I'm trying to show the following bound, $$\sum_{n=k}^{\infty} \frac{e^{-n\alpha}{(n\alpha)}^{n-1}}{n!} \leq \frac{1}{\alpha}\sum_{n=k}^{\infty} \frac{e^{-kI}}{\sqrt{2\pi n^3}} e^{-{(n-k)I}}$$
where $I=\alpha-1-\log(\alpha)$. I understand that I'm supposed to use Stirling's Formula but I'm lost as to how to do this and am not getting anywhere close. Any suggestion would be great.
Hint: For starting One may give simple bounds valid for all positive integers $n$, rather than only for large $n$ using this inequality:
$\sqrt{2\pi n}n^{n}e^{-n}\leq n!\leq e n^{n+1/2}e^{-n}$, you can use the LHS of that inequality which is :$\sqrt{2\pi n}n^{n}e^{-n}\leq n!$ implies :
$$\frac{1}{n!}\leq \frac{1}{\sqrt{2\pi n}n^{n}e^{-n}}\leq \frac{1}{\sqrt{2\pi n^3}e^{-n}}$$ , Mutltiplying both sides by your needed factors and do summation you will come up to your bounds for $n^n$ in your dominator wich it would be still valid for $n^3$ in dominator