5
$\begingroup$

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?

$\endgroup$
2
  • 7
    $\begingroup$ 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"... $\endgroup$
    – N. S.
    May 30, 2011 at 19:34
  • $\begingroup$ See also: math.stackexchange.com/q/638078, math.stackexchange.com/q/931829.. $\endgroup$ Dec 27, 2019 at 6:20

1 Answer 1

9
$\begingroup$

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

$\endgroup$
3
  • 1
    $\begingroup$ or equivalently $\dfrac{1}{k-1}-\dfrac{1}{k+1} = \dfrac{(k+1)-(k-1)}{(k-1)(k+1)} = \dfrac{2}{k^2-1}$ $\endgroup$
    – Henry
    May 30, 2011 at 19:36
  • 1
    $\begingroup$ So this should simplify to (n-1)(3n+2)/(4n(n+1)) $\endgroup$ May 30, 2011 at 20:16
  • $\begingroup$ @Christopher: That's what I got after doing it out, as well. $\endgroup$ May 30, 2011 at 21:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .