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

Did
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user 1591719
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  • something is funny - typo in the limit? –  Aug 05 '12 at 20:58
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    What does the limit as a approaches 1-0 mean? – Arthur Collé Aug 05 '12 at 20:58
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    It's still a little funny - seems like the limit is just 0, unless $n$ is also going to $\infty$. –  Aug 05 '12 at 21:12
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    I think the Op is rather asking for $\lim_{a \to 1} \sqrt{1-a} \sum_{n=0}^{+\infty} a^{n^2}$. – user10676 Aug 05 '12 at 21:23
  • @Chris'sister As your problem formulated my solution is correct. Maybe you should change problem as user10676 suggested – Norbert Aug 05 '12 at 21:28
  • With the help of Mathematica, I get $\frac{\sqrt{\pi}}2$, obviously taking the $n\to\infty$ limit first. – Fabian Aug 05 '12 at 21:38
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    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 05 '12 at 21:45

2 Answers2

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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 $c\geqslant1$, $$ \lim\limits_{a\to1,a\lt1}(1-a)^{1/c}\cdot\sum\limits_{n=0}^{+\infty}a^{n^c}=\Gamma\left(1+\frac1c\right). $$

Did
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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 $$\lim_{t \to 1^{-}}\sqrt{1 - t}\sum_{k = 0}^{\infty}t^{k^{2}} = \int_{0}^{\infty}e^{-x^{2}}\,dx$$

Paramanand Singh
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