Chapter 10, Problem 54RE

To determine

To find: Sum of the series ${\displaystyle \sum _{n=2}^{\infty }\frac{-2}{n\left(n+1\right)}}$

Answer to Problem 54RE

The sum of the given series is $S=-1$

Explanation of Solution

Given:

The given series is,

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

Calculation:

The $n^{th}$ term of the given series will be,

  $T_n=\frac{-2}{n\left(n+1\right)}$

Using partial fraction

  $T_n=\frac{-2}{n\left(n+1\right)}=\frac{-2}{n}+\frac{3}{n+1}$

Now the partial sums will be

  $\begin{array}{l}{S}_1=\frac{-2}{2}+\frac{2}{3}=-1+\frac{2}{3}\\ S_2=\left(-1+\frac{2}{3}\right)+\left(-\frac{2}{3}+\frac{2}{4}\right)=-1+\frac{2}{4}\\ S_3=\left(-1+\frac{2}{3}\right)+\left(-\frac{2}{3}+\frac{2}{4}\right)+\left(-\frac{2}{4}+\frac{2}{5}\right)=-1+\frac{2}{5}\end{array}$

In the same way

  $S_n=-1+\frac{2}{\left(n+1\right)}$

So,

The sum of the series will be

  $\begin{array}{l}S=\underset{n\to \infty }{\lim }{S}_n\\ =\underset{n\to \infty }{\lim }\left[-1+\frac{2}{\left(n+1\right)}\right]\\ \Rightarrow S=-1\end{array}$

Therefore, the sum of the given series is $S=-1$