Does anyone know of a closed-form formula for the sum $\sum_{n = 1}^\infty x^{2^n-1}$? We can assume that $0<x<1$.
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possible duplicate of Evaluating the sum of geometric series – Amr Jun 12 '13 at 20:02
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5@Amr, no, this is not a duplicate of that. This has $x^{2^n}$, not $x^n$ like a geometric series. – George V. Williams Jun 12 '13 at 20:03
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I think I saw a question extremely similar to this on this site (the exponent was $2^n$ instead of $2^n - 1$). The general consensus was that it is unlikely there is a closed form. – George V. Williams Jun 12 '13 at 20:04
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@GeorgeV.Williams OH. I thought that it had $x^{2n-1}$ and not $x^{2^n-1}$ – Amr Jun 12 '13 at 20:04
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For most $x$ (especially those much smaller than $1$) you can approximate that to very good accuracy because the tail is so small. – davin Jun 12 '13 at 20:05
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1This question is very similar to: Power Series Formula. However, this question should NOT be closed because it asks for an infinite sum, whereas the other one requests a partial sum. – George V. Williams Jun 12 '13 at 20:10
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2This is known as a lacunary function. I do not know much about them, but I understand they can exhibit very complicated behavior, so I would not expect a simple formula to be possible. – Jair Taylor Jun 13 '13 at 16:47
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Further discussion of this topic is available here (except that topic refers to partial sums). However, I did give an estimate in this answer, which gives:
$$ -\frac{\mathrm{Ei}\left(\log (x)\right)}{x\log (2)} \le \sum_{i=0}^\infty x^{2^i - 1} \le -\frac{\mathrm{Ei}\left(\frac{\log (x)}{2}\right)}{x\log (2)} $$
Where $\mathrm{Ei}(x)$ is the Exponential Integral.
George V. Williams
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Hmm... Seems like there's no hope :) Approximations are good, but not for what I want to do. Appreciate the comment though! – Spark Jun 13 '13 at 08:42