# Convergence of $\sum_{n=1}^{\infty}\frac{\sin^2(n)+n^n}{1+n^2e^nn!}$

Show whether the following series converges or not: $$\sum_{n=1}^{\infty}\frac{\sin^2(n)+n^n}{1+n^2e^nn!}$$

I know that $$n^n\leq n!e^n\;\forall n\in\mathbb{N}$$ but using this for a direct comparison test hasn't helped very much so far.

Hint: $$\sum \frac {\sin^{2}(c)} {1+n^{2}e^{n} n!}$$ is dominated by $$\sum \frac 1 {n^{2}}$$ so it is convergent. Show that $$\sum \frac {n^{n}} {n^{2}e^{n} n!}$$ is also convergent using the inequality you already know.
The given series is , in fact, dominated by $$\sum \frac 2 {n^{2}}$$.
$$\frac{\sin^2(n)+n^n}{1+n^2\mathrm ee^n\,n!}\sim_\infty\frac{n^n}{n^2\mathrm e^n\,n!}\sim_\infty\frac{n^n}{n^2\mathrm e^n\sqrt{2\pi n}\Bigl(\cfrac{n}{\mathrm e}\Bigr)^{\!n}}=\frac 1{\sqrt{2\pi}\,n^{5/2}},$$ which converges.