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?

You should proceed as follows:
$$P =\lim_{k\rightarrow\infty}\left|\frac{(k+1)!}{(k+1)^{k+1}}\cdot\frac{k^k}{k!}\right|$$ or
$$ P=\lim_{k\rightarrow\infty}\left|\frac{(k+1)\cdot k^k}{(k+1)^{k+1}}\right|$$ or
$$ P=\lim_{k\rightarrow\infty}\left|\frac{k^k}{(k+1)^{k}}\right|$$ or
$$ P=\frac{1}{\lim_\limits{k\rightarrow\infty}\left(1+\frac{1}{k}\right)^k}$$ or
$$ P=\frac{1}{e}$$ by definition.
