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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?

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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. – 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.… – 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
up vote 19 down vote accepted

I have checked with Introduction to Algebraic Independence Theory, where it is mentioned in the preface (p. V) that

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 $0 < |q| < 1$.

The precise references are not given but a little googling turned up the paper by D. Bertrand, "Theta Functions and Transcendence", The Ramanujan Journal, Vol. 1 (1997), pp. 339-350, which seems to be relevant. The second reference is DNNS, "Transcendence of Jacobi's theta series", Proc. Japan Acad. Ser. A Math. Sci., Vol. 72 (1996), pp. 202-203.

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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.

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