So i have this sum:
$$\sum_{n=2} ^{\infty} \frac{1} {n^3-n} $$
So for now i tried using this:
$$\sum_{n=2} ^{\infty} \frac{1} {n^3-n} = \frac{1} {6}(1+\frac{1} {4}+\frac{1} {10}+...$$
Since $n^3-n=n(n-1)(n+1) $ is divisible by 6.(The multiplication of k consecutive numbers are always divisible by k! )
But i am stuck as i dont identify any known series in here.
Any help eould be appreciated.
$$\displaystyle \frac 1 {(n-1)n(n+1)}$$ becomes $$\displaystyle \frac 1 {(n+1)-(n-1)}\left ( \frac 1 {n(n-1)}- \frac 1 {n(n+1)}\right)$$
And then, after you simplify it, you will be able to calculate it easily. Also, in case that you didn't know,
$$\displaystyle \frac 1{ABC}=\frac 1 {(C-A)}*\left(\frac 1{AB}- \frac 1{BC}\right)$$ Everyone knows $$\displaystyle \frac 1{AB}=\frac 1 {(B-A)}*\left(\frac 1A- \frac 1B\right)$$
, right? It is similar to that.