# Proving that $\sum_{n = 1}^\infty 2^{-4^n}$ is transcendental

I want to prove that the following number is transcendental: $$\alpha = \sum_{n = 1}^\infty 2^{-4^n}$$

Liouville's theorem states that if $$\alpha$$ is algebraic with minimal polynomial of degree $$d$$, then $$\exists c > 0$$ such that $$\forall \frac{p}{q} \in \mathbb{Q}$$ we have $$|\alpha - \frac{p}{q}| > \frac{c}{q^d}$$.

Here's my attempt: Assume $$\alpha \in \overline{\mathbb{Q}}$$ where $$\overline{\mathbb{Q}}$$ is the algebraic closure of $$\mathbb{Q}$$. I will look at $$\alpha$$ in base $$2$$ for simplicity. We will then have $$\alpha = 0.00.1010001...$$.
I defined $$\frac{p_r}{q_r} = \sum_{n = 1}^r2^{-4^n}$$. Then $$|\alpha - \frac{p}{q}| = \sum_{n = r + 1}^\infty 2^{-4^n} < 2\times2^{-4^{(r + 1)}} = \frac{2}{(2^{4^{r}})^4} = \frac{2}{q_r^4}$$. Now by Liouville's theorem, we have: $$\frac{c}{q_r^d} < |\alpha - \frac{p_r}{q_r}| < \frac{2}{q_r^4}$$ I don't know how to derive a contradiction from this. (If I haven't made any mistakes of course.) I'm also not sure if this approach works.

• Of possible relevance is my answer (including the references) to Does every normal number have irrationality measure $2$? @FShrike -- I'm pretty sure $\overline{\Bbb Q}$ denotes the (presumably real) algebraic closure of the rationals. The notation is fairly common, but probably "algebraic closure" should have prefaced its use. (For this reason, I often include "topological closure" when using this notation in MSE, at least when the context is not utterly clear to a casual reader -- example.) Commented Apr 6, 2022 at 13:31
• Perhaps instead you might want to look up Roth’s theorem, which is a huge improvement on Liouville’s… but is of course a lot harder to prove. Commented Apr 6, 2022 at 14:05
• @FShrike I meant the algebraic closure of $\mathbb{Q}$. I will edit the question to make it clear.
– Zara
Commented Apr 6, 2022 at 16:38
• Did you come up with this question or did you find it somewhere? This is likely quite hard, otherwise Liouville would have given this as an example of a transcendental number, rather than say $\sum 2^{-n!}$. Commented Apr 6, 2022 at 16:54
• I agree with @BartMichels ... You cannot prove $\alpha$ is transcendental using Liouville's theorem in this obvious way. See similar numbers in the list at en.wikipedia.org/wiki/… Commented Apr 6, 2022 at 17:10