Chapter 10.4, Problem 51E

To determine

To find: Sum of the given Telescoping series.

Answer to Problem 51E

Sum of Telescoping series is S = 1

Explanation of Solution

Given:

The series given is,

  $S={\displaystyle \sum _{n=1}^{\infty }\frac{2n+1}{n^2{\left(n+1\right)}^2}}$

Calculation:

As the series is,

  $S={\displaystyle \sum _{n=1}^{\infty }\frac{2n+1}{n^2{\left(n+1\right)}^2}}$

So, taking partial fraction for the $n^{th}$ term

  $\begin{array}{l}\frac{2n+1}{n^2{\left(n+1\right)}^2}=\frac{A}{n^2}+\frac{B}{{\left(n+1\right)}^2}\\ \Rightarrow A{\left(n+1\right)}^2+B{n}^2=2n+1\end{array}$

When n = 0 , A = 1 and n = -1 then B = -1

So,

  $\frac{2n+1}{n^2{\left(n+1\right)}^2}=\frac{1}{n^2}-\frac{1}{{\left(n+1\right)}^2}$

Now, calculating the partial sums of the given series,

  $\begin{array}{l}{S}_1=1-\frac{1}{4}\\ S_2=\left(1-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{9}\right)=1-\frac{1}{9}\\ S_3=\left(1-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{16}\right)=1-\frac{1}{16}\end{array}$

In the same way,

  $S_n=1-\frac{1}{{\left(n+1\right)}^2}$

So, sum of the series will be

  $\begin{array}{l}S=\underset{n\to \infty }{\lim }{S}_{n,}\\ \Rightarrow S=\underset{n\to \infty }{\lim }\left(1-\frac{1}{{\left(n+1\right)}^2}\right)\\ \Rightarrow S=\underset{n\to \infty }{\lim }\left(1\right)-\underset{n\to \infty }{\lim }\left(\frac{1}{{\left(n+1\right)}^2}\right)\\ \Rightarrow S=1\because \underset{n\to \infty }{\lim }\left(\frac{1}{{\left(n+1\right)}^2}\right)=0\end{array}$

Therefore, Sum of Telescoping series is S = 1