I have been trying to find the sum $\sum_{n=1}^{\infty}(-1)^{(n+1)}\frac{2n+1}{3^n} $. After some calculation, I got here: $\frac{-6}{8}+\frac{1}{4}+8\sum\frac{k}{9^k}$. I know the result is $\frac{5}{8}$ , and I verified it with Wolfram Alpha.I saw that $\sum_{k=1}^{\infty}\frac{k}{9^k}=\frac{9}{64}$. But I don't know how to prove the last equation: $\sum_{k=1}^{\infty}\frac{k}{9^k}=\frac{9}{64}$. I hope someone could help me or show me another method to find the sum for my initial series.
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You might be interested in the polylogarithm which is namely: $$\operatorname{Li}_s(z) = \sum_{k=1}^\infty {z^k \over k^s}.$$ You are looking for the special case $s=-1$, $z=\frac{1}{9}$. As illustrated here and this is probably what Prometheus wanted to point you at, you can find the value by derivation. |
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