# Study the convergence $\sum_{n=1}^\infty n^2/\exp(n)$

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............ – user17762 May 22 '12 at 6:06
Yes, the argument is perfectly fine. – André Nicolas May 22 '12 at 6:06
Please put the question in the body, not just the title. – Arturo Magidin May 22 '12 at 6:07
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 6:13
You can also show this series is convergent by using Cauchy's root test. – Kns May 22 '12 at 7:50

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.