# Compute $\lim_{a \to 1^{-}} \sqrt{1-a} \sum_{n=0}^{+\infty} a^{n^2}$

I need some suggestions, hints for the following limit:

$$\lim_{a \to 1^{-}} \sqrt{1-a} \sum_{n=0}^{+\infty} a^{n^2}$$

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something is funny - typo in the limit? –  user29743 Aug 5 '12 at 20:58
What does the limit as a approaches 1-0 mean? –  Arthur Collé Aug 5 '12 at 20:58
It's still a little funny - seems like the limit is just 0, unless $n$ is also going to $\infty$. –  user29743 Aug 5 '12 at 21:12
I think the Op is rather asking for $\lim_{a \to 1} \sqrt{1-a} \sum_{n=0}^{+\infty} a^{n^2}$. –  user10676 Aug 5 '12 at 21:23
Remark : using the fact that $\int e^{-t^2} dt = \sqrt{\pi}/2$, we get $\int_0^\infty a^{t^2} dt = \sqrt{\pi}/2\sqrt{-\log(a)} \sim \frac{\sqrt{\pi}}{2\sqrt{1-a}}$. –  user10676 Aug 5 '12 at 21:45
Note that $\frac1{\sqrt{1-a}}=\sum\limits_{k=0}^{+\infty}c_ka^k$ with $c_k=\frac1{4^k}{2k\choose k}\sim\frac1{\sqrt{\pi k}}$. Since $k\mapsto a^k$ is decreasing, $$b_{i,j}\cdot a^j\leqslant\sum_{k=i}^jc_ka^k\leqslant b_{i,j}\cdot a^i,\qquad b_{i,j}=c_i+\cdots+c_j.$$ Using this for $i=n^2+1$ and $j=(n+1)^2$ with $n\to\infty$, one gets $b_{i,j}\sim\frac2{\sqrt\pi}$. These estimates can be made rigorous to show that $$\frac1{\sqrt{1-a}}\sim\frac2{\sqrt\pi}\cdot\sum\limits_{n=0}^{+\infty}a^{n^2},$$ hence $$\lim\limits_{a\to1,a\lt1}\sqrt{1-a}\cdot\sum\limits_{n=0}^{+\infty}a^{n^2}=\frac{\sqrt\pi}2.$$ The same method shows more generally that, for every $c\geqslant1$, $$\lim\limits_{a\to1,a\lt1}(1-a)^{1/c}\cdot\sum\limits_{n=0}^{+\infty}a^{n^c}=\Gamma\left(1+\frac1c\right).$$