{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW13 - Math 191 11.3 2 converges a geometric series with r...

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

Math 191 HW 13 Solutions 11.3: 2 converges; a geometric series with r = 1 e < 1 11.3: 6 converges; X n =1 - 2 n n = - 2 X n =1 1 n 3 / 2 , which is a convergent p-series ( p = 3 2 ). 11.3: 18 diverges; lim n →∞ a n = lim n →∞ 1 + 1 n n = e 6 = 0 11.3: 28 diverges by the Integral Test; Z 1 x x 2 + 1 dx ; u = x 2 + 1 du = 2 xdx 1 2 Z 2 du 4 = lim b →∞ 1 2 ln u b 2 = lim b →∞ 1 2 (ln b - ln 2) = 11.3: 39 a. Z 2 dx x (ln x ) p ; u = ln x du = dx x Z ln 2 u - p du = lim b →∞ u - p +1 - p + 1 b ln 2 = lim b →∞ ( 1 1 - p )[ b - p +1 - (ln 2) - p +1 ] = 1 p - 1 (ln 2) - p +1 , p > 1 , p < 1 the improper integral converges if p > 1 and diverges if p < 1. For p = 1; Z 2 dx x ln x = lim b →∞ [ln(ln x )] b 2 = lim b →∞ [ln(ln b ) - ln(ln 2)] = , so the improper integral diverges if p = 1. b. Since the series and the integral converges or diverges together, X n =2 1 n (ln n ) p converges if and only if p > 1. 11.4: 4 converges by the Direct Comparison Test; 1+cos n n 2 2 n 2 and the p-series Σ 1 n 2 converges.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 2

HW13 - Math 191 11.3 2 converges a geometric series with r...

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

View Full Document
Ask a homework question - tutors are online