Sole14 - ANALYSIS EXERCISE 14–SOLUTIONS 1(a For the lower bound s(2 n = a(1 a(2 a(3 a(4 ·· a(8 ·· a(2 n ≥ a(1 a(2 2 a(4 4 a(8 ·· 2 n 1

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Unformatted text preview: ANALYSIS EXERCISE 14–SOLUTIONS 1. (a) For the lower bound, s (2 n ) = a (1) + a (2) + a (3) + a (4) + ··· + a (8) + ··· + a (2 n ) ≥ a (1) + a (2) + 2 a (4) + 4 a (8) + ··· + 2 n- 1 a (2 n ) ≥ 1 2 { a (1) + 2 a (2) + 4 a (4) + 8 a (8) + ··· + 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) + ··· + a (8) + ··· + a (2 n ) ≤ a (1) + 2 a (2) + 4 a (4) + 8 a (8) + ··· + 2 n- 1 a (2 n- 1 ) + a (2 n ) (b) Denote by ˜ s ( n ) the n-th partial sum of ∑ ∞ k =1 2 k a (2 k ). The result in (a) may be rewritten 1 2 ˜ s ( n ) ≤ s (2 n ) ≤ ˜ s ( n- 1) + a (2 n ) . Note that ( s ( n )) and (˜ s ( n )) are monotone increasing sequences. If ∑ ∞ n =1 a ( n ) converges to s then of course s ( n ) and hence s (2 n ) converge to s . By (a), the sequence (˜ s ( n )) is then bounded above by 2 s . So ∑ ∞ n =1 2 n a (2 n ) converges by Theorem 3.4.3. On the other hand, if ∑ ∞ n =1 2 n a (2 n ) converges to ˜ s , say, then the sequence ˜ s ( n ) converges to ˜ s and is bounded above by ˜ s . By (a), s (2 n ) ≤ ˜ 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.

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Sole14 - ANALYSIS EXERCISE 14–SOLUTIONS 1(a For the lower bound s(2 n = a(1 a(2 a(3 a(4 ·· a(8 ·· a(2 n ≥ a(1 a(2 2 a(4 4 a(8 ·· 2 n 1

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