Homework #4 Solutions - Homework 4 Solutions Math 131A-3 1(11.2 Well do all ve parts for each sequence in turn an =(1)n Two monotone subsequences are(x1

Homework #4 Solutions - Homework 4 Solutions Math 131A-3...

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Homework 4 Solutions Math 131A-3 1. (11.2) We’ll do all five parts for each sequence in turn. a n = ( - 1) n . Two monotone subsequences are ( x 1 , x 3 , x 5 , · · · ) = (1 , 1 , 1 , · · · ), and similarly for the even entries. The set of subsequential limits is 1 } . Therefore lim sup a n = 1 and lim inf a n = - 1. This sequence is bounded, | a n | < 1 for all n , but does not converge or diverge to ±∞ . b n = 1 n . The sequence ( b n ) is monotone decreasing, so any subsequence is as well; furthermore ( b n ) converges to 0, so any subsequence of ( b n ) also converges to 0. There- fore the set of subsequential limits is { 0 } , and lim sup b n = lim inf b n = 0. Finally ( b n ) is bounded, e.g. | b n | < 2 for all n . u n = ( - 1 2 ) n . Any subequence ( x n i ) where all the n i are even is monotone decreas- ing, and any subsequence ( x n i ) where all the n i are odd is monotone increasing. The sequence u n converges to 0, so the set of subsequential limits is { 0 } , and lim sup u n = lim inf u n = 0. Finally, u n is bounded, e.g. | u n | < 1 for all n . x n = 5 ( - 1) n . Possible monotone subsequences include ( x 1 , x 3 , x 5 , · · · ) = ( 1 5 , 1 5 , 1 5 , · · · ) and ( x 2 , x 4 , x 6 , · · · ) = (5 , 5 , 5 , · · · ). Note that (5 , 5 , · · · , 5 , 1 5 , 1 5 , · · · ) is also monotone decreasing, and ( 1 5 , 1 5 , · · · 1 5 , 5 , 5 , 5 , · · · ) is monotone increasing. The set of subsequen- tial limits is { 1 5 , 5 } , and lim sup x n = 5 while lim inf x n = 1 5 . The sequence x n neither
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  • Fall '08
  • hitrik
  • Limits, Order theory, Limit of a sequence, lim sup, subsequence

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