Chapter 10.4, Problem 35E

To determine

To find: the convergence or divergence of the series.

Answer to Problem 35E

Convergence according to the ratio test

Explanation of Solution

Given:

  ${\displaystyle \sum _{n=0}^{\infty }{n}^2{e}^{-n}}$

Calculation:

  $\begin{array}{l}{\displaystyle \sum _{n=0}^{\infty }{n}^2{e}^{-n}}={\displaystyle \sum _{n=0}^{\infty }\frac{n^2}{e^n}}\\ \underset{n\to \infty }{{\displaystyle \lim }}\frac{\frac{{\left(n+1\right)}^2}{e^{n+n}}}{\frac{n^2}{e^n}}\\ \underset{n\to \infty }{{\displaystyle \lim }}\frac{{\left(n+1\right)}^2}{e^{n+n}}\times \frac{e^n}{n^2}\\ \underset{n\to \infty }{{\displaystyle \lim }}\frac{1}{e}\\ r=\left|\frac{1}{e}\right|<1\end{array}$