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How the following series converges absolutely or conditionally or diverges:

$$\sum_{n \geq 1} (-1)^{n+1} \frac{n^n}{2^n n!} $$

Since it is alternating it is converging right? But I am not sure how to prove absolutely or conditionally.

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Using the ratio test: let $$u_n=(-1)^{n+1} \frac{n^n}{2^n n!}$$ we have $$\frac{|u_{n+1}|}{|u_n|}=\frac{(n+1)^{n+1}}{2^{n+1}(n+1)!}\times \frac{2^n n!}{n^n}=\frac{1}{2}\left(1+\frac 1 n\right)^n\sim_\infty\frac{e}{2}>1$$ so the series isn't convergent.

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  • $\begingroup$ I don't quite understand what you did in those last couple of steps. $\endgroup$ – Pat Green Dec 11 '13 at 14:28
  • $\begingroup$ Simplify and use that $\left(1+\frac 1 n\right)^n=e^{n\log\left(1+\frac 1 n\right)}\sim e^{n\frac 1 n}=e$ $\endgroup$ – user63181 Dec 11 '13 at 15:13
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    $\begingroup$ I indeed was busy today. Nice to see you my bro. :-) $\endgroup$ – mrs Dec 11 '13 at 15:18
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You can use Stirling's formula $$n! \sim \left(\frac{n}{e}\right)^n\sqrt{2 \pi n}$$

Thus

$$\frac{n^n}{2^nn!} \sim \left(\frac{e}{2}\right)^n\frac{1}{\sqrt{2 \pi n}}$$

And since $e > 2$, the general term of the series does not converge to zero, thus the series is divergent.

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