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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.

Ryan Goulden
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    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. – vadim123 Mar 01 '18 at 04:17
  • 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? – Ryan Goulden Mar 01 '18 at 04:18
  • 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. – vadim123 Mar 01 '18 at 04:19
  • @RyanGoulden Please remember that you can choose an aswer among the given is the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Mar 09 '18 at 22:43

2 Answers2

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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.

user
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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
superman
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  • 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. – probably_someone Mar 01 '18 at 04:53
  • @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. – superman Mar 01 '18 at 04:56
  • Then only post the first few digits, not 16. Do you know your margin of error on this partial sum? – probably_someone Mar 01 '18 at 04:57
  • @probably_someone without exact sum, how is it possible to calculate the error? – superman Mar 01 '18 at 05:06
  • There are several methods. Some are listed here: http://tutorial.math.lamar.edu/Classes/CalcII/EstimatingSeries.aspx – probably_someone Mar 01 '18 at 05:09
  • I'm fairly certain this is accurate to 4 places. – Ryan Goulden Mar 01 '18 at 05:17
  • @RyanGoulden yes, I tried it again with n=100000,m=100000, first 4 digits were the same – superman Mar 01 '18 at 05:18
  • The main issue is that it's possible to build sums that diverge arbitrarily slowly (see https://math.stackexchange.com/questions/452053/is-there-a-slowest-rate-of-divergence-of-a-series). How do you know that you won't see a change in the first digit $10^{20}$ terms from now? Without a margin of error, a partial sum is basically meaningless.

    And just to show you that these sums aren't even all that pathological, quoting my link above: $\sum \frac{1}{n\ln(n)(\ln(\ln(n)))^2}$ converges so slowly that it takes $10^{3.14\times 10^{86}}$ terms to get two-digit accuracy.

     – probably_someone Mar 01 '18 at 05:38
  • And, lo and behold, a proof emerges that this sum actually diverges. If you were able to compute many, many more terms, you would see that your answer isn't accurate to four digits after all. – probably_someone Mar 01 '18 at 05:42