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Calculate $\sum\limits_{i=0}^\infty(2^{2^{(-i)}}-1)$

By a Python program showed below, I can calculate that it is about $1.7818$($1.7818386318393353172743493971315840477111345642771002580086952600435098253141880734976160498115058888973$ by a comment). I've worked on this problem for days, but I can't solve it. Can anyone help me?

Update: I'm a student who is interested in math. I came out with this problem about one year ago, but I think it's too difficult for me.

Also, $2^{2^{(-i)}}-1\le2^{(-i)}$, so it is a convergent serie.

Update 2: Why $2^{2^{(-i)}}-1\le2^{(-i)}$?
Let $x$ be $2^{(-i)}$, $i\ge0$, so $x\ge1$, then $2^{2^{(-i)}}-1-2^{(-i)}=2^x-1-x$.
When $x=1$, $2^x-1-x=0$, and $\frac{d(2^x-1-x)}{dx}=2^x\ln-1>0$ for $x\ge1$, so $2^{2^{(-i)}}-1-2^{(-i)}=2^x-1-x>0$, that means $2^{2^{(-i)}}-1\le2^{(-i)}$.

Update 5 on 2021-07-02: Anyone who can show whether the sum is rational can win the bounty.

from math import *
s = 0
a = 2
while a != 1:
    s += a - 1
    a = sqrt(a)
print(s)
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    $\begingroup$ For purposes of context, please provide (1) The source of this question (What made you work on this question? Interest? Assignment? Part of a bigger problem?) (2) If need be, please include the program you have written : if it is small enough, then include it as text, otherwise a link to it would be great. Coming to your question itself, have you been able to prove that it is a convergent sequence, at least? $\endgroup$ Jun 18, 2021 at 4:53
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    $\begingroup$ I can give you more digits: Alpha gives $1.7818386318393353172743493971315840477111345642771002580086952600435098253141880734976160498115058888973$ and you can click on more digits if you want. The fact that it doesn't give a closed form says it is not easy to find one. $\endgroup$ Jun 18, 2021 at 5:15
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    $\begingroup$ @RossMillikan: Can you swear by each of the digits that "Alpha" gives? I doubt it. In any case, I agree that it is unlikely that a closed formula exists. $\endgroup$ Jun 18, 2021 at 5:22
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    $\begingroup$ @g.kov: I don'r doubt that with good multi precision algorithms you can get the 10^10 first digits with just a smartphone. the things is that once convergence has been established for this series (which by the looks of it, converges very fast) a simple double recision calculation in Fortran77 (or Python for that matter) would give you a decent approximation. The OP was asking for closed form, which very likely does not exists in terms of well known functions. So what is the point of producing 50 "significant" digits. $\endgroup$ Jun 18, 2021 at 17:27
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    $\begingroup$ @Alex-Github-Programmer: If you know that the series converges, and in fact converges very fast, numerical methods suffice to get more than descent approximation. Notice the you can bound your series $S$: $2\log(2)\leq S\leq 2\frac{2^{2^0}-1}{2^0}$ the check that you computer estimation is "fine". Your series is almost geometric so convergence is fast. $\endgroup$ Jun 24, 2021 at 2:06

1 Answer 1

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+25
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I don't think there is a closed form for this sum, but it can be rewritten to converge even more quickly $$ \begin{align} \sum_{k=0}^\infty\left(2^{2^{-k}}-1\right) &=\sum_{k=0}^\infty\left(e^{\log(2)\,2^{-k}}-1\right)\tag1\\ &=\sum_{j=1}^\infty\sum_{k=0}^\infty\frac{\left(\log(2)\,2^{-k}\right)^j}{j!}\tag2\\ &=\sum_{j=1}^\infty\frac{\log(2)^j}{j!}\left(1+\frac1{2^j-1}\right)\tag3\\ &=1+\sum_{j=1}^\infty\frac{\log(2)^j}{j!}\frac1{2^j-1}\tag4\\[6pt] &=1.7818386318393353172743493971315840477111\tag5 \end{align} $$ Explanation:
$(1)$: $2=e^{\log(2)}$
$(2)$: use the series for $e^x-1$
$(3)$: sum the geometric series in $k$
$(4)$: recognize the series for $e^{\log(2)}-1=1$
$(5)$: evaluate to $40$ places

The number of digits for $n$ terms of the original series is $$ n\log_{10}(2)+O(1)\sim n\log_{10}(2)\tag6 $$ whereas the number of digits for $n$ terms of the series in $(4)$ is $$ \left(n+\frac12\right)\log_{10}\left(\frac{2n}{e\log(2)}\right)+O(1)\sim n\log_{10}(n)\tag7 $$ and it is easy to see that $(7)$ grows faster than $(6)$.

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  • $\begingroup$ I put this approximation through the Inverse Symbolic Calculator and got nothing. $\endgroup$
    – robjohn
    Jun 26, 2021 at 9:35
  • $\begingroup$ I also found the general formula $$(a-1)+\sum_{j=1}^\infty\frac{\log(a)^j}{j!}\frac1{a^j-1}$$ However, how do you know this representation converges more quickly (than the initial representation)? $\endgroup$
    – NN2
    Jun 26, 2021 at 9:59
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    $\begingroup$ The terms of the initial representation are approximately $\log(2)2^{-j}$, whereas the terms of the series in this answer are approximately $\log(2)^j2^{-j}/j!$. Both the $\log(2)^j$ and the $1/j!$ contribute to the faster convergence (especially the $1/j!$). $\endgroup$
    – robjohn
    Jun 26, 2021 at 10:30
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    $\begingroup$ Just for the fun $$e+\frac 1{1000}\Big[e^{\pi } J_0(1)\Big]^{e^{\frac 1e}}-1=1.7818386$$ in absolute error of $2.16\times 10^{-8}$ $\endgroup$ Jun 26, 2021 at 15:09
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    $\begingroup$ @robjohn It might be worth noting that via Abel-Plana $$\sum\limits_{n=0}^{\infty }{{{2}^{{{2}^{-n}}}}-1}=\frac{1}{2}+\frac{1}{\alpha }\left\{ E{{i}_{1}}\left( \alpha \right)-\gamma -\log \left( \alpha \right) \right\}+2\int\limits_{0}^{\infty }{\frac{{{e}^{\alpha \cos \left( \alpha t \right)}}\sin \left( \alpha \sin \left( \alpha t \right) \right)}{{{e}^{2\pi t}}-1}dt}$$ The integral can be completed via a series which yields your result. Not sure if anything further can be squeezed from this though. $\endgroup$ Jun 28, 2021 at 0:11

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