Does $\sum_{n=1}^\infty(-1)^{n}\frac{n^n}{n!}$ converge or diverge $\sum_{n=1}^\infty(-1)^{n}\frac{n^n}{n!} $
I got that it diverges but I am not sure
 A: 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.
A: 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.
A: 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. 
A: 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 $a_n\to\infty$.
A: 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.
A: 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 $(-1)^n \frac{n^n}{n!}$?
