How does one go about exactly summing this series
$$\sum_{n=1}^{\infty}\frac{(-1)^nn^2}{3^n}$$
Stuck on this and not sure how to proceed. Appreciate any assistance!
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1If the $n^2$ wasn't there, would you know how to solve it? What if we keep the $n$ and remove only the $^2$? – Arthur May 13 '19 at 10:38
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1https://math.stackexchange.com/questions/593996/how-to-prove-sum-n-0-infty-fracn22n-6 – lab bhattacharjee May 13 '19 at 10:41
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$$ \sum\limits^\infty_{n=0} \frac{(-1)^n}{3^n}n^2 = (x\partial_{x})_{|x=1}^2\sum\limits^\infty_{n=0} \frac{(-1)^nx^n}{3^n} = (x\partial_{x})_{|x=1}^2\frac{1}{1+\tfrac{x}{3}} $$
You should be able to handle the rest...
denklo
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