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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, the argument is perfectly fine.2012-05-22
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    Please put the question in the body, not just the title.2012-05-22
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    Are we just pre-checking your series homework for you this evening before you turn it in? This is the fourth question in an hour in which you put down a series, give an argument for deciding whether it converges or does not converge, and then ask if you are correct. (And only properly formatted one of them, and not the latest one...)2012-05-22
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    You can also show this series is convergent by using Cauchy's root test.2012-05-22
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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.