Help calculating an infinite sum

Question

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.

Also: https://math.stackexchange.com/q/2149934/42969, https://math.stackexchange.com/q/2137750/42969 — Martin R, Apr 04 '18 at 20:02
I get an answer of 1/4. The series is telescoping. Use partial fraction decomp. — CogitoErgoCogitoSum, Apr 04 '18 at 22:57

score 2 · Answer 1 · edited Apr 04 '18 at 20:47

$$\displaystyle \frac 1 {(n-1)n(n+1)}$$ 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.

Help calculating an infinite sum

1 Answers1