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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.

Zara
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    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.) – Dave L. Renfro Apr 06 '22 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. – Aphelli Apr 06 '22 at 14:05
  • @FShrike I meant the algebraic closure of $\mathbb{Q}$. I will edit the question to make it clear. – Zara Apr 06 '22 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 $\sum 2^{-n!}$. – Bart Michels Apr 06 '22 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 https://en.wikipedia.org/wiki/Transcendental_number#Numbers_proven_to_be_transcendental – GEdgar Apr 06 '22 at 17:10
  • @BartMichels It was given to us as a problem in undergraduate algebraic number theory but I don't know the original source. We've only used Liouville's theorem to prove things like this for now. – Zara Apr 06 '22 at 17:47
  • this is an immediate consequence of Roth theorem (more generally if $f(n)>0$ and $\liminf_{n \to \infty}\frac{f(n+1)}{f(n)} >2$ then $\sum 2^{-f(n)}$ is transcendental as a consequence of Roth theorem – Conrad Apr 06 '22 at 18:31
  • Roth's Theorem night be a bit of a sledgehammer here. Have you looked at this nice article ? – dke Apr 14 '22 at 12:00

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