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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!

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3 Answers 3

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The sum $\sum _{n=0}^{\infty}t^{n^2}$ evaluates for $t<1$ to an elliptic theta function, and then taking the limit $t\rightarrow 1$ from below gives $$\lim_{t\nearrow 1}\sqrt{1-t}\sum _{n=0}^{\infty}t^{n^2}=\tfrac{1}{2}\sqrt{\pi}.$$

Alternatively, I can write $t=1-\epsilon$, with $(1-\epsilon)^{n^2}\rightarrow e^{-\epsilon n^2}$ for $\epsilon\rightarrow 0,n\rightarrow\infty$ at constant $\epsilon n^2$, and then approximate the sum by an integral, $$\lim_{t\nearrow 1}\sqrt{1-t}\sum _{n=0}^{\infty}t^{n^2}=\lim_{\epsilon\rightarrow 0}\sqrt{\epsilon}\int_0^\infty e^{-\epsilon x^2}\,dx=\tfrac{1}{2}\sqrt{\pi}.$$

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    $\begingroup$ ah nice for using $(1-\epsilon)^{n^2}$ goes to $e^{-\epsilon n^2}$. I didnt realize this...I am just using Taylor expansion to do this... $\endgroup$
    – Xu Shan
    Apr 5, 2023 at 13:49
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    $\begingroup$ It is an heuristic argument, not a proof. Writing $(1-\epsilon)^{𝑛^2} \to e^{-\epsilon n^2}$ does not make sense. $\endgroup$ Apr 5, 2023 at 19:02
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An alternative to Carlo Beenakker's argument, which explains the appearance of $\pi$ geometrically, is as follows: With $A(t)=\sum_{n=0}^{\infty}t^{n^2}$ we have \begin{equation} \frac{1}{1-t}A(t)^2=\sum_{n=0}^{\infty}p(n)t^n, \end{equation} where $p(n)$ is the number of solutions $a^2+b^2\le n$ for nonnegative integers $a$ and $b$. Thinking of these pairs $(a,b)$ as lattice points, we see that $p(n)=\frac{\pi}{4}n+O(\sqrt{n})=\frac{\pi}{4}(n+1)+O(\sqrt{n})$. Thus \begin{equation} \sum_{n=0}^{\infty}p(n)t^n=\frac{\pi}{4}\frac{1}{(1-t)^2}+\sum_{n=0}^{\infty}O(\sqrt{n})t^n. \end{equation} Therefore \begin{equation} (1-t)A(t)^2=\frac{\pi}{4}+(1-t)^2\sum_{n=0}^{\infty}O(\sqrt{n})t^n. \end{equation} The second sum goes to $0$ for $t\nearrow 1$, for instance because for every $\varepsilon>0$, $\sqrt{n}<\varepsilon n$ for all $n>1/\varepsilon^2$.

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  • $\begingroup$ thanks for the answer! Btw I was not familiar with the elliptic things...but it's interesting $\endgroup$
    – Xu Shan
    Apr 5, 2023 at 13:52
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    $\begingroup$ it's a very cool solution $\endgroup$
    – Conrad
    Apr 5, 2023 at 15:57
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    $\begingroup$ It’s also a cool way of evaluating $\int_0^\infty e^{-x^2}dx$, if one put it together with the other solutions $\endgroup$ Apr 5, 2023 at 19:25
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    $\begingroup$ It seems a discrete version of the Poisson's computation of the Gaussian integral from $\int_{\mathbb R^2}e^{-(x^2+y^2)}dxdy$ (via double integration vs integration in polar coordinates) $\endgroup$ Apr 6, 2023 at 7:53
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    $\begingroup$ @PietroMajer In kconrad.math.uconn.edu/blurbs/analysis/gaussianintegral.pdf Keith Conrad records 11 methods to compute the Gaussian integral, none uses this idea with the lattice point count combined with the discretization of the integral by a Riemann sum. So maybe it is worth a short note in the Amer. Math. Monthly? $\endgroup$ Apr 6, 2023 at 13:21
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The function $x \mapsto t^{x^2}$ decreases on $\mathbb{R}_+$, so for every $n \in \mathbb{N}$, $$t^{(n+1)^2} \le \int_n^{n+1}t^{x^2}dx \le t^{n^2}$$ By summation over $n$, $$\sum_{n=1}^{\infty}t^{n^2} \le \int_0^\infty t^{x^2}dx \le \sum_{n=0}^{\infty}t^{n^2}.$$ $$\int_0^\infty t^{x^2}dx \le \sum_{n=0}^{\infty}t^{n^2} \le \int_0^\infty t^{x^2}dx - 1.$$ But $$\int_0^\infty t^{x^2}dx = \int_0^\infty e^{x^2 \ln(t)}dx = \frac{1}{2}\sqrt\frac{\pi}{-\ln(t)}$$ Since $-\ln(t) \sim 1-t$ as $t \to 1-$, we derive $$\lim_{t \to 1-}\sqrt{1-t}\sum_{n=0}^{\infty}t^{n^2}=\frac{1}{2}\sqrt{\pi}.$$

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