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solutions2

# solutions2 - ³ ≤ n k 2 n = c n √ 2-n We know that c n...

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CM221A ANALYSIS I Solutions for Sheet 2 1. No. The formula lim n →∞ ( na n ) = n does not make sense. 2. We have - 2 - n (sin n ) 2 - n . Since 2 - n 0, the Sandwich Theorem implies that the sequence converges to 0. 3. lim n →∞ ± n + 1 - n ² = lim n →∞ 1 n + 1 + n = 0 because ± n + 1 - n ²± n + 1 + n ² = 1. 4. lim n →∞ c 2 n - 1 c 2 n + 1 ! = lim n →∞ c 2 n - 1 lim n →∞ c 2 n + 1 = - 1 if | c | < 1; lim n →∞ c 2 n - 1 c 2 n + 1 ! = 1 - lim n →∞ c - 2 n 1 + lim n →∞ c - 2 n = 1 if | c | > 1; lim n →∞ c 2 n - 1 c 2 n + 1 ! = 0 if | c | = 1 5(i). By the ratio test, the series n =1 ± 2 3 ² n converges. Since 0 < 2 n - n 3 n + n 2 ± 2 3 ² n , the comparison theorem implies that the series n =1 2 n - n 3 n + n 2 is also convergent. 5(ii). Since n n n ! does not converge to zero as n → ∞ , the series diverges. 5(iii). ³ ³ ³ ( - 1) n n k 1+2 n ³ ³
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Unformatted text preview: ³ ≤ n k 2 n = c n ( √ 2)-n . We know that c n → 0 as n → ∞ (see the previous exercise sheet). Therefore, by ratio test, the series ∑ ∞ n =1 c n ( √ 2)-n converges. Now the comparison theorem implies that the series ∑ ∞ n =1 (-1) n n k 1+2 n is also convergent. 5(iv). We have lim n-1 n +1 = 1. The ratio test implies that the series converges for all x such that | x + 1 | < 2. If | x + 1 | ≥ 2 then the sequence ( n-1) ( x +1) n ( n +1) 2 n does not converge to 0 and, consequently, the series diverges. 1...
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