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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

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

Page1 / 3

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online