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I am trying to compute the sum

$\sum_{i=0}^{\infty} \frac{i}{2^i}$

which I know should be equal to $2$, but I cannot prove it.

If I am not mistaken, it should be a arithmetico-geometric series (Wikipedia), hence the title.

Any help greatly appreciated!

Martin Sleziak
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Bernd
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  • See http://math.stackexchange.com/questions/92224/summation-of-an-infinite-sequence-sum-limits-k-1-infty-frack2k, http://math.stackexchange.com/questions/441481/why-does-sum-n-0-infty-fracn2n-converge-to-2, http://math.stackexchange.com/questions/87030/proving-sum-limits-i-0n-i-2i-1-n1-2n-1-by-induction and other similar questions. (Have a look at questions linked to those ones.) – Martin Sleziak Apr 18 '14 at 09:38
  • @MartinSleziak Yes, sorry, didn't see that. Should I delete my question? – Bernd Apr 18 '14 at 09:40
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    I don't think you can delete a question, if there is at least one upvoted answer. But even if you could delete it, I don't think it's best thing to do. – Martin Sleziak Apr 18 '14 at 09:42
  • This is a polylogarithm. – Lucian Apr 18 '14 at 14:26

1 Answers1

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Hint

Consider the series $$\sum_{i=0}^{\infty} {i}{x^i}=x \sum_{i=0}^{\infty} {i}{x^{i-1}}=x \frac {d}{dx}\sum_{i=0}^{\infty} {x^{i}}$$ Now you have a geometric series. Compute its sum, take its derivative, multiply by $x$ and replace $x$ by $\frac {1}{2}$.

I am sure that you can take from here.

Claude Leibovici
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