$\displaystyle\sum_{n=2}^{\infty}\frac{2}{(n^3-n)3^n} = -\frac{1}{2}+\frac{4}{3}\sum_{n=1}^{\infty}\frac{1}{n\cdot 3^n}$ Please help me, to prove that
$$
\sum_{n=2}^{\infty}\frac{2}{(n^3-n)3^n} = 
 -\frac{1}{2}+\frac{4}{3}\sum_{n=1}^{\infty}\frac{1}{n\cdot 3^n}.
$$

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)
