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I'm supposed to determine whether this sum diverges or converges and if it converges then find its value: $$ \sum_{n=2}^\infty \frac{1}{n^2-1}. $$

Using the comparison test I eventually showed that this converges. But I can't figure out how to show what this sum converges to. The only sums we actually found values for in my notes are geometric series which this clearly isn't.

I saw that I could use partial fraction decomposition to represent the terms as $$\frac{1/2}{n-1}- \frac{1/2}{n+1} $$ but that just gets me $\infty - \infty$, so this isn't the way to do it.

I'm not sure how to find the value of this sum. I don't need the full solution but a hint would be appreciated. Thanks. :)

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Use partial fraction decomposition to represent the sum and notice that it is a telescoping series.

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  • $\begingroup$ Got it. Thanks. :) $\endgroup$ – user307485 Jan 23 '16 at 22:47
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Since $$\frac{1}{n^2-1}=\frac{1}{(n-1)(n+1)}=\frac12\left(\frac{1}{n-1}-\frac{1}{n+1}\right)$$ as you already have, then write for $N\in \mathbb N$
\begin{align}\sum_{n=2}^{N}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)&=\left(1-\frac13\right)+\left(\frac12-\frac14\right)+\left(\frac13-\frac15\right)+\dots+\left(\frac1{N-1}-\frac1{N+1}\right)\\[0.2cm]&=1+\frac12+\left(\frac13-\frac13\right)+\dots+\left(\frac1{N-1}-\frac1{N-1}\right)-\frac1N-\frac1{N+1}\\[0.2cm]&=\frac32-\frac1N-\frac1{N+1}\end{align} In other words, this sum telescopes. Now let $N\to \infty$ (and of course do not forget 1/2 in front of the sum) to conclude that $$\sum_{n=2}^{+\infty}\frac{1}{n^2-1}=\lim_{N\to+\infty} \frac12\sum_{n=2}^N\left(\frac{1}{n-1}-\frac1{n+1}\right)=\frac12\lim_{N\to+\infty}\left(\frac32-\frac1N-\frac1{N+1}\right)=\frac34$$

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  • $\begingroup$ OP says he has this. $\endgroup$ – Ross Millikan Jan 23 '16 at 22:47
  • $\begingroup$ @RossMillikan Yes, sorry. I edited my answer. $\endgroup$ – Jimmy R. Jan 23 '16 at 22:53
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    $\begingroup$ @JimmyR. You didn't expand the telescoping sum correctly. The answer should be 3/4. $\endgroup$ – user307485 Jan 23 '16 at 22:55
  • $\begingroup$ @user307485 Yes, you are right. $\endgroup$ – Jimmy R. Jan 23 '16 at 22:57
  • $\begingroup$ I think the partial sum is actually $\dfrac32-\dfrac1N-\dfrac1{N+1}$. $\endgroup$ – Akiva Weinberger Jan 23 '16 at 23:12
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hint: $\dfrac{1}{n^2-1} = \dfrac{1}{2}\cdot \left(\dfrac{1}{n-1}-\dfrac{1}{n+1}\right) = \dfrac{1}{2}\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right) + \dfrac{1}{2}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)$. From this you see that there are $2$ sums you calculate, and using telescoping the first sum is $\dfrac{1}{2}$, and the second is $\dfrac{1}{4}$, thus the answer is $\dfrac{3}{4}$ as claimed.

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  • $\begingroup$ OP says he has proved convergence. $\endgroup$ – Ross Millikan Jan 23 '16 at 22:46

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