# Thread: series: integral test

1. ## series: integral test

Use the Integral Test to decide the convergence or divergence of the following series:
$\displaystyle \sum_{k=1}^{\infty} \dfrac{k^{2}}{e^{k}}$

I'm pretty much lost with this.

2. $\displaystyle \displaystyle \int_1^{\infty}\frac{x^2}{e^x}dx=\lim_{n\to\infty} \int_1^nx^2e^{-x}dx=\frac{5}{e}$ so the serie is convergent.

3. The integral test is as follows:
Theorem: The Integral Test
Let $\displaystyle f(x)$ be a positive continuous function that is eventually decreasing and let $\displaystyle f(n)=a_n$.
Then $\displaystyle \sum_{n=1}^\infty a_n$ converges if and only if $\displaystyle \int_1^\infty f(x)dx$ converges and diverges otherwise.
So to check $\displaystyle \sum_{k=1}^{\infty} \dfrac{k^{2}}{e^{k}}$ for convergence using the integral test, you would find its integral which, like red_dog said, is as follows:

$\displaystyle \int_1^\infty\frac{x^{2}}{e^x}dx=\lim_{b\to\infty} \int_1^bx^2e^{-x}dx=\lim_{b\to\infty} \left( -x^2e^{-x} - 2xe^{-x}-2e^{-x}\right)\big|_1^b$
$\displaystyle =\lim_{b\to\infty} \left( -b^2e^{-b} - 2be^{-b}-2e^{-b}-\left(-1^2e^{-1} - 2e^{-1}-2e^{-1}\right)\right)$$\displaystyle =0-0-0-\left(-e^{-1}-2e^{-1}-2e^{-1}\right)=\frac{5}{e}$

by using integration by parts twice and then evaluating the limit.

Therefore, by the Integral Test, because the integral is convergent, so is the series.