Is this number already known to be transcendental? Is there a survey about up-to-date trascendence results?

Is the number $\sum_{n=1}^\infty \frac{1}{2^{n^2}}$ known to be transcendental?

Is there a survey with up-to-date transcendence results?

• Would you please tell us what is your motivations?Thanks. – awllower Feb 14 '11 at 11:14
• I'm writing a paper in which I show that a variation of the number above is an l^2-Betti number of something (cf. arxiv.org/abs/1004.2030) – Łukasz Grabowski Feb 14 '11 at 12:44
• See also this answer of mine for a proof that this and other similar numbers are transcendental, and some related links. mathoverflow.net/questions/41609/… – George Lowther Feb 14 '11 at 15:18
• Sorry, made a mistake there. This series doesn't converge nearly fast enough for the results in the other question to apply. – George Lowther Feb 14 '11 at 16:11
• @George: I think you confuse the number from my question with $\sum \frac{1}{2^{2^n}}$. – Łukasz Grabowski Feb 14 '11 at 16:18

D. Bertrand and independently D. Duverney, Ke. Nishioka, Ku. Nishioka, I. Shiokawa (DNNS) deduced results on algebraic independence of the values of theta-functions at algebraic points and in particular derived the transcendence of the sums $\sum_{n=1}^\infty q^{n^2}$ for any algebraic $q$ satisfying $0 < |q| < 1$.
There is a family of functions called theta-functions. One of them - I think the standard notation for it might be $\theta_3$ - is given by $\theta_3(z)=\sum z^{n^2}$, so (modulo any mistakes in the definition I've given) your number is $\theta_3(1/2)$. Now the theta-functions are very well-studied, and I suspect there is a lot of information out there about the transcendence of their values at rational arguments. So I've given you a keyword to aid your searches.