I would like to know when $\frac{2 n f(n)}{n !}$ is $O (n^b)$ where $b$ is a constant. Here, $n$ is a positive integer.
My attempt:
$$ \frac{2 n f(n)}{n !} = \frac{2 n f(n)}{\sqrt{2 \pi n} (\frac{n}{e})^n}\\ = \sqrt{2 n} \pi^{\frac{-1}{2}} e^n n^{-n}f(n)\\ = \sqrt{2} \pi^{\frac{-1}{2}} e^n n^{\frac{1}{2} - n} f(n) $$
So, it becomes $O (n^b)$ when $f(n) = e^{-n} n^{n-\frac{1}{2}} n^b = O(n^n)$
Am I getting it right?