1 = 3 y n for n 1. Let lim y n = y. Since lim y n = lim y n+1, taking the limit across the recursive equation gives y = 3 y, so y = lim y n = 3/. (b) (4 points) Would the strategy in (a) work if the recursive sequence is y 1 = 1 and y n+1 = 3 1 y n for n 1? Explain your answer. (Abbott,.4.) 1

2 . (a) (4 points) Show that, +, + +,... converges. Find the limit. (b) (4 points) Does the sequence,,,... converge? If so, find the limit. (Abbott,.4.3) 3. (Calculating Square Roots). Let x 1 = and define +1 = 1 ( + ), n 1. (a) ( points) Prove that for any a, b R, a + b ab. (This is called the inequality of arithmetic-geometric means.) (b) ( points) Show that x n for all n N. (c) ( points) Use (b) to show that (d) ( points) Use (c) to prove that lim =. (e) ( points) How would you modify ( ) so that it converges to c for a fixed number c? (Abbott,.4.5) 4. For each series, find an explicit formula for the sequence of partial sums and determine if the series converges. (a) ( points) 1 n (b) ( points) 1 n(n + 1) (c) ( points) n + 1 log n (Abbott,.4.8)

3 5. Give an example or argue that such a request is impossible. (a) ( points) A sequence that has a bounded subsequence but no convergent subsequence. (b) ( points) A sequence that does not contain 0 or 1 as a term but contains subsequences converging to each of these values. (c) ( points) A sequence that contains subsequences converging to every point in the infinite set {1, 1/, 1/3, 1/4,...}. (d) ( points) A sequence that contains subsequences converging to every point in the infinite set {1, 1/, 1/3, 1/4,...}, and no subsequences converging to points outside of this set. (Abbott,.5.1) 6. (4 points) Assume (a n ) is a bounded sequence with the property that every convergent subsequence of (a n ) converges to the same limit a. Show that (a n ) must converge to a. (Abbott,.5.5) 7. (4 points) Show that lim b 1/n exists for all b 0. Find the value of the limit. (Abbott,.5.6) 8. (4 points) Show that lim(b n ) = 0 if and only if 1 < b < 1. (Abbott,.5.7) 3

4 Solutions Professional Problems (a) ( ) is decreasing. Since x 1 = 3 and x = 1, it suffices to show the inductive step: if x k+1 x k 0 then x k+ x k+1 0. Calculate x k+ x k+1 = 1 1 = (4 x k) (4 x k+1 ) = 4 x x+1 4 x k (4 x k+1 )(4 x k ) x k+1 x k (4 x k+1 )(4 x k ) which is 0 because x k+1 x k 0 and the denominator is clearly positive. ( ) is bounded below. It is obvious that > 0. Since ( ) is decreasing and bounded below, it converges by Monotone Convergence Theorem. (b) Assume ( ) L, then ɛ > 0, N 1 such that if n N 1 then L < ɛ. Let N = N 1 1. Then if n N = N 1 1, n + 1 N 1, and +1 L < ɛ. Therefore (+1 ) L. (b ) Another proof to (b): If ( ) converges to L then, by Theorem.5., all its subsequences converge to the same limit. Since (+1 ) is a subsequence, it must converge to L. (c) Let lim = lim +1 = L. By Algebraic Limit Theorem, L = 1 4 L. So Since L < 3, L = 3. L(4 L) = 1, L = ± 3. Written Problems 1. (a) We did not show that (y n ) is convergent. (b) We need to show the sequence is convergent. This can be done by showing it is increasing and bounded above, and by applying the Monotone Convergence Theorem. For increasing property, note that y 1 = 1 and y =. Prove the inductive step that if y k+1 y k 0 then y k+ y k+1 0. Calculate ( y k+ y k+1 = 3 1 ) ) (3 1yk = 1 1 = y k+1 y k y k+1 y k y k+1 y k y k+1 which is 0.. (a) The sequence is defined recursively by a 1 =, a n+1 = + a n, n. For convergence, it suffice to show that (a n ) is increasing and bounded above. ((a n ) is bounded.) Claim that a n. This can be proved by induction. Since a 1, we only need to prove the inductive step: if a k then a k+1 because a k+1 = + a n + = 4 =. 4

