Limit of $ \frac{e^k k!}{k^k} $ According to WolframAlpha we should have
$$
\lim_{k \to \infty} \frac{a^k k!}{k^k} = \infty
$$
whenever $a \geq e$. I know how to show that the limit is 0 when $a < e$ and that the limit is $\infty$ when $a > e$, but I struggle with the case $a = e$. Setting
$$
f(k) = \frac{e^k k!}{k^k}
$$
we have that
$$
\frac{f(k+1)}{f(k)} = e\left(1 - \frac{1}{k + 1}\right)^k
$$
converges to 1, but this is not sufficient.
 A: Hint
You can use Stirling formula:
$$n!\sim \sqrt{2\pi n} \frac {n^n}{e^n}.$$
A: Starting from your
$$ \dfrac{f(k+1)}{f(k)} = e \left( 1 - \dfrac{1}{k+1}\right)^k $$
take logarithms:
$$ \log f(k+1) - \log f(k) = 1 + k \log \left(1 - \dfrac{1}{k+1}\right) $$
Now if $1/2 < a < 1$,  there is $\epsilon > 0$ such that $\log(1-t) \ge -t - a t^2$ for $0 < t< \epsilon$, and thus for sufficiently large $k$,
$$\eqalign{ \log f(k+1) - \log f(k) &\ge 1 + k \left(- \dfrac{1}{k+1} - \dfrac{a}{(k+1)^2} \right)\cr &= \dfrac{(1-a)k+1}{(k+1)^2}\cr }$$
Now use the fact that $\displaystyle \sum_{k=1}^\infty \dfrac{k}{(k+1)^2} = \infty$.
A: You don't need Stirling's formula, you can use easier estimates, e.g. this one:
We have:
$$
\log n! = \sum_{x=1}^n \log x
$$
so, looking at the graph of $\log x$ we can easy bound it below and above by integrals:
$$
\int_1^n \log x \, dx \leq \sum_{x=1}^n \log x \leq \int_0^n \log (x+1) \, dx
$$
which gives us the estimate
$$
n\log\left(\frac{n}{e}\right)+1 \leq \log n! \leq (n+1)\log\left( \frac{n+1}{e} \right) + 1
$$
so
$$
e\left(\frac ne\right)^n \leq n! \leq e\left(\frac{n+1}e\right)^{n+1}
$$
This is clearly weaker estimate than Stirling's formula, but will work well enough for your problem.
