it implies s n t m for all n m N In particular taking m 3 we get s n b 3 256 81

It implies s n t m for all n m n in particular taking

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it implies s n t m for all n, m N . In particular, taking m = 3, we get s n b 3 = 256 81 < 4 . Therefore lim s n = e exists. Appendix. Some problems discussed in the class 4.4.1. True or false: (a) A sequence ( s n ) converges to s iff every subsequence of ( S n ) converges to s . (b) Every bounded sequence is convergent. (c) Let ( s n ) be a bounded sequence. if ( s n ) oscillates, then the set S of subse- quential limits of s n contains at least two points. (d) Let ( s n ) be a bounded sequence and let m = lim sup s n . Then for every > 0 , there exists N N such that n N implies that s n > m - . (e) If ( s n ) is unbounded above, then ( s n ) contains a subsequence that has + as a limit. Solution: (a) True. (= ) By Theorem 4.4.4. ( =) The ( s n ) itself is also a subsequence. (b) False. For example, s n = ( - 1) n . (c) True. lim sup s n and lim inf s n . (d) False. For example, s n = ( - 1) n , m = lim sup s n = 1, take = 1 2 . (e) True. We can choose ( s n k ) such that n 1 < n 2 < n 3 < ... and s n k k for all k N . 4.4.5. Find the limits: 85
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(a) s n = 1 + 1 2 n 2 n . (b) s n = 1 + 1 n 2 n . (c) s n = 1 + 1 n n - 1 . (d) s n = n n +1 n . (e) s n = 1 + 1 2 n n . (f) s n = n +2 n +1 n +3 . (g) s n = 1 - 1 n n . (h) s n = 1 - 2 n n . Solution: (a) s n = 1 + 1 2 n 2 n e. (b) s n = 1 + 1 n 2 n = 1 + 1 n n 2 e 2 . (c) s n = 1 + 1 n n - 1 = 1 + 1 n n 1 + 1 n e 1 = e. (d) s n = n n + 1 n = 1 1 + 1 n n e. (e) s n = 1 + 1 2 n n = s 1 + 1 2 n 2 n e. 86
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(f) s n = n + 2 n + 1 n +3 = 1 + 1 n + 1 n +1 · n + 2 n + 1 2 e · 1 = e. (g) 1 - 1 n n = n - 1 n n = 1 ( n n - 1 ) n = 1 (1 + 1 n - 1 ) n = 1 (1 + 1 n - 1 ) n - 1 · 1 1 + 1 n - 1 1 e · 1 = 1 e . (h) 1 - 2 n n = 1 - 2 n n 2 · 2 = 1 - 2 n n 2 2 1 e 2 = 1 e 2 . We used the result (g) in the last step. 4.4.8. If ( s n ) is a subsequence of ( t n ) and ( t n ) is a subsequence of ( s n ) , can we conclude that ( s n ) = ( t n )? Prove or give a counterexample. Proof: Not true. For example, ( s n ) = (0 , 1 , 0 , 1 , 0 , 1 , ... ) and ( t n ) = (1 , 0 , 1 , 0 , 1 , 0 , ... ). 4.4.13. Le ( s n ) and ( t n ) be bounded sequences. Prove lim sup ( s n + t n ) lim sup ( s n ) + lim sup ( t n ) , and find an example to show that the equality may not hold. Proof: By Corollary 4.4.12, there is a subsequence ( s n k + t n k ) of ( s n + t n ) such that lim k ( s n k + t n k ) = lim n sup ( s n + t n ) . Since ( s n k ) is bounded, there is a convergent subsequence ( s n km ). Since ( t n km ) is bounded, there is a convergent subsequence, still denoted by the same notation. Then we use Theorem 4.2.1 to imply lim m s n km + lim m t n km = lim m ( s n km + t n km ) .
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