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?

Hints:
$$n^3 - n = (n-1) n (n+1)$$
$$\frac{1}{n (n+1)} = \frac{1}{n} - \frac{1}{n+1}$$
$$\frac{1}{(n-1) n} = \frac{1}{n-1} - \frac{1}{n}$$
$$\frac{1}{(n-1)(n+1)} = \frac{1/2}{n-1} - \frac{1/2}{n+1}$$
$$\frac{1}{(n-1)n(n+1)} = \frac{1}{(n-1)n} - \frac{1}{(n-1)(n+1)} = \frac{1}{n-1} - \frac{1}{n} - \frac{1/2}{n-1} + \frac{1/2}{n+1} = \frac{1/2}{n-1} - \frac{1}{n} + \frac{1/2}{n+1}$$
$$\frac{2}{3^n (n-1) n (n+1)} = \frac{1}{3^n (n-1)} - \frac{2}{3^n n}
+ \frac{1}{3^n (n+1)}$$
Then try to combine terms when substituting this into the sum.

 There is probably a more elegant way to do the following computations... $$\begin{align}\sum_{n \ge 2} \frac{2}{3^n (n-1) n (n+1)}    &= \frac{1}{3^2}  + \left(\frac{1}{3^3} - \frac{2}{3^2}\right)\frac{1}{2} + \sum_{n \ge 3} \left(\frac{1}{3^{n-1}} - \frac{2}{3^{n}} + \frac{1}{3^{n+1}}\right) \frac{1}{n}     \\    &= \frac{1}{54} + \left(3 - 2 + \frac{1}{3}\right)\sum_{n \ge 3} \frac{1}{3^n n}    \\    &= \frac{1}{54} + \frac{4}{3} \sum_{n\ge 3} \frac{1}{3^n n}    \\    &= \frac{1}{54} - \frac{4}{3} \left(\frac{1}{3} + \frac{1}{9 \cdot 2}  \right)    + \frac{4}{3} \sum_{n\ge 1} \frac{1}{3^n n}    \\    &= - \frac{1}{2}     + \frac{4}{3} \sum_{n\ge 1} \frac{1}{3^n n}.    \end{align}$$

