Does $\sum_{k=1}^{\infty}\frac{k!}{k^k}$ converge? I have tried using ratio test:
$$P =\lim_{k\rightarrow\infty}\left|\frac{(k+1)!}{(k+1)^{k+1}}\cdot\frac{k^k}{k!}\right|$$
$$ P=\lim_{k\rightarrow\infty}\left|\frac{(k+1)\cdot k^k}{(k+1)^{k+1}}\right|$$
$$ P=\lim_{k\rightarrow\infty}\left|\frac{k^{k+1}+ k^k}{(k+1)^{k+1}}\right|$$
In the final expression, the highest degrees in the denominator and numerator are both $(k+1)$. So according to the L'Hospital's Rule, the limit would goes to $1$.
$$P=1$$
Thus the test failed.
Any suggestions on how to test the convergence of this series?

We have:
$$ \sum_{k=1}^{N}\frac{k!}{k^k} = \int_{0}^{+\infty}e^{-x}\sum_{k=1}^{N}\frac{x^k}{k^k}\,dx = \int_{0}^{+\infty}\sum_{k=1}^{N}k e^{-kz}z^k\,dz\tag{1}$$
hence the series is convergent since the function $$f(z)=\frac{z e^z}{(e^z-z)^2}=\sum_{k\geq 1}ke^{-kz}z^k\tag{2}$$ is integrable over $\mathbb{R}^+$, as a positive function bounded by $z\,e^{1-z}$.
