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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 $$\lim_{t\nearrow 1}\sqrt{1-t}\sum _{n=0}^{\infty}t^{n^2}=\tfrac{1}{2}\sqrt{\pi}.$$