Proof that $ \sum_{n=2}^{\infty} \frac{2}{3^n \cdot (n^3-n)} = \frac{-1}{2} + \frac{4}{3}\sum_{n=1}^{\infty} \frac{1}{n \cdot 3^n}$ Task
Proof that $ \sum_{n=2}^{\infty} \frac{2}{3^n \cdot (n^3-n)} = -\frac{1}{2} + \frac{4}{3}\sum_{n=1}^{\infty} \frac{1}{n \cdot 3^n}$
About
Hi, I have been trying to solve this task since yesterday. I have idea that I can evaluate both sides and show that they are the same. I supposed $ \sum_{n=1}^{\infty} \frac{1}{n \cdot 3^n} $ there is formula for that.
So I want to evaluate in the same way left side. So I compute that
$$ \sum_{n=2}^{\infty} \frac{2}{3^n \cdot (n^3-n)} = \sum_{n=2}^{\infty} \frac{1}{3^n \cdot n (n-1)} - \frac{1}{3^n \cdot n (n+1)} $$
I am trying to transform it to use that formula:
$$ \sum_{n=1}^{\infty} \frac{1}{n\cdot p^n} = ln\frac{p}{p-1} $$
but I still defeat.
So please tell me, there are better ways to proof that or should I consider to change my field of study?

hint
$$\frac{2}{n^3-n}=\frac{-2}{n}+\frac{1}{n-1}+\frac{1}{n+1}$$
