# What is an example of a telescoping series and how do you find its sum?

Mar 8, 2015

A telescopic serie is a serie which can be written

${\sum}_{k = 0}^{n} \left({a}_{k + 1} - {a}_{k}\right)$

This sum is equal to ${a}_{n + 1} - {a}_{0}$ because

${\sum}_{k = 0}^{n} \left({a}_{k + 1} - {a}_{k}\right) = \left({a}_{1} - {a}_{0}\right) + \left({a}_{2} - {a}_{1}\right) + \setminus \ldots + \left({a}_{n + 1} - {a}_{n}\right)$.

An easy example is ${\sum}_{k = 1}^{\infty} \frac{1}{n \left(n + 1\right)}$.

Remark that $\frac{1}{n \left(n + 1\right)} = \frac{1}{n} - \frac{1}{n + 1}$, so,

${\sum}_{k = 1}^{N} \frac{1}{n \left(n + 1\right)} = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \setminus \ldots + \left(\frac{1}{N} - \frac{1}{N + 1}\right)$

${\sum}_{k = 1}^{N} \frac{1}{n \left(n + 1\right)} = 1 - \frac{1}{N + 1}$.

When $N \to + \infty$, you get ${\sum}_{k = 1}^{\infty} \frac{1}{n \left(n + 1\right)} = 1$.