# Prove identity $\sum_{n=2}^{\infty}\frac{2}{(n^3-n)3^n}=-\frac{1}{2}+\frac{4}{3}\sum_{n=1}^{\infty}\frac{1}{n^3\cdot3^n}$

I have no idea how to prove following identity: $$\sum_{n=2}^{\infty}\frac{2}{(n^3-n)3^n}=-\frac{1}{2}+\frac{4}{3}\sum_{n=1}^{\infty}\frac{1}{n\cdot3^n}$$

My main problem concerned the fact, that I had not clear idea in what direction should my efforts go in order to find solution. I did try to use partial sums in similar vein as Stuart Gordon, but it was far from being successful.

• The identity is not true, there should be a misprint. – Start wearing purple Jan 1 '16 at 13:53
• See my answer below, it should be $n$, not $n^3$. – Ron Gordon Jan 1 '16 at 13:56
• In this community, I am ready to bet that, once more and not surprizingly, Ron Gordon is right. As Start wearing purple also commented, there is one more typo in a textbook. Happy New Year !! – Claude Leibovici Jan 1 '16 at 15:03

## 1 Answer

Use partial fractions...

\begin{align}\sum_{n=2}^{\infty} \frac{2}{(n^3-n) 3^n} &= \sum_{n=2}^{\infty} \frac{1}{(n-1) 3^n} + \sum_{n=2}^{\infty} \frac{1}{(n+1) 3^n} - \sum_{n=2}^{\infty} \frac{2}{n 3^n} \\ &=\frac13 \sum_{n=1}^{\infty} \frac{1}{n 3^n} + 3 \sum_{n=3}^{\infty} \frac{1}{n 3^n} - \sum_{n=2}^{\infty} \frac{2}{n 3^n} \\ &= \frac43 \sum_{n=1}^{\infty} \frac{1}{n 3^n} - 3 \left (\frac13 + \frac1{18} \right ) + \frac23 \\ &= \frac43 \sum_{n=1}^{\infty} \frac{1}{n 3^n} - \frac12 \end{align}

This is a log, so that the sum is actually

$$\sum_{n=2}^{\infty} \frac{2}{(n^3-n) 3^n} = \frac43 \log{\frac32} - \frac12$$