Since z n inf A n sup A n y n we have z n y n for all n 1 Then lim z n lim y n

Since z n inf a n sup a n y n we have z n y n for all

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Since z n = inf A n sup A n = y n , we have z n y n for all n 1. Then lim z n lim y n by Theorem 2.3.4(ii). Thus lim inf a n lim sup a n . If a n = (1) n , we clearly have z n = 1, y n = 1 for all n 1. Thus lim inf a n ¡ lim sup a n in this case. (d) Show that lim inf a n = lim sup a n if and only if lim a n exists. In this case, all three share the same value. We have z n a n y n for all n 1. If lim inf a n = lim sup a n , we have lim z n = lim y n = and then lim a n exists and is equal to by the Squeeze Theorem. Exercise 2.4.8 For each series, find an explicit formula for the sequence of partial sums and determine if the series converges. (b) n = 1 1 n ( n + 1 ) Let (s m ) denote the sequence of partial sums of n = 0 1 2 n Then compute s m = 1 1 x 1 2 + 1 2 x 1 3 + 1 3 x 1 4 +...+ 1 m x 1 m + 1 = ( 1 1 - 1 2 ) + ( 1 2 - 1 3 ) + ( 1 3 - 1 4 ) + . . . + ( 1 m - 1 m + 1 ) = 1 - 1 m + 1 . Therefore the Algebraic Limit Theorem implies that (s m ) converges and lim m s m = lim m (1- 1 m + 1 ) = 1. Exercise 2.4.9 Complete the proof of Theorem 2.4.6 by showing that if the series n = 1 2 n b 2 n diverges, then so does n = 1 b n Because n = 1 2 n b 2 n 2
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Hunter College MATH 351 Mathematical Analysis I diverges, its monotone sequence of partial sums (t k ) must be unbounded. To show that (s m ) is unbounded it is enough to show that for all k N , there is term s m satisfying s m t k /2. Given an arbitrary k, we focus our attention on s 2 k and observe that s 2 k = b 1 + b 2 + (b 3 + b 4 ) + (b 5 + b 6 + b 7 + b 8 ) + ... + (b 2 k - 1 + 1 + ... + b 2 k ) b 1 + b 2 + (b 4 + b 4 ) + (b 8 + b 8 + b 8 + b 8 ) + ... + (b 2 k + ... + b 2 k ) = b 1 + b 2 + 2b 4 + 4b 8 + ... + 2 k - 1 b 2 k = 1/2(2b1+ 2b2+ 4b4 + 8b8 + ... + 2 k b 2 k ) =b 1 /2 +t k /2. Because (t k ) is unbounded, the sequence (s m ) must also be unbounded and cannot converge. Therefore, n = 1 b n diverges. Exercise 2.5.1 Give an example of each of the following, or argue that such a request is impossible. (a) A sequence that has a subsequence that is bounded but contains no subsequence that converges. Impossible. Let (b n ) be a bounded sequence of (a n ) then by Bolzano-Weierstrass Theorem (b n ) contains a subsequence that converges. (b) A sequence that does not contain 0 or 1 as a term but contains subsequences converging to each of these values. Consider a n = (0.1, 0.9, 0.01, 0.99, 0.001, 0.999, . . .) a n = ( 1 - ( 0.1 ) m , i f n = 2 m , m N , ( 0.1 ) m , i f n = 2 m - 1, m N (1) (2) (c) A sequence that contains subsequences converging to every point in the infinite set { 1, 1/2, 1/3, 1/4, 1/5, . . . } . Consider a n = (1, 1, (1/2), 1, (1/2), (1/3), 1, (1/2), (1/3), (1/4), . . .). (d) A sequence that contains subsequences converging to every point in the infinite set { 1, 1/2, 1/3, 1/4, 1/5, . . . } , and no subsequences converging to points outside of this set. Impossible because we can build a subsequence approaching 0 which is outside the set. Choose a n 1 from the subsequence approaching 1 with | a n 1 -1 | < 1, (i.e. ε 1 = 1). For k 2, choose term a nk from the subsequence approaching 1/k with | a nk -(1/k) | < (1/k), (i.e. ε k = 1/k), and also require n k > n k 1 . Then (a n k ) is a subsequence with | a nk - 1 k | < 1 k , 1 k - 1 k < a nk < 1 k + 1 k , so | a nk -0 | < 2 k , which implies lim a nk = 0.
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