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Unformatted text preview: ANALYSIS EXERCISE 14â€“SOLUTIONS 1. (a) For the lower bound, s (2 n ) = a (1) + a (2) + a (3) + a (4) + Â·Â·Â· + a (8) + Â·Â·Â· + a (2 n ) â‰¥ a (1) + a (2) + 2 a (4) + 4 a (8) + Â·Â·Â· + 2 n 1 a (2 n ) â‰¥ 1 2 { a (1) + 2 a (2) + 4 a (4) + 8 a (8) + Â·Â·Â· + 2 n a (2 n ) } . For the upper bound, group terms 1, 2 to 3, 4 to 7, 8 to 15, etc. and take the smallest each time to get s (2 n ) = a (1) + a (2) + a (3) + a (4) + Â·Â·Â· + a (8) + Â·Â·Â· + a (2 n ) â‰¤ a (1) + 2 a (2) + 4 a (4) + 8 a (8) + Â·Â·Â· + 2 n 1 a (2 n 1 ) + a (2 n ) (b) Denote by Ëœ s ( n ) the nth partial sum of âˆ‘ âˆž k =1 2 k a (2 k ). The result in (a) may be rewritten 1 2 Ëœ s ( n ) â‰¤ s (2 n ) â‰¤ Ëœ s ( n 1) + a (2 n ) . Note that ( s ( n )) and (Ëœ s ( n )) are monotone increasing sequences. If âˆ‘ âˆž n =1 a ( n ) converges to s then of course s ( n ) and hence s (2 n ) converge to s . By (a), the sequence (Ëœ s ( n )) is then bounded above by 2 s . So âˆ‘ âˆž n =1 2 n a (2 n ) converges by Theorem 3.4.3. On the other hand, if âˆ‘ âˆž n =1 2 n a (2 n ) converges to Ëœ s , say, then the sequence Ëœ s ( n ) converges to Ëœ s and is bounded above by Ëœ s . By (a), s (2 n ) â‰¤ Ëœ s...
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This note was uploaded on 02/25/2010 for the course MATHEMATIC 11007 taught by Professor Spyros during the Winter '10 term at Bristol Community College.
 Winter '10
 Spyros

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