real_analysis_homework6 - Real Analysis II Homework 6 Hctor Guillermo Cullar R e e os 19.2(b The set of subsequential limits of a bounded sequence is

real_analysis_homework6 - Real Analysis II Homework 6 Hctor...

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Real Analysis II Homework 6 ector Guillermo Cu´ ellar R´ ıos March 30, 2006 19.2 (b) The set of subsequential limits of a bounded sequence is always non- empty. True by Corollary 19.12. (c) ( s n ) converges to s iff lim inf s n = lim sup s n = s True by third paragraph in Definition 19.9 and exercise 19.9. (d) Let ( s n ) be a bounded sequence and let m = lim sup s n . Then for every ε > 0 there are infinitely many terms in the sequence greater than m - ε True by Theorem 19.11(b) 19.3 For each sequence, find the set S of subsequential limits, the limit superior, and the limit inferior. (a) s n = ( - 1) n S = {- 1 , 1 } lim sup s n = 1 lim inf s n = - 1 (c) u n = n 2 [ - 1 + ( - 1) n ] S = {-∞ , 0 } lim sup u n = 0 lim inf u n = -∞ 19.4 Let ( s n ) be a bounded sequence and suppose that lim inf s n = lim sup s n = s . Prove that ( s n ) is convergent and that lim s n = s . (a) w n = ( - 1) n n S = { 0 } lim sup w n = 0 lim inf w n = 0 (d) z n = ( - n ) n S = {-∞ , + ∞} lim sup z n = + lim inf z n = -∞ 19.5 Use exercise 18.14 to find the limit of each sequence. 1
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(a) s n = ( 1 + 1 2 n ) 2 n s n = ( 1 + 1 2 n ) 2 n e since it is a subsequence of the sequence s n = ( 1 + 1 n ) n (b) s n = ( 1 + 1 n ) 2 n ( 1 + 1 n ) 2 n = ( 1 + 1 n ) n ( 1 + 1 n ) n since each sequence is converging to e , then by Theorem 17.1(c) ( 1 + 1 n ) 2 n e 2 (c) s n = n n +1 n n n +1 n = 1 ( n +1 n ) n = 1 ( 1+ 1 n ) n 1 e = e - 1 (e) s n = ( 1 + 1 2 n ) n ( 1 + 1 2 n ) n = ( 1 + 1 2 n ) 2 n 2 = (
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  • Fall '08
  • Akhmedov,A
  • Limits, Limit of a sequence, Sn, subsequence, lim sup sn, Snk

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