$\displaystyle (-1)^n n^n/ (n+1)to the power (n+1).$ I dont know how to show if it converges or diverges? Any ideas?
Last edited by hebby; Feb 13th 2010 at 07:20 PM.
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did you try the Leibniz test?
I need to use the AbeL test...i think, but im not sure how to use it
Originally Posted by hebby $\displaystyle (-1)^n n^n/ (n+1)to the power (n+1).$ I dont know how to show if it converges or diverges? Any ideas? $\displaystyle \frac{n^n(-1)^n}{(n+1)^{n+1}}=\frac{(-1)^n}{n+1}\cdot\left(1+\frac{1}{n}\right)^n\to 0\cdot e$?
so the series converges to 0?
Originally Posted by hebby so the series converges to 0? Uh...yes.
Originally Posted by Drexel28 Uh...yes. will let be conditional convergence as the abs. values will make 1/n+1 divergent?
Originally Posted by hebby will let be conditional convergence as the abs. values will make 1/n+1 divergent? Are we evaluating this? $\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{n+1}\cdot\left(1+\frac{1}{n}\right)^n$
yes
Originally Posted by hebby yes Then it converges conditionally, as you stated.
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