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?)
 A: 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
$$\sum\limits_{n = 0}^\infty  {\frac{{{x^{n + 1}}}}{{n + 1}}}  =  - \log \left( {1 - x} \right)$$
$$\sum\limits_{n = 0}^\infty  {{x^n}}  = \frac{1}{{1 - x}}$$
A: Note that $${n^2-2n\over n+1}=n-3+{3\over n+1}$$ by long division of polynomials, so your sum is $$\sum nx^{n-1}-3\sum x^{n-1}+3\sum{x^{n-1}\over n+1}$$ The third sum is $${3\over x^2}\sum {x^{n+1}\over n+1}$$ which you should recognize from its relation to $\log(1-x)$ (as in Peter's solution). The second sum is a geometric series, and the first sum is the derivative of the geometric series $\sum x^n$. 
