# Irrationality of $\sum\limits_{n=1}^{\infty} r^{-n^{2}}$ for every integer $r > 1$

In the preface to Introduction to Algebraic Independence Theory Yuri V. Nesterenko mentions the series $$f(r) = \sum_{n=1}^{\infty} \frac {1}{r^{n^{2}}}$$ which was introduced as an example by Joseph Liouville in 1851, who proved that $f(r)$ is irrational for all integers $r>1$.

It appears that the proof is elementary like Liouville's proofs for irrationality of $e^{2}$ and $e^{4}$ discussed in my blog posts. Is there any simple way to prove the irrationality of $f(r)$? Or perhaps a reference regarding Liouville's proof?

• Written in basis $r$, this number has a non periodic expansion (for example, because it has infinitely many ones, separated by portions of zeroes of increasing lengthes), hence it cannot be rational. Recall that the expansion of every rational number in any basis is ultimately periodic. – Did Feb 4 '17 at 11:03
• @Did: it appears that I was perhaps seduced by some non-trivial method of Liouville (see the linked blog posts) that I failed to see the simple solution offered in your comment. I don't know what more to say. – Paramanand Singh Feb 4 '17 at 11:40

## 1 Answer

I found this proof which sounds like something Liouville would have done. Let:

$$\mathcal{L}=\sum_{h=1}^\infty r^{-h^2}$$ $$\frac p{r^{n^2}} = \sum_{h=1}^n r^{-h^2}$$ $$r^{-(x-1)^2}\geq r^{\lfloor x \rfloor ^2}$$ $$\int_n^\infty r^{-(x-1)^2}dx\geq\int_n^\infty r^{\lfloor x \rfloor ^2}dx=\sum_{h=n}^\infty r^{-h^2}$$ $$\ln(r)^{-1/2}\int_{n\ln(r)^{1/2}}^\infty e^{-y^2}dy\geq \sum_{h=n+1}^\infty r^{-h^2}=\mathcal{L}-\frac p{r^{n^2}}$$ This limit is due to wolfram alpha: $$\lim_{n\to\infty}r^{n^2}\int_{n\ln(r)^{1/2}}^\infty e^{-y^2}dy=\lim_{x\to\infty}e^{x^2}\int_x^\infty e^{-y^2}dy=0$$ Then $$r^{n^2}\left(\mathcal L -\frac p{r^{n^2}}\right)\leq \ln(r)^{-1/2}r^{n^2}\int_{n\ln(r)^{1/2}}^\infty e^{-y^2}dy=\epsilon$$ Where $\epsilon$ can be made arbitrarily small. Then $$0<\mathcal L -\frac p{r^{n^2}}\leq\frac \epsilon {r^{n^2}}$$ Let $r^{n^2}=q$. If $\mathcal L$ were rational, say $\frac ab$ $$0<\frac ab-\frac pq=\frac{aq-bp}{bq}\leq \frac\epsilon q$$ $$aq-bp>0$$ $$aq-bp\leq\epsilon b$$ The LHS is a positive integer, and the RHS can be made arbitrarily small. Contradiction.