Prove the series identity 
Prove an identity:
   $$\sum_{n=2}^{ \infty } \frac{2}{(n^3-n)3^n}=- \frac{1}{2}+ \frac{4}{3} \cdot \sum_{n=1}^{ \infty } \frac{1}{n \cdot 3^n}$$

I've checked that the left-hand-side of this identity is convergent absolutely, hence I can write it as:
$$\sum_{n=2}^{ \infty } \frac{2}{(n-1)(n+1)} \cdot \frac{1}{n3^n}$$ 
I've also calculated the sum $$\sum_{n=2}^{ \infty } \frac{2}{(n-1)(n+1)}= \frac{3}{2}
$$
But now, I don't see what can I do with the right-hand-side and what to do with  $$\sum_{n=2}^{ \infty }\frac{1}{n3^n}$$

Using Partial Fraction Decomposition,
$$\frac1{n^3-n}=\frac An+\frac B{n-1}+\frac C{n+1}$$
multiply either sides $n^3-n=n(n-1)(n+1)$ and compare the coefficients of the different powers of $n$ to find $A,B,C$

Alternatively,
$$\frac2{n(n-1)(n+1)}=\frac{n+1-(n-1)}{n(n-1)(n+1)}=\frac1{n(n-1)}-\frac1{(n+1)n}$$
Again, $\displaystyle \frac1{n(n-1)}=\frac{n-(n-1)}{n(n-1)}=\frac1{n-1}-\frac1n$
and $\displaystyle \frac1{n(n+1)}=\frac{n+1-n}{n(n+1)}=\frac1n-\frac1{n+1}$
So, the $n$th term $\displaystyle T_n=\left(\frac1{n-1}-\frac2n+\frac1{n+1}\right)\frac1{3^n}=\frac13\cdot\frac1{(n-1)3^{n-1}}-\frac2{n3^n}+\frac3{(n+1)3^{n+1}}$ 
$$\implies\sum_{n=2}^\infty T_n=\sum_{n=2}^\infty\left(\frac13\cdot\frac1{(n-1)3^{n-1}}-\frac2{n3^n}+\frac3{(n+1)3^{n+1}}\right)$$
$$=\frac13\sum_{n=2}^\infty\frac1{(n-1)3^{n-1}}-2\sum_{n=2}^\infty\frac1{n3^n}+3\sum_{n=2}^\infty\frac1{(n+1)3^{n+1}}$$
Adjust each summation so that $n$ ranges between $[1,\infty)$
