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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.

TMM
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NumLock
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  • try to use $(x^k)'=kx^{k-1}$ – Ofir Aug 10 '11 at 10:26
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    See http://math.stackexchange.com/questions/50919/calculate-the-sum-of-the-infinite-series-sum-n-0-infty-fracn4n – Hans Lundmark Aug 10 '11 at 10:28
  • Thank you! The link helped me. – NumLock Aug 10 '11 at 10:48
  • @Arjang: Why did you remove part of the title? – TMM Apr 08 '13 at 11:56
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    @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? – jimjim Apr 08 '13 at 18:54
  • @Arjang: If you think titles of such questions should be stripped from all text, I suggest you start a topic about this on meta first, and see what the responses are. – TMM Apr 09 '13 at 08:32
  • @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. – jimjim Apr 09 '13 at 13:17

2 Answers2

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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.

Listing
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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

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

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

$$=\frac58$$

Simply Beautiful Art
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  • If the person doesn't already know the sum of nx^(n-1), I doubt they know the justification needed for differentiating the series term by term. – Anshuman Agrawal Mar 15 '23 at 13:53