Solving $\lim\limits_{n\to\infty}\frac{n^n}{e^nn!}$ I was solving a convergence of a series and this limit popped up:
$$\lim\limits_{n\to\infty}\frac{n^n}{e^nn!}$$
I needed this limit to be $0$ and it is in fact (according to WolframAlpha), but I just don't see how to get the result.

The ratio of two consecutive values is
$$
\left.\frac{(n+1)^{n+1}}{e^{n+1}(n+1)!}\middle/\frac{n^n}{e^nn!}\right.=\frac{\left(1+\frac1n\right)^n}{e}\tag{1}
$$
Taking the log of $(1)$ gives
$$
n\left(\frac1n-\frac1{2n^2}+O\left(\frac1{n^3}\right)\right)-1
=-\frac1{2n}+O\left(\frac1{n^2}\right)\tag{2}
$$
Since
$$
\int_1^{n+1}\frac{\mathrm{d}x}{x}\le\sum_{k=1}^n\frac1k\le1+\int_1^n\frac{\mathrm{d}x}{x}\tag{3}
$$
we have
$$
\sum_{k=1}^n\frac1k=\log(n)+O(1)\tag{4}
$$
Therefore, summing $(2)$ using $(4)$ yields
$$
\log\left(\frac{n^n}{e^nn!}\right)=-\frac12\log(n)+O(1)\tag{5}
$$
Thus, we have
$$
\frac{n^n}{e^nn!}\le\frac{c}{\sqrt{n}}\tag{6}
$$
The limit sought is therefore, $0$.
