Calculate $\lim _{n\to \infty }\left(\frac{e^nn!}{n^n}\right)$

Question

$$\lim _{n\to \infty }\left(\frac{e^nn!}{n^n}\right)$$

According to wolfram, the limit is $\infty$.

Both ratio and Cauchy tests yield $1$, so don't don't help much. I tried to bound the sequence from below, as follows

$$\frac{e^n}{n^n}n!\ge \frac{e^n}{n^n}\left(\frac{n}{2}\right)^{\frac{n}{2}}$$

but then I get

$$\lim _{n\to \infty }\left(\frac{e^n}{n^n}\left(\frac{n}{2}\right)^{\frac{n}{2}}\right)=0$$

Answer 1

By Stirling's approximation, $n!\geqslant\sqrt{2\pi}\,n^{n+\frac12}e^{-n}$ and therefore$$\frac{e^nn!}{n^n}\geqslant\sqrt{2\pi n}.$$

Answer 2

1. The sequence $\left\{\left(1+\frac{1}{n}\right)^{n+1/2}\right\}_{n\geq 1}$ is decreasing towards $e$ by the concavity of the logarithm function. We may prove this by considering $$ \left(n+\frac{1}{2}\right)\log\left(1+\frac{1}{n}\right)=\left(n+\frac{1}{2}\right)\int_{0}^{1}\frac{dx}{n+x}=\int_{0}^{1}\frac{dx}{1-\left(\frac{x}{2n+1}\right)^2} $$ where the RHS is clearly decreasing with respect to $n$.
2. It follows that $$ e^N>\prod_{n=1}^{N}\left(1+\frac{1}{n}\right)^{n+1/2} = \sqrt{N+1}\prod_{n=1}^{N}\left(1+\frac{1}{n}\right)^{n}=\sqrt{N+1}\frac{(N+1)^N}{N!} $$ and $N! e^N > \sqrt{N+1}(N+1)^N$. The full power of Stirling's approximation is not needed.