{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# 112 - MATH 153 Selected Solutions for 11.2 Exercise 18...

This preview shows pages 1–3. Sign up to view the full content.

MATH 153 Selected Solutions for § 11.2 Exercise 18 : Determine whether the geometric series is convergent or diver- gent. If it is convergent, find its sum. X n =0 1 ( 2) n Solution : We first rewrite the series in ar n form: X n =0 1 · 1 2 n . This converges if and only if 1 2 < 1, which is true. Since it converges, we know that the limit of a convergent geometric series is a 1 - r , where a is the first term and r is the ratio. Thus, X n =0 1 ( 2) n = 1 1 - 1 2 = 2 2 - 1 . Exercise 24 : Determine whether the series is convergent or divergent. If it is convergent, find the sum. X k =1 k ( k + 2) ( k + 3) 2 Solution : We show that the series diverges by the Test for Divergence; the terms of the sequence don’t go to 0, so the series cannot converge. lim k →∞ k ( k + 2) ( k + 3) 2 = lim k →∞ k 2 + 2 k k 2 + 6 k + 9 = lim k →∞ 1 + 2 /k 1 + 6 /k + 9 /k 2 = 1 + 0 1 + 0 + 0 = 1 . 1

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

View Full Document
2 Exercise 32 : Determine whether the series is convergent or divergent. If it is convergent, find the sum. X n =1 3 5 n + 2 n Solution : We can break up the series in the following way: 3 X n =1 1 5 n + 2 X n =1 1 n . The first series is geometric, so it converges. The second is the harmonic series, so it diverges. Then it cannot be that the full series converges, because
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}