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

Question

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?

Answer 1

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.