5 ((a n ) is increasing.) Consider a n+1 a n = + a n a n = ( a n )(1 + a n ) which is 0 because a n and a n 0. For the limit, assume lim a n = L = lim a n+1. Then Since L 0, L =. (b) The sequence is defined recursively by L = + L, L = + L, L =, 1. a 1 =, a n+1 = a n, n. For convergence, it suffice to show that (a n ) is increasing and bounded above. ((a n ) is bounded.) Claim that a n. This can be proved by induction. Since a 1, we only need to prove the inductive step: if a k then a k+1 because ((a n ) is increasing.) Consider a k+1 = a n = 4 =. a n+1 a n = a n a n = a n ( a n ) which is 0 because a n and a n 0. For the limit, assume lim a n = L = lim a n+1. Then Since L 0, L =. L = L, L = L, L =, (a) (b) a + b ab = a + ab + b 4 x n+1 = 4ab 4 = a ab + b 4 xn + =. a b = 0. (c) So ( ) is decreasing. +1 = 1 ( + ) ( = ) = 0. (d) By Monotone Convergence Theorem, ( ) converges because it is decreasing and bounded below. Let L = lim = lim +1, then L = 1 ( L + ), L = L + L L, L =, L = ±. Since L 0, L =. (e) x 1 = c, +1 = 1 + c, n 1. 5

6 4. (a) Partial sum k 1 n = (1/) 1 (1/) 1 (1/) k; limit 1. (b) Partial sum (c) Partial sum k k 1 n(n + 1) = 1 1 ; limit 1. k + 1 n + 1 log = log(k + 1); limit does not exist. n 5. (a) This is impossible. If the sequence (a n ) has a bounded subsequence (a nk ), then (a nk ) itself is a bounded sequence, which has a convergent subsequence by Bolzano-Weierstrass Theorem. The subsequence of (a nk ) is also a subsequence of (a n ) because it consists of terms in (a n ). (b) This is possible. Consider (c) This is possible. For example, (a n ) = (0.1, 0.9, 0.01, 0.99, 0.001, 0.999,...), i.e. { 1 (0.1) a n = m if n = m, m N, (0.1) m if n = m 1, m N. (a n ) = (1, 1, (1/), 1, (1/), (1/3), 1, (1/), (1/3), (1/4),...). (d) This is impossible because we can build a subsequence approaching 0, which is outside of the set. Proof: Choose a n1 from the subsequence approaching 1 with a n1 1 < 1, (i.e. ɛ 1 = 1). For k, choose term a nk from the subsequence approaching 1/k with a nk (1/k) < (1/k), (i.e. ɛ k = 1/k), and also require n k > n k 1. Then (a nk ) is a subsequence with a 1 nk k < 1 k, 1 k 1 k < a n k < 1 k + 1 k, so a nk 0 < k, which implies lim a n k = Assume (a n ) does not converge to a, then there exists ɛ > 0 so that for any N there exists n, n N and a n a ɛ. Let the first term be a n1. For k, let N = n k and there exists n, n N and a n a ɛ. Let this value be n k+1. Construct inductively, and (a nk ) is a subsequence of (a n ) with all terms at least ɛ away from a. Since (a nk ) is bounded, by the Bolzano-Weierstrass Theorem, it has a convergent subsequence. This subsequence of (a nk ) is also a subsequence of (a n ) and its limit is at least ɛ away from a, which is a contradiction. 7. (a) If b = 1, b 1/n = 1 for all n, and the limit is 1. (b) If b > 1, it is clear that b 1/n is decreasing and bounded below by 1. By the Monotone Convergence Theorem, lim b 1/n exists. Assume lim b 1/n = L. Then consider the subsequence a nk = b 1/k, i.e. a n1 = b, a n = 4 b, a n3 = 8 b, etc. For this subsequence, a nk+1 = a nk. Since the subsequence must converge to the same limit as the sequence, let k and we have L = L, so L = 0 or L = 1. (c) If 0 < b < 1, it is clear that b 1/n is increasing and bounded above by 1. By the Monotone Convergence Theorem, lim b 1/n exists. Assume lim b 1/n = L. Then consider the subsequence 6

7 a nk = b 1/k, i.e. a n1 = b, a n = 4 b, a n3 = 8 b, etc. For this subsequence, a nk+1 = a nk. Since the subsequence must converge to the same limit as the sequence, let k and we have L = L, so L = 0 or L = This proposition is equivalent to lim b n = 0 if and only if b < 1. ( ). Assume b < 1 then b n = b n is decreasing and bounded below by 0, so it converges by Monotone Convergence Theorem and we denote lim b n = L. Let a n = b n, then a n+1 = b a n. Since lim a n+1 = lim a n, we have Since b < 1, L = 0. L = bl, (b 1)L = 0, B = 1 or L = 0. ( ). We prove the contrapositive. Assume b 1 then b n 1, so b n 0. 7