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Is the number \( \displaystyle \sum_{n=1}^{\infty} \frac{1}{2^{n^2}} \) rational?

No. In more detail: \begin{align*}
\displaystyle\mathop{\sum}\limits_{n=1}^{+\infty} \frac{1}{2^{n^2}}&=\frac{1}{2}\bigl({\vartheta_3\bigl({\tfrac{1}{2}}\bigr)-1}\bigr)\\
&=-\frac{1}{2}+\frac{1}{2}\mathop{\prod}\limits_{n=1}^{+\infty}\Bigl({1-\frac{1}{2^{2n}}}\Bigr)\Bigl({1+\frac{1}{2^{2n-1}}}\Bigr)^2\,,
\end{align*} where \[\vartheta_3(z,q)=\displaystyle\mathop{\sum}\limits_{n=-\infty}^{+\infty}{q^{n^2}{\mathrm{e}}^{2\pi{\mathrm{i}}z}}\] is one of the Jacobi theta functions. For \(z=0\) we have \[\vartheta_3(q)=\vartheta_3(z,q)=\displaystyle\mathop{\sum}\limits_{n=-\infty}^{+\infty} q^{n^2}\] and it is proven that for \(|q|<1\) the number \(\vartheta_3(q)\) is transcendental. So, the real number \(\vartheta_3\bigl({\tfrac{1}{2}}\bigr)\) is transcendental and because all rational numbers are algebraic, it follows that \(\vartheta_3\bigl({\tfrac{1}{2}}\bigr)\) is irrational. Finally \(\sum_{n=1}^{+\infty}{2^{-n^2}}\) is irrational.

Of course the above is not a direct proof. It comes from the general theory of Jacobi theta functions. I'll be glad to see a direct proof.

Hi Grigoris...here is a second proof.

First of all we see that series is convergent, because it is bounded from above from the geometric series that has ratio \( 1/2 \). Let us assume that the sum is rational \( \frac{a}{b} \) and we will be led to a contradiction. Choose \( n \) such that \( b< 2^n \). Then $$ \frac{a}{b}-\sum_{k=1}^{n}\frac{1}{2^{k^2}}=\sum_{k\geq n+1}\frac{1}{2^{k^2}} $$ However, the sum \( \displaystyle \sum_{k=1}^{n}\frac{1}{2^{k^2}} \) is equal to \( \dfrac{m}{2^{n^2}} \) for some integrer \( n \). Hence:

$$ \frac{a}{b}-\sum_{k=1}^{n }\frac{1}{2^{k^2}}=\frac{a}{b}-\frac{m}{2^{n^2}}>\frac{1}{2^{n^2}b}>\frac{1}{2^{n^2+n}}>\frac{1}{2^{(n+1)^2+1}}=\sum_{k\geq \left ( n+1 \right )^2}\frac{1}{2^k}>\sum_{k\geq n+1}\frac{1}{2^{k^2}} $$

leading us to a contradiction. Thus the series converges to an irrational number.

In fact the number is transcedental.
The above is the official solution given to the problem by Titu Andreescu.

Actually, there is a much simpler solution.

It is well known that a number written as a decimal is rational if and only if it is a repeating decimal. (We view terminating decimals as repeating.) The same result holds if instead of a decimal we write the number in base 2 or in fact any other base.

The particular number is evidently non-repeating in base 2. So it is not rational. Q.E.D.