The sum $\sum _{n=0}^{\infty}t^{n^2}$ evaluates for $t<1$ to an elliptic theta function, and then taking the limit $t\rightarrow 1$ from below gives $$\lim_{t\nearrow 1}\sqrt{1-t}\sum _{n=0}^{\infty}t^{n^2}=\tfrac{1}{2}\sqrt{\pi}.$$
Alternatively, I can write $t=1-\epsilon$, with $(1-\epsilon)^{n^2}\rightarrow e^{-\epsilon n^2}$ for $\epsilon\rightarrow 0,n\rightarrow\infty$ at constant $\epsilon n^2$, and then approximate the sum by an integral, $$\lim_{t\nearrow 1}\sqrt{1-t}\sum _{n=0}^{\infty}t^{n^2}=\lim_{\epsilon\rightarrow 0}\sqrt{\epsilon}\int_0^\infty e^{-\epsilon x^2}\,dx=\tfrac{1}{2}\sqrt{\pi}.$$