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

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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
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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
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