Chapter 8.2, Problem 29E

To determine

To find: The series is convergent or divergent and find its sum if it is convergent.

Answer to Problem 29E

The series is convergent and the sum of the series is $\frac{e}{e-1}$ .

Explanation of Solution

Given information: The series is ${\displaystyle \sum _n^{\infty }(\frac{1}{e^n}}+\frac{1}{n(n+1)}$

given, ${\displaystyle \sum _n^{\infty }\frac{1}{e^n}}+\frac{1}{n(n+1)}$

  $\begin{array}{l}={\displaystyle \sum _n^{\infty }\frac{1}{e^n}}+{\displaystyle \sum _n^{\infty }\frac{1}{n(n+1)}}\\ ={\displaystyle \sum _n^{\infty }{u}_n+}{\displaystyle \sum _n^{\infty }{v}_n}\end{array}$

Where $u_n=\frac{1}{e^n},v_n=\frac{1}{n(n+1)}$

  $v_n=\frac{1}{n(n+1)}\le \frac{1}{n^2}$

Here ${\displaystyle \sum u_n}$ is geometric series with common ratio $\frac{1}{e}(<1)$ and its convergent.

Again ${\displaystyle \sum _n^{\infty }\frac{1}{n^2}}$ is ap series and so it is convergent and so by comparison test ${\displaystyle \sum v_n}$

Now we find the sum of the series , we have

  $\begin{array}{l}{\displaystyle \sum _n^{\infty }\frac{1}{e^n}}+{\displaystyle \sum _n^{\infty }\frac{1}{n(n+1)}}={\displaystyle \sum _n^{\infty }\frac{1}{e^n}}+{\displaystyle \sum _n^{\infty }\frac{1}{n(n+1)}}\\ =\frac{1}{e-1}+1\\ \text{ =}\frac{1+e-1}{e-1}\\ \text{ =}\frac{e}{e-1}\end{array}$

Hence, the sum of the series is $\frac{e}{e-1}$ .