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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?

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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

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

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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
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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

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