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.

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    $\begingroup$ The identity is not true, there should be a misprint. $\endgroup$ – Start wearing purple Jan 1 '16 at 13:53
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    $\begingroup$ See my answer below, it should be $n$, not $n^3$. $\endgroup$ – Ron Gordon Jan 1 '16 at 13:56
  • $\begingroup$ 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 !! $\endgroup$ – Claude Leibovici Jan 1 '16 at 15:03

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

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


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