Is the number rational?
- Tolaso J Kos
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Is the number rational?
Is the number \( \displaystyle \sum_{n=1}^{\infty} \frac{1}{2^{n^2}} \) rational?
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- Grigorios Kostakos
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Re: Is the number 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.
\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.
Grigorios Kostakos
- Tolaso J Kos
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Re: Is the number rational?
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.
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.
Imagination is much more important than knowledge.
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Re: Is the number rational?
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.
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.
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