Studying convergence of $\sum\limits_{k=1}^{\infty}\frac{k!\exp(k)}{(k+1)^k}$ I have the following series:
$$\sum_{k=1}^{\infty}\frac{k!\exp(k)}{(k+1)^k}.$$
I want to prove that it diverges.
I  have already tried Alembert's quotient test and square root test and they are inconclusive. I have not found any suitable series for a comparison test or $k^a$ test.
I think $$\lim_{k\to\infty} \frac{k!\exp(k)}{(k+1)^k}= +\infty,$$ but I cannot prove it.
Any help or suggestions?
 A: $\def\e{\mathrm{e}}$By Stirling's formula,$$
k! \sim \sqrt{2πk} \left(\frac{k}{\e}\right)^k, \quad (k \to \infty)
$$
thus$$
\frac{k!\,\e^k}{(k + 1)^k} \sim \sqrt{2πk} \left(\frac{k}{\e}\right)^k \cdot \frac{\e^k}{(k + 1)^k} = \sqrt{2πk} \left(1 - \frac{1}{k + 1}\right)^k \sim \frac{\sqrt{2πk}}{\e}, \quad (k \to \infty)
$$
which implies that $\sum\limits_{k = 1}^\infty \dfrac{k!\,\e^k}{(k + 1)^k}$ diverges.
A: Without Stirling:
$\frac{a_k}{a_{k-1}}=\frac{e}{\left(1+\frac1k\right)^k}>1$ and so $a_k$ doesn't go to $0.$  
A: Just added for you curiosity.
As Alex Francisco answered, using Stirling approximation shows that the summation diverges.
We could even approximate the partial sum
$$S_p=\sum_{k=1}^{p}\frac{k!\, e^k}{(k+1)^k}=\sum_{k=1}^{p}a_k$$ Using again Stirling approximation,
$$\log(a_k)=\log \left(\frac{\sqrt{2 \pi }}{e}\right)+\frac{1}{2} \log
   \left({k}\right)+\frac{7}{12
   k}+O\left(\frac{1}{k^2}\right)$$ Continuing with Taylor
$$a_k=e^{\log(a_k)}=\frac{\sqrt{2 \pi } }{e}\sqrt{k}+\frac{7 \sqrt{\frac{\pi }{2}}
   }{6 e}\sqrt{\frac{1}{k}}+O\left(\frac{1}{k}\right)$$ making
$$S_p=\frac{\sqrt{\frac{\pi }{2}}}{6 e}\left(12 H_p^{\left(-\frac{1}{2}\right)}-7 \zeta \left(\frac{1}{2},p+1\right)+7 \zeta
   \left(\frac{1}{2}\right)\right)$$ where appear generalized harmonic numbers and the Hurwitz zeta function.
Using asymptotics,
$$S_p=\frac{2 \sqrt{2 \pi }}{3 e} p^{3/2}+\frac{5 \sqrt{2 \pi } }{3
   e}p^{1/2}+\frac{\sqrt{\frac{\pi }{2}} \left(12 \zeta \left(-\frac{1}{2}\right)+7 \zeta
   \left(\frac{1}{2}\right)\right)}{6 e}+O\left(\frac{1}{p^{1/2}}\right)$$ in which the constant term is almost $-1$ ($\approx -0.977244$).
So, let us use 
$$S_p \approx \frac{2 \sqrt{2 \pi }}{3 e} p^{3/2}+\frac{5 \sqrt{2 \pi } }{3
   e}p^{1/2}-1$$ and below are given some results
$$\left(
\begin{array}{ccc}
 p & \text{approximation} & \text{exact} \\
 10 & 23.3004 & 23.1897 \\
 100 & 629.127 & 628.889 \\
 1000 & 19488.0 & 19487.7 \\
 10000 & 614911. & 614910.
\end{array}
\right)$$
