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Let $\displaystyle f(x) = \sum_{i=0}^{n}\dfrac{x^i}{\left(1-x^2\right)^i}$

While solving a problem I came up with this function which requires me to solve this function into a closed form. How do I solve this?

I have tried geometric series.

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$f(x) = \displaystyle \sum_{i=0}^n \left(\dfrac{x}{1-x^2}\right)^i = \dfrac{1-r^{n+1}}{1-r}$, with $r=\dfrac{x}{1-x^2}$

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Put $$u=\frac{x}{1-x^2}$$ Then $$\begin{align} \sum_{i=0}^n\frac{x^i}{(1-x^2)^i}&=\sum_{i=0}^nu^i\\ &=\frac{\ \ \quad 1-u^{n+1}}{1-u}\\ &=\frac{\quad \ \ 1-\left(\frac{x}{1-x^2}\right)^{n+1}}{1-\left(\frac{x}{1-x^2}\right)}\end{align}$$

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  • $\begingroup$ Edited - thanks. $\endgroup$ – hypergeometric Nov 15 '14 at 4:31

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