What is an example of telescopic series and how do you find its sum?

Hint: A telescoping series is a series where each term \[{{a}_{k}}\]​ can be written as \[{{a}_{k}}={{t}_{k}}-{{t}_{k+1}}\] for some series \[{{t}_{k}}\]. It is a series whose partial sums eventually only have a finite number of terms after cancellation. The cancellation technique, with part of each term cancelling with part of the next term, is known as the method of differences that helps to determine the sum of the telescopic series.

Complete step-by-step solution:
According to the question, example of telescopic series is as follows:
\[\sum\limits_{n=1}^{\infty }{\dfrac{1}{n\left( n+1 \right)}}...\left( 1 \right)\]
It can be written as \[{{a}_{k}}={{t}_{k}}-{{t}_{k+1}}\] for some series \[{{t}_{k}}\]. In order to find its sum, we will use the cancellation technique, with part of each term cancelling with part of the next term, is known as the method of differences.
We know that \[\dfrac{1}{n\left( n+1 \right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\], applying it in the equation \[\left( 1 \right)\], we get:
\[\Rightarrow \sum\limits_{n=1}^{\infty }{\dfrac{1}{n\left( n+1 \right)}}=\sum\limits_{n=1}^{\infty }{\left( \dfrac{1}{n}-\dfrac{1}{n+1} \right)}\]
Now opening the right-hand side, we get:
\[\Rightarrow \sum\limits_{n=1}^{\infty }{\dfrac{1}{n\left( n+1 \right)}}=\displaystyle \lim_{N \to \infty }\left[ \left( 1-\dfrac{1}{2} \right)+\left( \dfrac{1}{2}-\dfrac{1}{3} \right)+...+\left( \dfrac{1}{N}-\dfrac{1}{N+1} \right) \right]\]
We know that according to Commutative property of addition: Changing the order of addends does not change the sum., so we can shift the brackets of the terms in the right-hand side as follows:
\[\Rightarrow \sum\limits_{n=1}^{\infty }{\dfrac{1}{n\left( n+1 \right)}}=\displaystyle \lim_{N \to \infty }\left[ 1+\left( -\dfrac{1}{2}+\dfrac{1}{2} \right)+\left( -\dfrac{1}{3}+\dfrac{1}{3} \right)+...+\left( -\dfrac{1}{N}+\dfrac{1}{N} \right)-\dfrac{1}{N+1} \right]\]
As we can see some terms are cancelling each other. So, after cancellation, we get:
\[\Rightarrow \sum\limits_{n=1}^{\infty }{\dfrac{1}{n\left( n+1 \right)}}=\displaystyle \lim_{N \to \infty }\left[ 1-\dfrac{1}{N+1} \right]\]
We know that as \[N \to +\infty \], \[\displaystyle \lim_{N \to \infty }\left[ 1-\dfrac{1}{N+1} \right]=1\], therefore applying this result in the above equation, we get:
\[\Rightarrow \sum\limits_{n=1}^{\infty }{\dfrac{1}{n\left( n+1 \right)}}=1\]
Therefore, example of telescopic series is \[\sum\limits_{n=1}^{\infty }{\dfrac{1}{n\left( n+1 \right)}}\] and its sum is equal to \[1\].

Note: Students may find it hard to find the sum of the telescopic series but this is not the case. If we know the approach to the problem, we can easily solve it. So, the trick here is in most of the cases we can find the sum by rationalizing the terms and then except the first and the last terms, most of the terms get cancelled out.