# Compute the limit of $\sqrt{1-a}\sum\limits_{n=0}^{+\infty} a^{n^2}$ when $a\to1^-$

I need some suggestions, hints for the limit when $a \to 1^{-}$ of $$\sqrt{\,1 - a\,}\,\sum_{n = 0}^{\infty}a^{n^{2}}.$$

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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 $c\geqslant1$, $$\lim\limits_{a\to1,a\lt1}(1-a)^{1/c}\cdot\sum\limits_{n=0}^{+\infty}a^{n^c}=\Gamma\left(1+\frac1c\right).$$
Would you similarly proceed for $$\lim_{x\to 1^{-}} (1-x) \sum_{n=1}^{\infty} 2^n x^{2^n}$$? –  Chris's sis Sep 11 '14 at 20:45
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 $$\lim_{t \to 1^{-}}\sqrt{1 - t}\sum_{k = 0}^{\infty}t^{k^{2}} = \int_{0}^{\infty}e^{-x^{2}}\,dx$$