What is $\sum_{n=1}^{\infty}\frac{1}{2^{n^2}}$ equal to? So we know that 
$$
\displaystyle\sum_{n=1}^{\infty} \frac{1}{2^n} = 1
$$
Is there any information on what 
$$
\displaystyle\sum_{n=1}^{\infty} \frac{1}{2^{n^2}}
$$
equals?
 A: HINT: Try writing out the first few partial sums in base 2 . . .

If you're not satisfied by this description of the result, note that it's not hard (EDIT: Whoops, this is bonkers! Hey, it was late, I was tired. :P See below.) to show that the number is transcendental, ruling out any nice description in terms of polynomials. With more effort you can rule out other nice kinds of description. I don't think this number does have a snappy description, other than the above.
A: Maybe it's interesting to see that sums like this are described by the Jacobi theta functions. Recalling that we have $$\theta_{3}\left(q\right)=\sum_{n=-\infty}^{\infty}q^{n^{2}},\,\left|q\right|<1$$ in our case we have, by symmetry, $$\sum_{n=1}^{\infty}\frac{1}{2^{n^{2}}}=\frac{1}{2}\left(\sum_{n=-\infty}^{\infty}\frac{1}{2^{n^{2}}}-1\right)=\frac{1}{2}\theta_{3}\left(\frac{1}{2}\right)-\frac{1}{2}\approx0.56445.$$
A: For what it's worth, this number expressible in terms of one of the theta funcitons.
$$
\vartheta_3(0,1/2)=1+2\sum_{n=1}^\infty\frac{1}{2^{n^2}}\text{.}
$$
So your number is $$\frac{\vartheta_3(0,1/2)-1}{2}\text{.}$$
