Is the maximum entry in the simple continued fraction $6$?

Question

I came across the sum $$S:=\large \sum_{j=1}^{\infty} \frac{1}{2^{2^j}}$$ which I think is a Liouville-number and therefore transcendental (is this correct ?). To clarify : the denominator is $2^{2^j}$ , not $2^{2j}$ as it might look.

I expected large entries in the simple continued fraction of $S$ , but in fact, numerical analysis revealed that the first $8\cdot 10^5$ entries are not larger than $6$. Are all entries in the simple continued fraction of $S$ bounded by $6$ ? If yes, how can we prove it ?

Answer 1

The article "Simple Continued Fractions for Some Irrational Numbers" by Jeffrey Shallit , Theorem 8 applied to this sum stablishes that the only coefficients after the first two are $2$,$4$ and $6$.

Answer 2

Partial answer, assuming that your series is a Liouville number.

Let $\alpha$ be any irrational. It can be shown that: $$\mu(\alpha)=2+\limsup_{n\to\infty}\frac{\ln(a_n)}{\ln(q_{n-1})}$$Where $\mu$ is the irrationality measure, $\alpha=[a_0;a_1,a_2,\cdots]$ as an infinite simple continued fraction (SCF) (this representation is unique) with convergents $p_n/q_n$, indexing these by the recurrence $q_{n+1}=a_{n+1}q_n+q_{n-1}$.

If the entries of the SCF of: $$\alpha=\sum_{j=1}^\infty\frac{1}{2^{2^j}}$$Were bounded (nevermind bounded by $6$, just bounded at all!) then it would follow from this formula that: $$\mu(\alpha)=2$$

If $\alpha$ is a Liouville number, then $\mu(\alpha)=+\infty$, which stands in contradiction. We could go further to say that the $a_n$ asymptotically grow strictly faster than $e^{o(n)}$.