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17Sept2008QuizSol

# 17Sept2008QuizSol - ± k(1 1/k k ² k = | x | lim k →∞...

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Partial Solutions for 17 September 2008 Quiz 1) Determine whether the following series converges: k k/ 2 k ! Since we have a k ! term, we will use the ratio test: lim k →∞ ( k + 1) ( k +1) / 2 ( k + 1)! · k ! k k/ 2 = lim k →∞ ± ( k + 1) k/ 2 ( k + 1) 1 / 2 k !( k + 1) · k ! k k/ 2 ² = lim k →∞ ± ( k + 1) k/ 2 ( k + 1) 1 / 2 · 1 k k/ 2 ² = lim k →∞ ± ( k + 1) k ² k/ 2 · 1 ( k + 1) 1 / 2 ! = lim k →∞ ± k (1 + 1 k ) k ² k/ 2 ! · ± lim k →∞ 1 ( k + 1) 1 / 2 ² = lim k →∞ ± 1 + 1 k ² k ! 1 / 2 · ± lim k →∞ 1 ( k + 1) 1 / 2 ² = e 1 / 2 · 0 = 0 Since 0 < 1, the ratio test tells us that the series converges. 3) Find the radius of convergence of X k =1 k k k ! x k . To ﬁnd the radius of convergence, we look at when the series converges absolutely. We will use the ratio test, so we examine: lim k →∞ ( k + 1) ( k +1) | x | k +1 ( k + 1)! · k ! k k | x | k = lim k →∞ | x | ( k + 1) k ( k + 1) k !( k + 1) · k ! k k = | x | lim k →∞ ( k + 1) k 1 · 1 k k = | x | lim k →∞ ± ( k + 1) k ² k = | x | lim k →∞
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Unformatted text preview: ± k (1 + 1 /k ) k ² k = | x | lim k →∞ ± 1 + 1 k ² k = e | x | This means that the series converges when e | x | < 1, that is, when | x | < 1 /e . This means the radius of convergence is 1 /e . NOTE: In particular, this means that when x = 1, the ratio test gives e as the limit and so ∑ ∞ k =1 k k k ! does NOT converge. For those of you that I talked to today, this means you can’t just do a comparison test with this series in problem 1 (since it is less than something that diverges the comparison does not give you any information). 1...
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