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