TITLE: Compute the limit of $\sqrt{1-a}\sum\limits_{n=0}^{+\infty} a^{n^2}$ when $a\to1^-$
QUESTION [10 upvotes]: I need some suggestions, hints for the limit when $a \to 1^{-}$ of $$\sqrt{\,1 - a\,}\,\sum_{n = 0}^{\infty}a^{n^{2}}.$$

REPLY [8 votes]: This problem was asked in Mathematical Tripos 1932 and here is an answer based on a hint given in Hardy's Pure Mathematics.
Let $h$ be positive and $n$ be a positive integer. Since $e^{-x^{2}}$ is decreasing it is easy to see that $$\int_{h}^{(n + 1)h}e^{-x^{2}}\,dx < h \sum_{k = 1}^{n}e^{-k^{2}h^{2}} < \int_{0}^{nh}e^{-x^{2}}\,dx$$ If we let $n \to \infty$ we get $$\int_{h}^{\infty}e^{-x^{2}}\,dx \leq h\sum_{k = 1}^{\infty}e^{-k^{2}h^{2}} \leq \int_{0}^{\infty}e^{-x^{2}}\,dx$$ Now we put $t = e^{-h^{2}}$ so that as $h \to 0^{+}$ we have $t \to 1^{-}$ and also note that $h/\sqrt{1 - t} \to 1$ as $h \to 0^{+}$ and thus on taking limits as $h \to 0^{+}$ or as $t \to 1^{-}$ we get $$\lim_{t \to 1^{-}}\sqrt{1 - t}\sum_{k = 1}^{\infty}t^{k^{2}} = \int_{0}^{\infty}e^{-x^{2}}\,dx$$ Note that the question given by OP requires the lowest index in $\sum t^{k^{2}}$ to be $k =  0$ but above we have it as $k = 1$. This does not make any difference as the term corresponding to $k = 0$ is $1$ and hence a term $\sqrt{1 - t}$ gets added up which also tends to $0$. Hence we have $$\lim_{t \to 1^{-}}\sqrt{1 - t}\sum_{k = 0}^{\infty}t^{k^{2}} = \int_{0}^{\infty}e^{-x^{2}}\,dx$$