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Study the convergence of $\displaystyle \sum_{n=1}^\infty \frac{n^{2}}{e^{n}}$.

Use the ratio test.

$r = \lim_{n\to\infty}\frac{\quad\frac{(n+1)^2}{e^{n+1}}\quad}{\frac{n^2}{e^n}} = \frac{1}{e}\lim_{n\to\infty}\frac{(n+1)^2}{n^2} = \frac{1}{e}.$

Since $r = 1/e < 1$, this series converges.

Is this right?

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    Yes, it is perfect but I will suggest you to do the same problem with Comparison Test using Integration, then you will learn something new.2012-05-22

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Alternative way: $e^n\geq \frac{n^4}{4!}$ hence $e^{-n}\leq \frac{24}{n^4}$ and $0\leq \frac{n^2}{e^n}\leq \frac{24}{n^2} $, and we get the convergence since $\sum_{n\geq 1}\frac 1{n^2}$ is convergent.

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Yes. As you have computed using the ratio test, the series converges.