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Everybody knows that $\sum_{k=0}^\infty{\frac{1}{2^{2^k}}}$ is transcendental. Is number $\sum_{k=0}^\infty{\frac{1}{2^{k^2}}}$ algebraic or not?
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Everybody knows that $\sum_{k=0}^\infty{\frac{1}{2^{2^k}}}$ is transcendental. Is number $\sum_{k=0}^\infty{\frac{1}{2^{k^2}}}$ algebraic or not?
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Yes, it is, and even the three numbers $\sum_{k \geq 0}{2^{-k^2}}$, $\sum_{k \geq 1}{k^22^{-k^2}}$ and $\sum_{k \geq 1}{k^42^{-k^2}}$ are algebraically independent. This results from algebraic independence results for theta functions, see Waldschmidt's excellent survey https://webusers.imj-prg.fr/~michel.waldschmidt/articles/pdf/SurveyTrdceEllipt2006.pdf, Corollary 52, and the reference given there: Nesterenko & Philippon, Introduction to algebraic independence theory, Lecture Notes in Mathematics, vol. 1752, Springer-Verlag, Berlin, 2001, Chapter 3.
I can't classify this question as a duplicate, but Is this number already known to be transcendental? Is there a survey about up-to-date trascendence results?
asks the same thing.
Introduction to algebraic Independence Theory, this number is transcendental.