# Showing that $\sum_{n=2}^\infty \frac{2}{n^2-1}$ is convergent

by James odare   Last Updated April 20, 2019 13:20 PM

I am trying to show that $$\sum_{n=2}^\infty \frac{2}{n^2-1}$$ is convergent by using telescoping sum.

So far I have reduced $$\frac{2}{n^2-1}$$ via partial fractions to $$\frac{1}{n-1} - \frac{1}{n+1}$$.

Then in an attempt to reduce I wrote $$\sum_{n=2}^\infty \frac{1}{n-1} - \frac{1}{n+1} = (1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4}).....$$

I then noticed that they all cancel apart from $$1+\frac{1}{2}$$ however I am not sure what happens generally at the end? So whether $$\frac{1}{n-1}$$ will cancel or stay?

Thank you!

Tags :

Your answer is correct, the sum dose converge to $$\frac{3}{2}$$. But it is possible duplicate, see:

yousef magableh
April 20, 2019 12:51 PM

\begin{align} \sum_{k=2}^n\frac{2}{k^2-1} &=\sum_{k=2}^n\left(\frac1{k-1}-\frac1{k+1}\right)\\ &=\left(\frac11-\frac13\right)+\left(\frac12-\frac14\right)+\left(\frac13-\frac15\right)+\dots+\left(\frac1{n-2}-\frac1{n}\right)+\left(\frac1{n-1}-\frac1{n+1}\right)\\ &=\frac32-\frac1{n}-\frac1{n+1}\\ \end{align} So we have that $$\sum_{k=2}^\infty\frac{2}{k^2-1}=\lim_{n\to\infty}\left(\frac32-\frac1{n}-\frac1{n+1}\right)=\frac32$$

Peter Foreman
April 20, 2019 12:54 PM

$$\sum_{k=2}^n\frac{2}{k^2-1}=\sum_{k=2}^n\left(\frac1{k-1}-\frac1{k+1}\right)$$ $$=\sum_{k=2}^n\bigg[\left(\frac{1}{k-1}-\frac{1}{k}\right)+\left(\frac{1}{k}-\frac1{k+1}\right)\bigg]$$ $$=\sum_{k=2}^n\left(\frac{1}{k-1}-\frac{1}{k}\right)-\sum_{k=2}^{n}{\left(\frac{1}{k+1}-\frac1{k}\right)}$$ $$=\big(1-\frac1n\big)-\big(\frac1{n+1}-\frac12\big)=\frac32-\frac{1}{n}-\frac1{n+1}$$

HAMIDINE SOUMARE
April 20, 2019 13:17 PM

## Related Questions

Updated April 22, 2017 16:20 PM

Updated December 24, 2017 09:20 AM

Updated November 22, 2018 06:20 AM

Updated June 10, 2019 18:20 PM