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I was wondering if we know what this sum converges to? How can we show it? It's just an odd looking sum I came across in some work.

It does converge.

$$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{(2n)^{2^m} -1}$$

And/or it's alternating series.

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    $\begingroup$ For $m=1$, the inner sum is $\frac{1}{2}$. For $m=2$, the inner sum is $\frac{1}{2}-\frac{\pi}{8}coth(\frac{\pi}{2})$, and it gets more and more complicated as $m$ increases. $\endgroup$ – vadim123 Mar 1 '18 at 4:17
  • $\begingroup$ Yeah, I've seen the first few cases where m=1. Do we have any ways if evaluating this? Maybe even by switching the order? $\endgroup$ – Ryan Goulden Mar 1 '18 at 4:18
  • $\begingroup$ Reversing the order of summation, Alpha can't find the sum even for $n=1$; estimate is $0.404$. It doesn't look good to get a closed form for this double sum. $\endgroup$ – vadim123 Mar 1 '18 at 4:19
  • $\begingroup$ @RyanGoulden Please remember that you can choose an aswer among the given is the OP is solved, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – gimusi Mar 9 '18 at 22:43
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Note that by Cauchy condensation test

$$ 0 \ \leq\ \sum_{n=1}^{\infty} f(n)\ \leq\ \sum_{n=0}^{\infty} 2^{n}f(2^{n})\ \leq\ 2\sum_{n=1}^{\infty} f(n)$$

the series

$$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{(2n)^{2^m} -1}$$

converges if and only if the following converges

$$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m}\frac{1}{(2n)^{m} -1}$$

which diverges since for m=1

$$ \sum_{n=1}^{\infty} \frac{1}{2n -1}$$

diverges.

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Ok, lets assume it converges, I wrote a small code which takes few seconds to evaluate n=10000, and m=10000. the partial sum is

 0.5757556130311484

I will post the code here:

format long e
x=0;

for m=1:10000
    for n=1:10000
        x=x+1/((2*n)^(2^m)-1);
    end
end
x
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  • $\begingroup$ This isn't even close to showing that it converges. For example, the harmonic series is still finite even after 10000 terms, but it diverges. You need to show that it increases more slowly than a sequence which has already been shown to converge, if you're going to try to do things numerically. $\endgroup$ – probably_someone Mar 1 '18 at 4:53
  • $\begingroup$ @probably_someone I didn't mean to show it converges, OP asked what does it converge to, so I assumed it converges, I said it in the beginning. 10000 terms is sufficient to show first few digits of the sum. $\endgroup$ – superman Mar 1 '18 at 4:56
  • $\begingroup$ Then only post the first few digits, not 16. Do you know your margin of error on this partial sum? $\endgroup$ – probably_someone Mar 1 '18 at 4:57
  • $\begingroup$ @probably_someone without exact sum, how is it possible to calculate the error? $\endgroup$ – superman Mar 1 '18 at 5:06
  • $\begingroup$ There are several methods. Some are listed here: tutorial.math.lamar.edu/Classes/CalcII/EstimatingSeries.aspx $\endgroup$ – probably_someone Mar 1 '18 at 5:09

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