Find the sum for $\sum_{n=1}^{\infty}(-1)^{(n+1)}\frac{2n+1}{3^n} $

Question

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.

Answer 1

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.

Answer 2

$$\frac1{1+x}=-\sum_{n=1}^\infty(-1)^nx^n$$

$$\frac x{(1+x)^2}=\sum_{n=1}^\infty(-1)^{n+1}nx^n\tag1$$

We then have

$$\sum_{n=1}^\infty(-1)^{n+1}\frac{2n+1}{3^n}=2\sum_{n=1}^\infty(-1)^{n+1}n(1/3)^n-\sum_{n=1}^\infty(-1)^n(1/3)^n$$

$$=\frac{2/3}{(1+(2/3))^2}+\frac1{1+(2/3)}$$

$$=\frac58$$