What is a telescoping infinite series?

1 Answer
Sep 8, 2014

Here is an example of a telescoping series
sum_{n=1}^infty(1/n-1/{n+1})n=1(1n1n+1)
=(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+cdots=(1112)+(1213)+(1314)+
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_nSn first.
S_n=(1/1-1/2)+(1/2-1/3)+cdots+(1/n-1/{n+1})Sn=(1112)+(1213)++(1n1n+1)
by cancelling the overlapping terms,
=1-1/{n+1}=11n+1
Hence, the sume of the infinite series can be found by
sum_{n=1}^infty(1/n-1/{n+1})=lim_{n to infty}S_n=lim_{n to infty}(1-1/{n+1})=1