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$\sum_{n=1}^\infty(-1)^{n}\frac{n^n}{n!} $

I got that it diverges but I am not sure

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    $\begingroup$ ${n^n\over n!}={n\over n}\cdot {n\over n-1}\cdot{n\over n-2}\cdots {n\over 2}\cdot {n\over1}>1$ for each $n$. $\endgroup$ – David Mitra May 16 '13 at 19:14
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There are already four answers on the page, but all of them refer to either Stirling's approximation, or to the famous limit for $e$. Here is a totally elementary argument. Compare two terms with indexes $n$ and $n+1$: $$ a_n=\frac{n^n}{n!}\quad ?\quad a_{n+1}=\frac{(n+1)^{n+1}}{(n+1)!} $$ Note that $$ a_{n+1}=\frac{(n+1)^{n+1}}{(n+1)!}=\frac{(n+1)^n(n+1)}{(n+1)!}>\frac{n^n(n+1)}{(n+1)!}=\frac{n^n}{n!}=a_n, $$ which proves that the terms of the series do not tend to zero. Hence the series diverge.

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    $\begingroup$ Everyone is so quick to appeal to Stirling. $\endgroup$ – David Mitra May 16 '13 at 19:15
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By Stirling formula $$n!\sim_\infty \left(\frac{n}{e}\right)^n\sqrt{2\pi n}$$ we have $$\left|(-1)^{n}\frac{n^n}{n!}\right|\sim\frac{e^n}{\sqrt{2\pi n}}\not\to0$$ hence your series is divergent.

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The ratio of the absolute value of consecutive summands is $$ \frac{(n+1)^{n+1}/(n+1)!}{n^n/n!} = \bigg(1+\frac1n\bigg)^n, $$ the limit of which is $e$ as $n\to\infty$. Since $e>1$, the Ratio Test tells us that the series diverges.

It is true that the summand does not tend to $0$ (which also implies divergence of the series); but establishing that the summand does not tend to $0$ is more complicated than this Ratio Test argument, I believe.

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If a series $\sum_{n = 0}^\infty a_n$ converges then $\displaystyle \lim_{n\to \infty}a_n = 0$ is necessary condition. Calculate $\lim_{n\to \infty}a_n$. Stirling approximation might be helpful to calculate the limit.

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  • $\begingroup$ +1 for offering a hint and not just throwing out the answer for problems like this. $\endgroup$ – 6005 May 16 '13 at 19:03
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By Stirling $$\frac{{{n^n}}}{{n!}} \approx \frac{{{n^n}}}{{{n^n}{e^{ - n}}\sqrt {2\pi n} }} = \frac{{{e^n}}}{{\sqrt {2\pi n} }}$$ so the series cannot converge.

Alternatively, $${a_n} = \frac{{{n^n}}}{{n!}} \to \frac{{{a_{n + 1}}}}{{{a_n}}} = {\left( {1 + \frac{1}{n}} \right)^n} \to e > 1$$ so $a_n\to\infty$.

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Simply expand the numerator and the denominator as products. Can you find a lower bound for $\frac{n^n}{n!}$? Does this lower bound tend to $0$ as $n\rightarrow\infty$? What does this say about the terms of $(-1)^n \frac{n^n}{n!}$?

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