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!

Answers 3

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

Hint on computing the series $\sum_{n=2}^\infty \frac{1}{n^2-1}$..

yousef magableh
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
Peter Foreman
April 20, 2019 12:54 PM

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

April 20, 2019 13:17 PM

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