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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 ?

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

2 Answers 2

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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 $e^{o(n)}$.

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  • $\begingroup$ Nice (+1) . So probably this is no Liouville-number , right ? Is it at least transcendental ? $\endgroup$
    – Peter
    Dec 12, 2022 at 17:12
  • $\begingroup$ @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 $(h_m,k_m)$. $\endgroup$
    – FShrike
    Dec 12, 2022 at 17:16
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    $\begingroup$ @Peter Oh, and unfortunately $\mu(\alpha)=2$ (if this is indeed the case) gives no information as to whether or not $\alpha$ is transcendental $\endgroup$
    – FShrike
    Dec 12, 2022 at 17:35
  • $\begingroup$ Someone claimed in another question where this sum occured that the number is transcendental, maybe we need other methods like Roth's theorem. $\endgroup$
    – Peter
    Dec 12, 2022 at 17:37
  • $\begingroup$ @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 $\endgroup$
    – FShrike
    Dec 12, 2022 at 17:41
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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 $6$.

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  • $\begingroup$ Wow (+1 and accept). Is it also mentioned whether it is a transcendental number ? $\endgroup$
    – Peter
    Dec 12, 2022 at 18:29
  • $\begingroup$ I provided a link where you can download the article. $\endgroup$
    – jjagmath
    Dec 12, 2022 at 18:30
  • $\begingroup$ The very first page refers to $u\ge3$. It may be that this does not apply to $u=2$ $\endgroup$
    – FShrike
    Dec 12, 2022 at 21:04
  • $\begingroup$ @FShrike Yes, but the sum begins at $k=0$. So the problem in this question is equivalent to $u=4$ in the article: $$\sum_{j=1}^{\infty} \frac{1}{2^{2^j}} = \sum_{k=0}^{\infty} \frac{1}{4^{2^j}}$$ $\endgroup$
    – jjagmath
    Dec 12, 2022 at 22:13

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