Question

What is telescoping series ? Give an example of convergent telescoping series and a example of divergent telescoping series.

Short answer

$\sum _{k=1}^{\mathrm{\infty }} \left(\frac{1}{k}-\frac{1}{k+2}\right).$ is telescopic and convergent in nature .

$\sum _{k=1}^{\mathrm{\infty }} \ln \frac{k}{k+1}$ is telescopic and also divergent in nature .

Step-by-step solution

Step 1. Given information 

We have to define the telescopic series and examples of convergent and divergent series .

Step 2. Defining telescopic series .

The series is telescopic if the each term of the series can be written as sum of two or more summands and after that there will be cancellation of terms in partial sums .

Step 3. Expanding the series 

Consider the series

The nth term in the sequence of partial sums

$S_n=\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+\dots +\left(\frac{1}{n}-\frac{1}{n+2}\right)$

In each two consecutive pairs the terms are being cancelled hence the series is telescopic .

Step 4.The general term of series and checking for convergent nature of series .

$\begin{array}{r}{S}_n=\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+\dots +\left(\frac{1}{n}-\frac{1}{n+2}\right)\\ =\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+\dots \left(\frac{1}{n-1}-\frac{1}{n+1}\right)+\left(\frac{1}{n}-\frac{1}{n+2}\right)\\ =1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}\\ =\frac{3}{2}-\frac{1}{n+2}\end{array}$$\frac{3}{2}-\frac{1}{n+2}$

The general term in the sequence is $\frac{3}{2}-\frac{1}{n+2}$

When this general term limit approach to zero then

$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim} S_n=\underset{n\to \mathrm{\infty }}{lim} \left[\frac{3}{2}-\frac{1}{n+2}\right]\\ =\frac{3}{2}-\underset{n\to \mathrm{\infty }}{lim} \frac{1}{n+2}\\ =\frac{3}{2}\end{array}$

Therefore the series$\sum _{k=1}^{\mathrm{\infty }} \left(\frac{1}{k}-\frac{1}{k+2}\right)$is convergent to$\frac{3}{2}$

Step 5. Checking for other telescopic series 

Consider the series$\sum _{k=1}^{\mathrm{\infty }} \ln \frac{k}{k+1}$

The series can be written as $\sum _{k=1}^{\mathrm{\infty }} \ln \frac{k}{k+1}$

$\sum _{k=1}^{\mathrm{\infty }} \ln \frac{k}{k+1}=\sum _{k=1}^{\mathrm{\infty }} [\ln k-\ln (k+1\left)\right]\left(\text{Because}\ln \frac{a}{b}=\ln a-\ln b\right)$

The nth term in partial sums is

$S_n=\ln \left(1\right)-\ln \left(2\right)+(\ln 2-\ln 3)+\dots (\ln n-\ln (n+1\left)\right)$

In each pair the terms are being cancelled hence this is telescopic series .

Step 6. General term of series 

the general term of given series is

$\begin{array}{r}{S}_n=\ln \left(1\right)-\ln \left(2\right)+(\ln 2-\ln 3)+\dots (\ln n-\ln (n+1\left)\right)\\ =\ln \left(1\right)-\ln \left(2\right)+(\ln 2-\ln 3)+(\ln 3-\ln 4)\dots +(\ln n-\ln (n+1\left)\right)\\ =\ln \left(1\right)-\ln (n+1)\\ =-\ln (n+1)(\text{Because}\ln 1=0)\\ =\ln \left(\frac{1}{n+1}\right)\end{array}$

The general term in its sequence of partial sum is$\ln \left(\frac{1}{n+1}\right)$

Step 7. Checking for divergent series .

The limit of series as$n\underset{}{\to }\infty$

$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim} S_n=\underset{n\to \mathrm{\infty }}{lim} \left[\ln \left(\frac{1}{n+1}\right)\right]\\ =\mathrm{\infty }\end{array}$$\sum _{k=1}^{\mathrm{\infty }} \ln \frac{k}{k+1}$This series is divergent .