Solution 1:

As mentioned in the comments, this sum is related to one of the Jacobi theta functions. $$\begin{eqnarray*} \sum_{n=0}^\infty e^{-n^2} &=& \frac{1}{2}\left(1 + \sum_{n=-\infty}^\infty \left(\frac{1}{e}\right)^{n^2}\right) \\ &=& \frac{1}{2}\left[1+\vartheta_3\left(0,\frac{1}{e}\right)\right] \\ &\simeq& 1.386 \end{eqnarray*}$$