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G. for n → ∞. We get by a decomposition, 1 1 1 1 1 = · − · . (n − 2)n 2 n−2 2 n Here n occurs in the denominator, so we are aiming at a logarithmic series. We get for |x| < 1, ∞ ∞ ∞ (−1)n−1 n (−1)n−1 n 1 (−1)n−1 n 1 x = x − x (n − 2)n 2 n=3 n − 2 2 n=3 n n=3 ∞ ∞ = (−1)n−1 n 1 (−1)n−1 n 1 1 2 x x − x + 2 n=1 n 2 n=1 n 2 = 1 1 1 x − x2 + (x2 − 1) ln(1 + x). 1 Since ∞ ∞ n=3 (−1)n−1 n 1 = lim x→1− (n − 2)n n=3 1 1 1 x − x2 + (x2 − 1) ln(1 + x) 2 4 2 = 1 , 4 and ∞ 1 1 (−1)n−1 1 (−1)n = lim x− x2 + (x2 −1) ln(1+x) x→−1+ 2 (n−2)n 4 2 n=3 3 =− , 4 because we get by the laws of magnitudes (x2 − 1) ln(1 + x) = (x − 1){(1 + x) ln(1 + x)} → 0 for 1 + x → 0+.

An (x) n+1 n+1 |x|n n+1)2 We conclude from the criterion of quotients that we have convergence for |x| < 1 and divergence for |x| > 1, hence the radius of convergence is 1. com 30 Calculus 3c-3 Power series; radius of convergence and sum b) The criterion of roots. It follows from 1≤ n n+1 = n n 1+ √ 1 n ≤ 2 → 1 for n → ∞, n 1+ 1 · |x| → |x| for n → ∞. n that n n+1 n |x| = n n We conclude from the criterion of roots that we have convergence for |x| < 1 and divergence for |x| > 1, and the radius of convergence must be 1.

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