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The series is convergent for $|x|<1$ and divergent for $|x|>1$.

I can't find the sum. Integrating three times gives $$\frac{n^2}{(n^2-1)(n+3)(n+2)^2}x^{n+3}$$

that should have a closed form.

How to find the sum of this series?

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You can write $$ \frac{n^2}{(n-1)(n+2)}=1+\frac{-n+2}{(n-1)(n+2)}= 1+\frac{1}{3}\frac{1}{n-1}-\frac{4}{3}\frac{1}{n+2} $$ with partial fraction decomposition. So you reduce the problem to finding the sum of $$ \sum_{n=2}^{\infty}x^n, \quad \sum_{n=2}^{\infty}\frac{x^n}{n-1}, \quad \sum_{n=2}^{\infty}\frac{x^n}{n+2} $$

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  • $\begingroup$ This is much more efficient than the method I was going to try. $\endgroup$ – Lubin Jan 11 '16 at 1:31

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