what's the limit of
$\sqrt{1-t}\sum \_{n=0}^{\infty}t^{n^2}$ as $t$ goes to the left of $1$? i.e. $t\to 1^{-}$? I tried several times but failed. Here is my thought:


This is a $0\cdot\infty$ problem, so I just tried with $\frac{\sum \_{n=0}^\infty t^{n^2}}{\frac{1}{\sqrt{1-t}}}$. Then it becomes a $\frac{\infty}{\infty}$ one which we could apply L'hospital rules. But it seems that the $n$-order derivative of $\frac{1}{\sqrt{1-t}}$ is always $\infty$ for $t\to 1^{-}$...so what else can I do? Thanks!



Solving a limit about sum of series
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,
$$\lim\_{t\nearrow 1}\sqrt{1-t}\sum \_{n=0}^{\infty}t^{n^2}=\lim\_{\epsilon\rightarrow 0}\sqrt{\epsilon}\int\_0^\infty e^{-\epsilon x^2}\,dx=\tfrac{1}{2}\sqrt{\pi}.$$


