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?

10$\begingroup$ I asked my mother if she knows that $\sum_{k=0}^\infty{\frac{1}{2^{2^k}}}$ is transcendental. She didn't know. You lied. $\endgroup$– Red BananaSep 22 '16 at 3:46

3$\begingroup$ may be she lied $\endgroup$– userdedSep 23 '16 at 11:43
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.imjprg.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, SpringerVerlag, Berlin, 2001, Chapter 3.

3$\begingroup$ I don't know if it is courageous to answer a question like "Is *** algebraic or not" with "yes it is"  mathematically the answer is correct, plus it's a nice Beatles song. $\endgroup$ Sep 21 '16 at 22:14
I can't classify this question as a duplicate, but Is this number already known to be transcendental? Is there a survey about uptodate trascendence results?
asks the same thing.
Introduction to algebraic Independence Theory, this number is transcendental.