hw6[1] - HOMEWORK 6: SOLUTIONS/HINTS 14.2 (a) As N X n =1 n...

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

View Full Document Right Arrow Icon
HOMEWORK 6: SOLUTIONS/HINTS 14.2 (a) As N X n =1 n ± 1 n 2 = N X n =1 1 n ± N X n =1 1 n 2 ; we see that the series diverges. (d) We have lim ± ± ± a n +1 a n ± ± ± = 1 3 , hence it converges by the ratio test. (e) As lim ± ± ± a n +1 a n ± ± ± = 0 the series converges. 14.3 The series in (e) does not converge, as sin( 9 ) is not a zero sequence. All the other series converge: (a) and (f) by the ratio test, (b)-(d) by the comparison test. 14.4 The series in (a) and (c) converge by the comparison test respectively the ratio test. The series in (b) diverges by the comparison test. 14.6 As ( b n ) is bounded, there is an M > 0 such that j b n j < M for all n: Then P M j a n j converges (by exc. 14.5 (b)) and j a n b n j < M j a n j , therefore P a n b n converges by the comparison test. For part (b), just take b n = 1 in (a) to get the statement of Corollary 14.7. 14.12 Since lim inf j a n j = 0, there is a subsequence ( a n i ) of ( a n ) such that lim a n i = 0. Set b i := a n i , then as lim b i = 0 there exists an i 1 2 N such that j b i j < 1 1 2 for all i ² i 1 . Now by induction we ±nd an increasing sequence of integers
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.

Page1 / 2

hw6[1] - HOMEWORK 6: SOLUTIONS/HINTS 14.2 (a) As N X n =1 n...

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