# find $\sum_{k=2}^{\infty}\frac{1}{k^2-1}$

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

I found that:

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

and $\sum_{k=2}^{\infty}\frac{1}{2(k-1)}-\frac{1}{2(k+1)}$ is a telescopic series so we need $lim_{n \to \infty} \frac{1}{2}-\frac{1}{2(n+1)}=\frac{1}{2}$ but the answer is $\frac{3}{4}$

Notice that the first few terms are:

$(1/2-1/6)+(1/4-1/8)+(1/6-1/10)+(1/8-1/12)\cdots$

which gives: $1/2+1/4=3/4$.

It helps to write out the first few terms of the series to see what's happening. Setting aside the multiplicative factor ${1\over2}$, we have

$$\sum_{k=2}^\infty\left({1\over k-1}-{1\over k+1}\right)=\left(1-{1\over 3}\right)+\left({1\over2}-{1\over4}\right)+\left({1\over3}-{1\over 5}\right)+\left({1\over4}-{1\over6}\right)+\cdots$$

Can you see from this what's left over from the telescoping?

Starting from where you left off: $S = \displaystyle \sum_{k=2}^\infty \left(\left(\dfrac{1}{2(k-1)} - \dfrac{1}{2k}\right)+\left(\dfrac{1}{2k} - \dfrac{1}{2(k+1)}\right)\right)$. Can you find the $2$ telescoping sums and their values?

• Why to add $\frac{1}{k}$?
– gbox
Commented Mar 15, 2016 at 17:43
• To make 2 telescoping series. Commented Mar 15, 2016 at 17:44
• Sorry, I did not understand
– gbox
Commented Mar 15, 2016 at 17:45
• Your hint is correct but pretty subtle. The OP clearly does not understand. Could you add more detail so the OP can understand? Commented Mar 15, 2016 at 18:02

The telescoping series cancel off itself after two terms, since $k-1$ and $k+1$ are 2 aparr

• This response is correct but fairly subtle. Could you explain more clearly? Commented Mar 15, 2016 at 18:06