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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.

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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 $\sum \frac 2 {n^{2}}$.

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With Stirling's formula, you have an equivalent of the (positive) general term of the series:

$$\frac{\sin^2(n)+n^n}{1+n^2\mathrm ee^n\,n!}\sim_\infty\frac{n^n}{n^2\mathrm e^n\,n!}\sim_\infty\frac{n^n}{n^2\mathrm e^n\sqrt{2\pi n}\Bigl(\cfrac{n}{\mathrm e}\Bigr)^{\!n}}=\frac 1{\sqrt{2\pi}\,n^{5/2}}, $$ which converges.

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