Chapter 10.4, Problem 53E

To determine

To find: Sum of the Telescopic series.

Answer to Problem 53E

The sum of the Telescopic series is $S=-\frac{1}{\ln 2}$

Explanation of Solution

Given:

The series given is,

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

Calculation:

First calculate the sum of general terms by substituting the value of n .

So,

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

In the same way,

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

Now, the sum of the series will be,

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

Therefore, the sum of the Telescopic series is $S=-\frac{1}{\ln 2}$