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 $0*\infty$ 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!

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!