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Unformatted text preview: SOLUTIONS TO HOMEWORK # 10 15.1 (a) ∑ (- 1) n n satisfies the conditions of the alternating series test and thus converges. (b) As we proved in the class, using the Ratio Test, the series ∑ 2 n n ! converges and hence lim 2 n n ! = 0. Therefore (- 1) n n ! 2 n does not converge to 0, and ∑ (- 1) n n ! 2 n diverges by Corollary 14.5. A 1) The partial sums s n = 1 if n is even and s n = 0 if n is odd. Thus the series diverges. Another approach: (- 1) n does not converge to 0 so the series cannot converge. 2) This is an alternating series. But the main condition of the Alternating Series Theorem (AST), namely a n ↓ 0, is not fulfilled here so there is no contradiction to this Theorem. B 1) The series ∑ ∞ n =2 1 n (log n ) p converges for every real p . Proof a) lim 1 n (log n ) p = 0. Case 1. If p ≥ 0 this follows from the fact that logn → + ∞ and hence lim n (log n ) p = + ∞ . Therefore our statement holds. Case 2. If p < 0 we have 1 n (log n ) p = (log n ) | p | n ; consider the functions f ( x ) := log x x 1 | p | ,g ( x ) := (log x ) | p | x . By the L’Hospital Rule, since | p | > 0, lim x →∞ f ( x ) = lim x →∞ 1 x 1 | p | x 1 | p |- 1 = | p | lim x →∞ 1 x 1 | p | = 0 ....
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- Spring '08
- #, 2 m, 0 K, series A, 2 -M