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

Question

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)?

Answer 1

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

Answer 2

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.

Answer 3

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.