Find ∑∞k=112k+1−1\sum_{k=1}^{\infty}\frac{1}{2^{k+1}-1}

Calculate ∞∑k=112k+1−1.\sum_{k=1}^{\infty}\frac{1}{2^{k+1}-1}.

I used Wolfram|Alpha to compute it and got it to be approximately equal to 0.60.6. How to compute it? Can someone give me a hint or a suggestion to do it?

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@JackD’Aurizio: While I did NOT downvote, perhaps it was because of not “answering” the question. In your defense, however, I feel like you made the right decision to post it as an answer instead of a long comment. Anyways, it seems like good work. +11 🙂
– Clayton
2 days ago

  

 

About an explicit form, that is quite hopeless. It is not difficult to prove that SS is an irrational number (the infinitude of primes has a precise effect on the binary representation of SS), and probably it is a trascendental number, too.
– Jack D’Aurizio
2 days ago

  

 

@Jack D’Aurizio I am very surprised also by the downvote to your excellent answer. There is a general problem with downvoting freed on SE. Downvoters should not be anonymous. Besides, why have you deleted your answer ?
– JeanMarie
2 days ago

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@Clayton: it looks to me that I actually provided a way for computing seven figures. I do not care about the silly downvote, it is enough the OP is aware of the contents of my comment.
– Jack D’Aurizio
2 days ago

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@JeanMarie: the principle of anonymity of upvotes/downvotes cannot be violated, in order to prevent voting wars. But I think that anonymous explanations should be made compulsory. I do not know the reason behind the downvote, such phenomenon occurs to me quite often since a while. (I undeleted my answer, by the way)
– Jack D’Aurizio
2 days ago

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2 Answers
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Maybe it is interesting to see that there is a “closed formâ€‌ of this series in terms of the qq-polygamma function. We have that S=∑k≥112k+1−1=−1−∑k≥111−2kS=\sum_{k\geq1}\frac{1}{2^{k+1}-1}=-1-\sum_{k\geq1}\frac{1}{1-2^{k}}
and recalling the definition of the qq-polygamma function we have S=−1−ψ1/2(1)+log(1/2)log(2)≈0.60669515.S=\color{red}{-1-\frac{\psi_{1/2}\left(1\right)+\log\left(1/2\right)}{\log\left(2\right)}}\approx0.60669515.

  

 

nice observation, +1
– Bacon
2 days ago

It is a fast-convergent series, and a Lambert series, too, since

S=∑k≥112k+1−1=−1+∑k≥112k−1=−1+∑k≥1∑m≥112mk=−1+∑n≥1d(n)2n S=\sum_{k\geq 1}\frac{1}{2^{k+1}-1}=-1+\sum_{k\geq 1}\frac{1}{2^k-1}=-1+\sum_{k\geq 1}\sum_{m\geq 1}\frac{1}{2^{mk}}=-1+\sum_{n\geq 1}\frac{d(n)}{2^n}
where d(n)d(n) is the number of divisors of nn. Since d(n)≤nd(n)\leq n (this is a very crude bound)
S+1−N∑n=1d(n)2n≤∑n>Nn2n=N+22N S+1-\sum_{n=1}^{N}\frac{d(n)}{2^n}\leq \sum_{n>N}\frac{n}{2^n}=\frac{N+2}{2^N}
hence by choosing N=30N=30 we get that
−1+30∑n=1d(n)2n=0.606695149… -1+\sum_{n=1}^{30}\frac{d(n)}{2^n} = \color{red}{0.6066951}49\ldots
is an extremely good approximation of SS, with the correct red digits.

As suggested by Yves Daoust, another good strategy comes from noticing that 12k−1\frac{1}{2^k-1} is pretty close to 12k\frac{1}{2^k} if kk is large, hence we may perform a series acceleration in the following way:

S=∑k≥212k−1=122−1+∑k≥312k+∑k≥312k(2k−1) S=\sum_{k\geq 2}\frac{1}{2^k-1}=\frac{1}{2^2-1}+\sum_{k\geq 3}\frac{1}{2^k}+\sum_{k\geq 3}\frac{1}{2^k(2^k-1)}
turning SS into
S=712+123(23−1)+∑k≥414k+∑k≥414k(2k−1) S = \frac{7}{12}+\frac{1}{2^3(2^3-1)}+\sum_{k\geq 4}\frac{1}{4^k}+\sum_{k\geq 4}\frac{1}{4^k(2^k-1)}
or
S=8151344+144(24−1)+∑k≥518k+∑k≥518k(2k−1) S = \frac{815}{1344}+\frac{1}{4^4(2^4-1)}+\sum_{k\geq 5}\frac{1}{8^k}+\sum_{k\geq 5}\frac{1}{8^k(2^k-1)}
so that SS equals \frac{260927}{430080}\frac{260927}{430080} plus \sum_{k\geq 5}\frac{1}{8^k(2^k-1)}\sum_{k\geq 5}\frac{1}{8^k(2^k-1)}. With just three iterations of this technique we get already S=\color{red}{0.60669}41\ldotsS=\color{red}{0.60669}41\ldots, and and the fourth step we get S\approx \frac{1391613}{2293760}=\color{red}{0.6066951}20\ldotsS\approx \frac{1391613}{2293760}=\color{red}{0.6066951}20\ldots

In a compact form, this acceleration technique leads to:
S = \sum_{k\geq 1}\left(\frac{1}{2^{(k^2-1)}(2^{k+1}-1)}+\frac{1}{2^{(k^2+k)}(2^k-1)}\right) S = \sum_{k\geq 1}\left(\frac{1}{2^{(k^2-1)}(2^{k+1}-1)}+\frac{1}{2^{(k^2+k)}(2^k-1)}\right)
collapsing to:
\boxed{S=\frac{1}{4}+\sum_{k\geq 2}\frac{8^k+1}{(2^k-1)\,2^{k^2+k}}=0.6066951524152917637833\ldots} \boxed{S=\frac{1}{4}+\sum_{k\geq 2}\frac{8^k+1}{(2^k-1)\,2^{k^2+k}}=0.6066951524152917637833\ldots}

with a significant convergence boost. Now the main term of the sum behaves like 2^{-k^2}2^{-k^2} instead of 2^{-k}2^{-k}.

  

 

What is wrong with this answer?
– Jack D’Aurizio
2 days ago

  

 

Explain the downvote.
– Jack D’Aurizio
2 days ago

  

 

I really don’t understand the downvote so (+1).
– Marco Cantarini
2 days ago

  

 

By ignoring the -1-1 at the denominator, the tail of the summation of the initial series after nn terms is close to 2^{1-n}2^{1-n}. This gives a very fast converging estimate (1212 exact decimals with 1919 terms).
– Yves Daoust
2 days ago

  

 

@YvesDaoust: that is a very good remark. For instance, it is not difficult to design a tailor-made acceleration technique for our series based on that. Answer updated, many thanks.
– Jack D’Aurizio
2 days ago