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.
Find the sum for $\sum_{n=1}^{\infty}(-1)^{(n+1)}\frac{2n+1}{3^n} $
2
$\begingroup$
sequences-and-series
-
0@TMM : Already did that, consensus varies, and each side has it point. Personally I do not believe that it "must" be some way, but that pro grammatically it should be possible to have different views concise, expanded and even maybe customized. By trying it every now and then I hope to get some responses to guide me and come up with some ideas to propose. I really can't see myself waging a text massaging campaign without feedback. – 2013-04-09
2 Answers
2
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.
-
0Thank you for your help too! – 2011-08-10
0
$\frac1{1+x}=-\sum_{n=1}^\infty(-1)^nx^n$
$\frac x{(1+x)^2}=\sum_{n=1}^\infty(-1)^{n+1}nx^n\tag1$
We then have
$\sum_{n=1}^\infty(-1)^{n+1}\frac{2n+1}{3^n}=2\sum_{n=1}^\infty(-1)^{n+1}n(1/3)^n-\sum_{n=1}^\infty(-1)^n(1/3)^n$
$=\frac{2/3}{(1+(2/3))^2}+\frac1{1+(2/3)}$
$=\frac58$