Finding sum of sequence $\sum\limits_{k=2}^n \frac{1}{k^2-1}$

Question

Here is a problem I need to solve:

$$ \sum_{k=2}^n \frac{1}{k^2-1} $$

It came with another one alreay done in the same task:

$$ \sum_{k=1}^n \frac{1}{k(k+1)} = \sum_{k=1}^n \frac{(k+1)-k}{k(k+1)} = \sum_{k=1}^n \frac{1}{k} - \sum_{k=1}^n \frac{1}{k+1} = \sum_{k=1}^n \frac{1}{k} - \sum_{k=2}^{n+1} \frac{1}{k} = 1 - \frac{1}{n+1} $$

So, the problem have to have similar solution. I can't find a proper change for the numerator. I think the numeratore could be find using factorised dominator:

$$ \sum_{k=2}^n \frac{1}{(k-1)(k+1)} $$

Basically it's the same idea: try to write $1/(k^2-1)$ as a sum of simple fractions like $1/(k+a)$. — Tom-Tom, Jan 14 '14 at 13:22
See also: https://math.stackexchange.com/q/931829, https://math.stackexchange.com/q/42205. — Martin Sleziak, Dec 27 '19 at 06:20

score 3 · Answer 1 · answered Jan 14 '14 at 13:22

3

Hint

$$\frac{1}{k^2-1}=\frac 1 2\left(\frac{1}{k-1}-\frac{1}{k+1}\right)=\frac 1 2\left(\frac{1}{k-1}-\frac 1 k+\frac 1 k-\frac{1}{k+1}\right)$$

answered Jan 14 '14 at 13:22

score 0 · Accepted Answer · answered Jan 14 '14 at 13:23

0

$\frac{1}{(k-1)(k+1)}=\frac{1}{2}\frac{1}{k-1}-\frac{1}{2}\frac{1}{k+1}$

answered Jan 14 '14 at 13:23

kmitov

4,731

Finding sum of sequence $\sum\limits_{k=2}^n \frac{1}{k^2-1}$

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