Is the number $\sum\limits_{n=1}^\infty2^{-n^2}$ rational?

I could prove that the series is convergent (as it is bounded above by the geometric series with common ratio $\frac{1}{2}$. But how do I prove that it is rational (or not)?


Assume that the sum is a rational number given by $\frac{p}{q}$ where $p$ and $q$ are coprime. Choose $n$ such that $q < 2^n$. Then we have:

$$\frac{p}{q} - \sum_{k=1}^{n}\frac{1}{2^{k^2}} = \sum_{k \geqslant n+1}\frac{1}{2^{k^2}}$$

But we know that the sum $\sum_{k=1}^{n}\frac{1}{2^{k^2}} = \frac{m}{2^{n^2}}$ for some integer $n$. Therefore,

$$\frac{p}{q}- \sum_{k=1}^{n}\frac{1}{2^{k^2}} =\frac{p}{q}- \frac{m}{2^{n^2}} > \frac{1}{2^{n^2}q} > \frac{1}{2^{n^2 + n}}>\frac{1}{2^{(n+1)^2-1}} = \sum_{k \geqslant(n+1)^2}\frac{1}{2^k} > \sum_{k \geqslant n+1}\frac{1}{2^{k^2}},$$

which is a contradiction! Hence the sum is irrational, infact, I believe it is trancendental.

EDIT: Thanks to Robert Israel for pointing out the mathoverflow link which asserts the transcendence of the number.

Link: https://mathoverflow.net/questions/55397/is-this-number-already-known-to-be-transcendental-is-there-a-survey-about-up-to

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    $\begingroup$ See this Math overflow question for the (not at all obvious) fact that it is transcendental. $\endgroup$ – Robert Israel Apr 5 '17 at 6:43
  • $\begingroup$ But it is not a Liouville number. The error in the $n$'th partial sum is on the order of $e_n \sim 2^{-(n+1)^2}$ while the denominator of that partial sum is $d_n = 2^{n^2}$. Since $e_n$ is much bigger than $1/d_n^2$, it's not Liouville, nor can you use Thue-Siegel-Roth. $\endgroup$ – Robert Israel Apr 5 '17 at 6:49
  • $\begingroup$ Can someone explain the first step,how choosing q to be less than 2^n makes the sum go from k+1 up to infinity? $\endgroup$ – Daniel Evans Apr 5 '17 at 7:13
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    $\begingroup$ There must be a typo in the first equality, as the LHS is deemed to be $0$. The sum should be to $n$. $\endgroup$ – Yves Daoust Apr 5 '17 at 8:23
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    $\begingroup$ It seems that typo is continued in several different places. For example, the sentence "we know that the sum $\sum_{k=1}^{\infty} \frac{1}{2^{k^2}} = \frac{m}{2^{n^2}}$" does not make any sense unless we change to $\sum_{k=1}^n$. $\endgroup$ – Michael Apr 5 '17 at 8:43

A real number is rational if and only if its positional representation (in any base) either terminates, or repeats. (To see this, perform long division of a rational $a/b$ in a given base, and notice that the calculation at step $i$ depends only on the remainder at step $i-1$. For the converse, you can compute the rational number directly by noting that the positional representation is a geometric series.)

Now your series converges to a number which, when written in binary, clearly does not repeat, so is irrational.


No, if it were it would have ultimately periodic exapnsion for every base $b$. Now in binary we have the binary expansion being $1$ at position $n^2$ and $0$ otherwise. This is clearly not periodic.


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