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Inspired by this question, the series $\dfrac{1}{2^{2^{0}}}+\dfrac{1}{2^{2^{1}}}+\dfrac{1}{2^{2^{2}}}+\dfrac{1}{2^{2^{3}}}+\dots$ is clearly irrational.

But is it algebraic or transcendental?

I was thinking of answering this question by checking whether or not it can be represented as a periodic continued fraction:

  • If no, then (as far as I know) it is transcendental
  • If yes, then (as far as I know) we cannot infer the answer

But how do I determine whether or not it can be represented as a periodic continued fraction?

Is there a better way for tackling this question, or is the answer already known by any chance?

UPDATE:

Based on @Wojowu's comment:

  • If it can be represented as a periodic continued fraction, then it is algebraic
  • If it cannot be represented as a periodic continued fraction, then we cannot infer the answer
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barak manos
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    Continued fraction seems to link to the "Ask Question" page. Is there any reason for this? – Element118 Dec 27 '15 at 10:11
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    Aperiodicity of continued fraction would only tell you the number is not a quadratic irrational (it could still be algebraic). Periodicity would tell you it's a quadratic irrational, hence algebraic. – Wojowu Dec 27 '15 at 10:12
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    This is relevant: http://mathoverflow.net/a/24197/30186 – Wojowu Dec 27 '15 at 10:14
  • @Element118: Ooops, sorry. Thanks :) – barak manos Dec 27 '15 at 10:18
  • @Wojowu: Wait, I thought $e$ has a periodic CF, yet it is transcendental... Which part did I get wrong? – barak manos Dec 27 '15 at 10:19
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    CF of $e$ isn't periodic. It has the form, iirc, $[2;1,2,1,1,4,1,1,6,...]$ in which arbitrarily high coefficients appear. – Wojowu Dec 27 '15 at 10:20
  • @Wojowu: Oh, right. I guess I've mixed "periodicity" with "an observable pattern" (which, for example, $\pi$ is not known to have as far as I know). – barak manos Dec 27 '15 at 10:21
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    @Wojowu: Regarding your "this is relevant" comment above - doesn't this link essentially answers my question? – barak manos Dec 27 '15 at 10:27
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    It does :) ${}{}{}$ – Wojowu Dec 27 '15 at 10:27
  • Periodic continued fractions represent irrational quadratic numbers (elements of $\mathbb{Q}(\sqrt{a})$ with $a\in \mathbb{Q}$ that is not a perfect square). Unless you found some clue that might indicate it is quadratic I really doubt testing for such a small family of algebraic numbers is going to be of any help. – Darth Geek Dec 27 '15 at 10:33

1 Answers1

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It is a general theorem proven by Mahler that given an integer $d>1$ and a nonzero algebraic number $z\in(-1,1)$ then the sum of the series $\sum_{n=0}^\infty z^{d^n}$ is a transcendental number. I can't find a direct reference, but you can find this theorem in this MO answer.

Transcendence of the number in your question follows by taking $d=2,z=\frac{1}{2}$.

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