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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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  • $\begingroup$ try to use $(x^k)'=kx^{k-1}$ $\endgroup$ – Ofir Aug 10 '11 at 10:26
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    $\begingroup$ See math.stackexchange.com/questions/50919/… $\endgroup$ – Hans Lundmark Aug 10 '11 at 10:28
  • $\begingroup$ Thank you! The link helped me. $\endgroup$ – NumLock Aug 10 '11 at 10:48
  • $\begingroup$ @Arjang: Why did you remove part of the title? $\endgroup$ – TMM Apr 8 '13 at 11:56
  • $\begingroup$ @YMM : Looking at the Linked and related links on right hand side, things like "EValuate ", "How to Find", "What is the the value of", "help me find". ... are totally redundant. Look at the linked "Calculate the sum of the infinite series ..." the main thing is last part. How often do we want to see repeated non information? $\endgroup$ – Arjang Apr 8 '13 at 18:54
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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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  • $\begingroup$ Thank you for your help too! $\endgroup$ – NumLock Aug 10 '11 at 10:48
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$$\frac1{1+x}=-\sum_{n=1}^\infty(-1)^nx^n$$

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

We then have

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

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

$$=\frac58$$

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