# What is a collapsing infinite series?

Oct 17, 2014

Here is an example of a collapsing (telescoping) series

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

$= \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots$

As you can see above, terms are shifted with some overlapping terms, which reminds us of a telescope. In order to find the sum, we will its partial sum ${S}_{n}$ first.

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

by cancelling ("collapsing") the overlapping terms,

$= 1 - \frac{1}{n + 1}$

Hence, the sume of the infinite series can be found by

${\sum}_{n = 1}^{\infty} \left(\frac{1}{n} - \frac{1}{n + 1}\right) = {\lim}_{n \to \infty} {S}_{n} = {\lim}_{n \to \infty} \left(1 - \frac{1}{n + 1}\right) = 1$

I hope that this was helpful.