6
$\begingroup$

Does anyone know of a closed-form formula for the sum $\sum_{n = 1}^\infty x^{2^n-1}$? We can assume that $0<x<1$.

Thanks!

$\endgroup$
7
  • $\begingroup$ possible duplicate of Evaluating the sum of geometric series $\endgroup$
    – Amr
    Jun 12, 2013 at 20:02
  • 5
    $\begingroup$ @Amr, no, this is not a duplicate of that. This has $x^{2^n}$, not $x^n$ like a geometric series. $\endgroup$ Jun 12, 2013 at 20:03
  • $\begingroup$ 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. $\endgroup$ Jun 12, 2013 at 20:04
  • 1
    $\begingroup$ This 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. $\endgroup$ Jun 12, 2013 at 20:10
  • 2
    $\begingroup$ This 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. $\endgroup$ Jun 13, 2013 at 16:47

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Hmm... Seems like there's no hope :) Approximations are good, but not for what I want to do. Appreciate the comment though! $\endgroup$
    – Spark
    Jun 13, 2013 at 8:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.