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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?)

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somehow the range of x isnt displaying clearly: x is greater than -1 and less than 1 – mello Sep 6 '12 at 23:14

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

$$\sum\limits_{n = 0}^\infty {\frac{{{x^{n + 1}}}}{{n + 1}}} = - \log \left( {1 - x} \right)$$

$$\sum\limits_{n = 0}^\infty {{x^n}} = \frac{1}{{1 - x}}$$

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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 $\sum x^n$.

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