Proving the convergence of $\sum_{n=1}^{\infty}\frac{n^n}{e^n (n+1)!}$ I'm tasked with the simple problem of finding the interval of convergence of
$$\sum_{n=1}^{\infty}\frac{x^n n^n}{(n+1)!}.$$
Using the ratio test, I've found that it certainly converges on $|x|<\frac{1}{e}$; however, I haven't been able to test the endpoints of that interval. (It's too much work to put all the $\LaTeX$ here.) When $x=\frac{1}{e}$, the series becomes 
$$\sum_{n=1}^{\infty}\frac{n^n}{e^n (n+1)!},$$
which WolframAlpha claims converges by the comparison test. But I can't find another series to compare with even after careful thinking. Would someone be able to give me a hint in the right direction? Thanks!
 A: Stirling's approximation states that
$$
n!\sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n
$$
where $\sim$ means that the ratio of the two sides tends to $1$ as $n\to \infty$. Hence
$$
a_n=\frac{n^n}{e^n (n+1)!}\sim \frac{n^n}{e^n \sqrt{2\pi (n+1)}\left(\frac{n+1}{e}\right)^{n+1}} =\frac{e}{\sqrt{2\pi}(n+1)^{3/2}}\left(1-\frac{1}{n+1}\right)^n\leq C \frac{e}{\sqrt{2\pi}(n+1)^{3/2}}
$$
for some $C$ since $\left(1-\frac{1}{n+1}\right)^n$ is a convergent sequence. It follows that the original series converges.
A: In the same spirit as in Foobaz John's answer, consider
$$\sum_{n=1}^{p}\frac{n^n}{e^n (n+1)!}=\sum_{n=1}^{\infty}\frac{n^n}{e^n (n+1)!}+\sum_{n=p+1}^{\infty}\frac{n^n}{e^n (n+1)!}$$
Now, for large values of $p$
$$a_n=\frac{n^n}{e^n (n+1)!}\implies \log(a_n)=n \log(n)-n-\log((n+1)!)$$
Using Stirling approximation and Taylor series
$$\log(a_n)=-\frac{3}{2} \log
   \left({n}\right)-\frac{1}{2} \log (2 \pi )-\frac{13}{12
   n}+O\left(\frac{1}{n^2}\right) < -\frac{3}{2} \log
   \left({n}\right)-\frac{1}{2} \log (2 \pi )$$
$$a_n< \frac{1}{\sqrt{2 \pi }\, n^{3/2}}\implies \sum_{n=p+1}^{\infty}\frac{n^n}{e^n (n+1)!} <\sqrt{\frac{2}{\pi p }}  $$ and the second sum converges.
A: The first three terms of the series for $\log(1+x)$ is an overestimate, when $x\le1$, so
$$
\begin{align}
(n+1)\log\left(1+\frac1n\right)
&\le(n+1)\left(\frac1n-\frac1{2n^2}+\frac1{3n^3}\right)\\
&=1+\frac1{2n}-\frac1{6n^2}+\frac1{3n^3}\tag1\\
\frac12\log\left(1-\frac1n\right)
&\le-\frac1{2n}-\frac1{4n^2}-\frac1{6n^3}\tag2\\
(n+1)\log\left(1+\frac1n\right)+\frac12\log\left(1-\frac1n\right)
&\le1-\frac5{12n^2}+\frac1{6n^3}\\[6pt]
&\lt1\tag3
\end{align}
$$
Therefore,
$$
\sqrt{\frac{n-1}n}\left(1+\frac1n\right)^{n+1}\lt e\tag4
$$
Using $(4)$ yields
$$
\begin{align}
\frac{\frac{(n+1)^{n+1}}{e^{n+1}(n+2)!}}{\frac{n^n}{e^n(n+1)!}}
&=\frac{n}{e(n+2)}\left(1+\frac1n\right)^{n+1}\\
&\le\frac{n}{n+2}\sqrt{\frac{n}{n-1}}\tag5
\end{align}
$$
Applying telescoping products gives
$$
\begin{align}
\frac{n^n}{e^n(n+1)!}
&\le\frac4{e^2}\frac{\sqrt{n-1}}{n(n+1)}\\
&\lt\frac4{e^2n^{3/2}}\tag6
\end{align}
$$
so the series converges by comparison to the $p$-series for $p=\frac32$.
A: We have
$$
\frac{{n^n }}{{e^n (n + 1)!}}\frac{{e^{n + 1} (n + 2)!}}{{(n + 1)^{n + 1} }} = \frac{e}{{\left( {1 + \frac{1}{n}} \right)^n }}\frac{{n + 2}}{{n + 1}}.
$$
By Maclaurin series expansion
\begin{align*}
\left( {1 + \frac{1}{n}} \right)^{ - n}  = \exp \left( { - n\log \left( {1 + \frac{1}{n}} \right)} \right) &= \exp \left( { - 1 + \frac{1}{{2n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)} \right) \\ & = \frac{1}{e}\left( {1 + \frac{1}{{2n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)} \right)
\end{align*}
and
$$
\frac{{n + 2}}{{n + 1}} = \frac{{1 + 2/n}}{{1 + 1/n}} = 1 + \frac{1}{n} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right).
$$
Consequently
$$
\frac{{n^n }}{{e^n (n + 1)!}}\frac{{e^{n + 1} (n + 2)!}}{{(n + 1)^{n + 1} }} = 1 + \frac{3}{{2n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right).
$$
Hence,
$$
\lim_{n\to+\infty} n\!\left( {\frac{{n^n }}{{e^n (n + 1)!}}\frac{{e^{n + 1} (n + 2)!}}{{(n + 1)^{n + 1} }} - 1} \right) = \frac{3}{2} > 1,
$$
and so the series converges by Raabe's test.
