Consider the sequence $r_n = n!e^n/n^{n+1/2}$ for positive integers $n$. It's straightforward to show that $r_n$ converges to some finite limit (applying the monotone convergence theorem to this decreasing sequence). Call the limit value $C$ and show that for all $n$:
$$r_n \leq e^{1/(4n)}C$$
I'm working on a proof by induction, but I'm having trouble with the induction step.
Note, this is used in a proof of Stirling's approximation.
As mentioned in the comments above, consider $\log C/r_n$ and note that $C = \sum^\infty_{i=1} \log r_{i+1}/r_i$. So we have: $$\log C/r_n = \sum^\infty_{i=n} \log r_{i+1}/r_i$$
Now we will bound this sum term-wise: Use Jensen's inequality to show $\frac{2x}{2+x} < \log(1+x) < \frac{2x+x^2}{2+2x}$ for $x>0$
$$\frac{2x}{2+x} < \log(1+x) < \frac{2x+x^2}{2+2x}$$
And substituting $x = 1/n$ we get: $$(n+1/2)\log(1+1/n) - 1 < \frac{1}{4n(n+1)}$$ We also have: $\log r_{n+1}/r_n=1-(n+1/2)\log(1+1/n)$ So now: $$\log C/r_n = \sum^\infty_{i=n} \log r_{i+1}/r_i > -\sum^\infty_{i=n} \frac{1}{4i(i+1)}=-\frac{1}{4n}$$
Exponentiating each side and rearranging gives the desired result.