Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have to find the formula for sum $$\sum_{i=2}^{n} \frac1{i^2-1}$$ I remember reading somewhere that $\displaystyle \frac1{i^2-1}$ can be shown as $\displaystyle\frac1{i+1}$ and $\displaystyle \frac1{i-1}$.

If I can express it as this, then I should be able to create a telescopic sum, and produce the formula easily. But how can I get to those two fractions in the first place so that I can build the proof?

share|cite|improve this question
Writing $\frac{1}{i^2-1}$ as a combination of $\frac{1}{i-1}$ and $\frac{1}{i+1}$ is usually called "partial fraction decomposition". You can easely find sites covering this, and it is also covered in any Calculus textbook, in the section called something like "integrals of rational functions"... – N. S. May 30 '11 at 19:34
up vote 6 down vote accepted

$$\frac1{k^2-1} = \frac1{2} \left( \frac1{k-1} - \frac1{k+1} \right)$$ Now let the telescopic summation take over.

share|cite|improve this answer
or equivalently $\dfrac{1}{k-1}-\dfrac{1}{k+1} = \dfrac{(k+1)-(k-1)}{(k-1)(k+1)} = \dfrac{2}{k^2-1}$ – Henry May 30 '11 at 19:36
So this should simplify to (n-1)(3n+2)/(4n(n+1)) – Christopher May 30 '11 at 20:16
@Christopher: That's what I got after doing it out, as well. – Michael Chen May 30 '11 at 21:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.