hw7_sol

# hw7_sol - Math 311 HW 7 Solutions April 11, 2008 7.3.2 Let...

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Math 311 HW 7 Solutions April 11, 2008 7.3.2 Let | b n | < B for all n . By hypothesis, such a B exists. Then, | a n b n | < | a n | B for all n, and so | a n b n | converges by comparison with | a n | B = B | a n | (linearity theorem). Then by absolute convergence theorem, a n b n converges. 7.3.3 a) | a n | converges by hypothesis. Then let S N = N i =1 | a n i | . We have S N n N i =1 | a i | (since S N is some but not necessrily all of the positive terms on the right). And n N i =1 i =1 | a i | , so S N is bounded. Clearly S N is increasing, hence by completeness property i | a n i | converges. By absolute convergence theorem, i a n i must converge. 7.3.3 b) Let a n = ( - 1) n 1 n . Then a 2 n = 1 2 n and a n = ln 2 is convergent but a 2 n = 1 2 n = 1 2 1 n is divergent. 7.4.1 b) 1 n 2 2 n . Ratio Test: lim n →∞ ± ± ± ± a n +1 a n ± ± ± ± = lim n →∞ ( n + 1) 2 2 n n 2 2 n +1 = lim n →∞ 1 2 ² 1 + 2 n + 1 n 2 ³ = 1

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## This note was uploaded on 09/29/2011 for the course MATH 3110 at Cornell.

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hw7_sol - Math 311 HW 7 Solutions April 11, 2008 7.3.2 Let...

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