TITLE: Evaluation of limit without using Stirling's approximation
QUESTION [0 upvotes]: I'm trying to find a way to show that $\lim_{n\to+\infty} \frac{n!e^{n}}{n^{n}}=+\infty$ without using Stirling's approximation. I tried a few things that didn't work, any help would be appreciated. (I came across this limit the following way: we all know that $\lim_{n\to+\infty} \frac{n!}{n^{n}}=0$ the next obvious thing to try is $\lim_{n\to+\infty} \frac{n!a^{n}}{n^{n}}$. By the ratio criterium for sequences $\frac{n!a^{n}}{n^{n}}$ goes to $+\infty$ if $a>e$ and to $0$ if $a<e$ so it's natural to wonder what happens with $a=e$. Using Stirling it's obvious but I wanted to be able to present this in a more elementary fashion if needed)

REPLY [1 votes]: Since
$$
\begin{align}
(k-1)\log\left(1+\frac1{k-1}\right)
&=(k-1)\int_0^{\frac1{k-1}}\frac{\mathrm{d}x}{1+x}\\
&\le(k-1)\int_0^{\frac1{k-1}}\left(1-x+x^2\right)\,\mathrm{d}x\\
&=1-\frac1{2(k-1)}+\frac1{3(k-1)^2}\tag1
\end{align}
$$
we have
$$
1+(k-1)\log\left(1-\frac1k\right)
\ge\frac1{2(k-1)}-\frac1{3(k-1)^2}\tag2
$$
Therefore, since
$$
\frac{\frac{n!\,e^n}{n^n}}{\frac{(n-1)!\,e^{n-1}}{(n-1)^{n-1}}}
=e\left(\frac{n-1}n\right)^{n-1}\tag3
$$
we have
$$
\begin{align}
\frac{n!\,e^n}{n^n}
&=e\prod_{k=2}^ne\left(1-\frac1k\right)^{k-1}\tag4\\
&=\exp\left[1+\sum_{k=2}^n\left(1+(k-1)\log\left(1-\frac1k\right)\right)\right]\tag5\\[3pt]
&\ge\exp\left[1+\frac12H_{n-1}-\frac13\frac{\pi^2}6\right]\tag6\\[7pt]
&\ge e^{1+\frac\gamma2-\frac{\pi^2}{18}}\sqrt{n-1}\tag7\\[18pt]
&\ge2\sqrt{n-1}\tag8
\end{align}
$$
Explanation:
$(4)$: induction on $(3)$
$(5)$: write the product as the exponential of a sum
$(6)$: apply $(2)$
$(7)$: $H_n\ge\log(n)+\gamma$ where $\gamma$ is the Euler-Mascheroni Constant
$(8)$: $e^{1+\frac\gamma2-\frac{\pi^2}{18}}\gt2$
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