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

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 ?

• I am very curious whether the downvoter can explain what is wrong with this question ! Dec 12, 2022 at 16:47
• The sequence of coefficients is in the OEIS. Dec 12, 2022 at 17:43
• @jjagmath I guess less than $8\cdot 10^5$ entries :) Unbelievable what we can find in OEIS. Dec 12, 2022 at 17:45

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 $$e^{o(n)}$$.

• Nice (+1) . So probably this is no Liouville-number , right ? Is it at least transcendental ? Dec 12, 2022 at 17:12
• @Peter Intuition dictates this should be a Liouville number, I'm using the criterion: "for all $m\in\Bbb N$, there exist $h_m,k_m\in\Bbb Z$ coprime such that $|\alpha-h_m/k_m|<(k_m)^{-m}$". However I wasn't yet able to exhibit such $(h_m,k_m)$. Dec 12, 2022 at 17:16
• @Peter Oh, and unfortunately $\mu(\alpha)=2$ (if this is indeed the case) gives no information as to whether or not $\alpha$ is transcendental Dec 12, 2022 at 17:35
• Someone claimed in another question where this sum occured that the number is transcendental, maybe we need other methods like Roth's theorem. Dec 12, 2022 at 17:37
• @Peter Yes, if we could prove $\mu(\alpha)>2$ strictly, then Roth tells us $\alpha$ is transcendental. But the converse does not hold. And with regards to your question, $\mu(\alpha)>2$ iff. the continued fraction has asymptotically very large peaks Dec 12, 2022 at 17:41

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 $$6$$.

• Wow (+1 and accept). Is it also mentioned whether it is a transcendental number ? Dec 12, 2022 at 18:29
• The very first page refers to $u\ge3$. It may be that this does not apply to $u=2$ Dec 12, 2022 at 21:04
• @FShrike Yes, but the sum begins at $k=0$. So the problem in this question is equivalent to $u=4$ in the article: $$\sum_{j=1}^{\infty} \frac{1}{2^{2^j}} = \sum_{k=0}^{\infty} \frac{1}{4^{2^j}}$$ Dec 12, 2022 at 22:13