Find the sum function of $\sum_{n=0}^{\infty}\frac{n(n-2)}{n+1}x^{n-1}$ series summation:
$$\sum_{n=0}^{\infty}\frac{n(n-2)}{n+1}x^{n-1}$$
where $-1 <x <1$
is there a convinient function that sums the above series?
(unsure but this may be an expanded taylor series?)

Write that as
$$\sum\limits_{n = 0}^\infty  {\frac{{{n^2}}}{{n + 1}}{x^{n - 1}}}  - \sum\limits_{n = 0}^\infty  {\frac{{2n}}{{n + 1}}{x^{n - 1}}} $$
Now think about primitives and derivatives.
$$\eqalign{
  & \sum\limits_{n = 0}^\infty  {\frac{2}{{n + 1}}n{x^{n - 1}}}  = f'\left( x \right) = \frac{d}{{dx}}\left[ {\sum\limits_{n = 0}^\infty  {\frac{2}{{n + 1}}{x^n}} } \right]  \cr 
  & \sum\limits_{n = 0}^\infty  {\frac{n}{{n + 1}}n{x^{n - 1}}}  = g'\left( x \right) = \frac{d}{{dx}}\left[ {\sum\limits_{n = 0}^\infty  {\frac{n}{{n + 1}}{x^n}} } \right] \cr} $$
and $$\sum\limits_{n = 0}^\infty  {\frac{n}{{n + 1}}{x^n}}  = \sum\limits_{n = 0}^\infty  {\frac{{n + 1 - 1}}{{n + 1}}{x^n}}  = \sum\limits_{n = 0}^\infty  {{x^n}}  - \sum\limits_{n = 0}^\infty  {\frac{{{x^n}}}{{n + 1}}} $$
Now use
$$\sum\limits_{n = 0}^\infty  {\frac{{{x^{n + 1}}}}{{n + 1}}}  =  - \log \left( {1 - x} \right)$$
$$\sum\limits_{n = 0}^\infty  {{x^n}}  = \frac{1}{{1 - x}}$$
