$\displaystyle\sum_{n=2}^{\infty}\frac{2}{(n^3-n)3^n} = -\frac{1}{2}+\frac{4}{3}\sum_{n=1}^{\infty}\frac{1}{n\cdot 3^n}$ Please help me, to prove that
$$
\sum_{n=2}^{\infty}\frac{2}{(n^3-n)3^n} = 
 -\frac{1}{2}+\frac{4}{3}\sum_{n=1}^{\infty}\frac{1}{n\cdot 3^n}.
$$
 A: Hint: Use partial fractions, 
$$\sum_{n=2}^{\infty} \frac {2}{(n^3-n)3^n} = \sum_{n=2}^\infty \frac {1}{3^n}  \left( \frac {1}{n-1} - \frac {2}{n} + \frac {1}{n+1}\right)$$
Now, shift the indexing up/down 1 as necessary. Remember to check the power of 3.
A: \begin{align*}
  \sum_{n=2}^\infty \frac{2}{3^n( n^3 - n) } &=  \sum_{n=2}^\infty  \frac{1}{3^n}\left( \frac{-2}{n} + \frac{1}{n+1} + \frac{1}{n-1} \right) \\ 
 &= \sum_{n=1}^\infty \frac{1}{3^{n+1}n} - \sum_{n=1}^\infty \frac{2}{3^{n+1} (n+1)}+\sum_{n=1}^\infty \frac{1}{(n+2)3^{n+1}} \\
 &= \frac{1}{3}\sum_{n=1}^\infty\frac{1}{n 3^n} - 2 \left ( \sum_{n=1}^\infty \frac{1}{n 3^n} - \frac{1}{3} \right ) + 3 \left( \sum_{n=1}^\infty \frac{1}{n 3^n} - \frac{1}{3} -\frac{1}{3^2 \cdot 2}\right ) \\
&= \sum_{n=1}^{\infty}\frac{1}{n 3^n}\left( \frac{1}{3} - 2 + 3 \right ) + \left( \frac{2}{3} - 1 - \frac{1}{6} \right ) \\
&= \frac{4}{3}\sum_{n=1}^{\infty}\frac{1}{n\cdot 3^n} -\frac{1}{2}
\end{align*}
A: Maybe we wanna use the fact that $\displaystyle \sum_{n=1}^{\infty}\frac{1}{ns^n}=\ln\frac{s}{s-1}, \space s>1$. Then
$$\sum_{n=2}^{\infty} \frac {2}{(n^3-n)3^n}=\frac{1}{3}\sum_{n=2}^\infty \frac {1}{(n-1)3^{n-1}}    - 2\sum_{n=2}^\infty \frac {1}{n3^n} + 3\sum_{n=2}^\infty \frac {1}{(n+1)3^{n+1}}=\frac{4}{3} \log\frac{3}{2}-\frac{1}{2}$$
Chris. (the auxiliary limit is straightforward by Taylor series)
