solutions2

solutions2 - n k 2 n = c n ( 2)-n . We know that c n 0 as n...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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...
View Full Document

This document was uploaded on 01/31/2011.

Ask a homework question - tutors are online