TITLE: How to compute the series $\sum_{n=0}^\infty q^{n^2}$?
QUESTION [25 upvotes]: Let $q\in (0,1)$. Is there a way of computing the series
$$
\sum_{n=0}^\infty q^{n^2}
$$
explicitly? Is there at least a nice accurate estimate?
All I could get is the estimate
$$\sqrt{\frac{\pi}{4\cdot\mathrm{ln}\frac{1}{q}}}\leq\sum_{n=0}^\infty q^{n^2}\leq 1+\sqrt{\frac{\pi}{4\cdot\mathrm{ln}\frac{1}{q}}}$$
via integration (quite possibly flawed).
For $q=\frac{1}{2}$, Maple gives the values
$$
1.064467020\leq 1.564468414\leq 2.064467020,
$$
showing that my estimate is not very precise. (Of course, the sum of the two errors will always be $1$. Here both errors are coincidentally almost exactly $\frac{1}{2}$.)

REPLY [18 votes]: As Qiaochu says, the Euler-Maclaurin formula is useful here.  It gives the error in the approximation 
$$\sum_{n=0}^{\infty} q^{n^2} \approx \sqrt{\frac{\pi}{4 \log (1/q)}} + \frac{1}{2}$$
to be exactly 
$$ \lim_{a \to \infty}  \int_0^a 2xq^{x^2} (\log q) \left(x - \lfloor x \rfloor - \frac{1}{2}\right) dx.$$
Since $q^{x^2} \to 0$ quite rapidly, the value of this integral can be approximated closely by using small values of $a$.
For example, if $q = 1/2$, the integral (i.e., the error in the approximation) evaluates to $1.39417 \times 10^{-6}$ (via Mathematica).