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?
 A: 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.
A: Note that
$$
\frac{(k+1)k^k}{(k+1)^{k+1}}=\frac{k^k}{(k+1)^k}=
\left(\left(1+\frac{1}{k}\right)^{\!k}\right)^{-1}
$$
A: 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 $z\,e^{1-z}$.
A: Stirling's approximation gives
$$k!\approx k^k\mathrm e^{-k}\sqrt{2\pi k}\;\left(1+O\Big(\frac1k\Big)\right)$$
therefore the generic term is equivalent to
$$\frac{k!}{k^k}\sim\mathrm e^{-k}\sqrt{2\pi k}$$
and the series converges.
EDIT I have added a precision concerning Stirling approximation.
Usually Stirling approximation is given in logarithmic form
$$\ln k!\approx k\ln k-k+\frac12\ln k+\frac12\ln(2\pi)+O(k^{-1})$$
A: For $n\ge 3$ we have
$$\frac{n!}{n^n}\le\frac{1\cdot 2}{n^2}$$
Since the series $\sum n^{-2}$ converges, your series converges, too.
A: Notice:
$$\left|\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}\right|=\left|\left(\frac{n}{n+1}\right)^n\right|=$$
$$\left|\left(\frac{1}{1+\frac{1}{n}}\right)^n\right|=\left|\frac{1}{\left(1+\frac{1}{n}\right)^n}\right|=$$
$$\frac{1}{\left|\left(1+\frac{1}{n}\right)^n\right|}=\frac{1}{\left|1+\frac{1}{n}\right|^n}$$
