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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?

Daniela del Carmen
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    Yes, the argument is perfectly fine. – André Nicolas May 22 '12 at 06:06
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    Please put the question in the body, not just the title. – Arturo Magidin May 22 '12 at 06:07
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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...) – Arturo Magidin May 22 '12 at 06:13
  • You can also show this series is convergent by using Cauchy's root test. – Kns May 22 '12 at 07:50
  • 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. – users31526 May 22 '12 at 08:05

2 Answers2

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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.

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

Willie Wong
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