What is a collapsing infinite series?

1 Answer
Oct 17, 2014

Here is an example of a collapsing (telescoping) series

#sum_{n=1}^infty(1/n-1/{n+1})#

#=(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+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=(1/1-1/2)+(1/2-1/3)+cdots+(1/n-1/{n+1})#

by cancelling ("collapsing") the overlapping terms,

#=1-1/{n+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#


I hope that this was helpful.