Since $\displaystyle\sum_{n=1}^\infty{x^{n^2}}=\dfrac{\operatorname{\vartheta}_3\left(0,x\right)-1}2$ for $x\in\left(0,1\right)$ (just in case), it suffices to consider the former below. (Another relevant identity is that $\displaystyle\sum_{n=1}^\infty{x^{n^2}}=\left(1-x\right)\mspace{-2mu}\sum_{n=1}^\infty{\!\left\lfloor\mspace{-1mu}\sqrt{n}\right\rfloor\!x^n}$.)
Many a post (e.g.,
- What is $\lim_{x\to1^{-}} (1-x)\left(\sum_{i=0}^{\infty} x^{i^2}\right)^{2}$?,
- Why is $\lim_{x\to 1}\sqrt{1-x}\sum_{n=0}^\infty x^{n^2}=\sqrt{\pi}/2\,\,$?,
- Compute the limit of $\sqrt{1-a}\sum\limits_{n=0}^{+\infty} a^{n^2}$ when $a\to1^-$,
- $\lim_{x\to 1^-}\sqrt{1-x}\ \left(1+x+x^4+x^9+x^{16}+x^{25}+\cdots\right)=\sqrt{\pi}/2$ is true?, and
- How to prove that $\lim_{x\to 1^-} \left(\left(\sum_{n=1}^{\infty}x^n \right)\cdot \log\left(\frac{1}{x}\right) \right)= 1$ WITHOUT computing the sum)
has shown that
$$\lim_{\mspace{-1mu}x\to1^-}{\sqrt{1-x\mspace{2mu}}\sum_{n=1}^\infty{\mspace{-1.5mu}x^{n^2}}}=\frac{\mspace{-1.5mu}\sqrt\pi\mspace{1mu}}2\text{.}$$ Nevertheless, none of them evaluated the limit $$\lim_{\mspace{-1mu}x\to1^-}{\mspace{-3mu}\left({\sqrt{\mspace{-0.5mu}\frac\pi{1-x}}-2\mspace{-0.75mu}\sum_{n=1}^\infty{\mspace{-1.75mu}x^{n^2}}}\right)\mspace{-1mu}}\text{,}$$ which is equal to $1$. how to find the asymptotic expansion of the following sum: sketched out a possible approach to finding this kind of limit. Some other related posts are as follows:
- Asymptotics of a recurrence relation,
- Asymptotic behavior of $\sum\limits_{n=0}^{\infty}x^{b^n}$ when $x\to1^-$,
- Evaluate $\lim_{x\to1^-}\left(\sum_{n=0}^{\infty}\left(x^{(2^n)}\right)-\log_2\frac{1}{1-x}\right)$ and
- What's the limit of the series $\log_2(1-x)+x+x^2+x^4+x^8+\cdots$..
In Analysis of Series and Products. Part 1: The Euler–Maclaurin Formula and Analysis of Series and Products. Part 2: The Trapezoidal Rule, the authors show that $${{\sum_{j=1}^\infty\mathrm{e}^{-j^2/x^2}-\frac{x\sqrt\pi}2+\frac12}\to0}\quad\text{as}\quad{x\to{+\infty}}\text{,}$$ and the magnitude of the left-hand side decreases exponentially as $x$ increases. (Then set $u=\exp\left(-x^{-2}\right)$. How about the Abel–Plana formula?)
In addition, in On the asymptotics of some partial theta functions and Some new asymptotic expansions of certain partial theta functions, each author showes that $$\dfrac{\operatorname{\vartheta}_3\left(0,x\right)+1}2=\sum_{n\in\mathbb{N}}{x^{n^2}}=\frac12\sqrt{\frac{-\pi}{\ln{x}}}+\frac12+\operatorname{\mathcal{O}}\left(\ln^q{\!x}\right)\text{,}\quad\forall{q\in\mathbb{N_+}}\text{,}$$ as $x\to1^-$ (after making some substitutions).
It is generally known that $\ln{x}\sim{x-1}$ as $x\to1$; hence, as $x\to1^-$, $\displaystyle\sum_{n=1}^\infty{x^{n^2}}\sim\frac12\sqrt{\frac{\pi}{1-x}}$ just means that $\displaystyle\sum_{n=1}^\infty{x^{n^2}}\sim\frac12\sqrt{\frac{-\pi}{\ln{x}}}$. Yet, if one uses the former in the asymptotic expansion, what about the remainder term?
Via a number of numerical experiments, it seems that $$\frac{\operatorname{\vartheta}_3\left(0,x\right)}{\sqrt{\pi\!\left(1-x\right)}}=\frac1{\sqrt{\pi\!\left(1-x\right)}}\sum_{n\in\mathbb{Z}}{x^{n^2}}=\frac1{1-x}-\frac14+\operatorname{\mathcal{o}}\left(1\right)\quad\text{as}\quad{x\to1^-}\text{.}$$ Unfortunately, I have no idea how to prove it. Any idea? Many thanks!
Best Answer
Considering$$f(x)=\frac{\operatorname{\vartheta}_3\left(0,x\right)}{\sqrt{\pi\!\left(1-x\right)}}$$ Adaptating an approximation nwork done in my group more than thirty years ago, ws have $$g(x)=-\frac{1+2 \left(t+t^4+t^9\right)}{x-1}-\frac{1}{4} \left(1+2 \left(t+t^4+t^9\right)\right)+$$ $$\frac 1{96}\left(7+14 t+96 t^4+14 t^9\right) (x-1)-\frac{5}{128} \left(1+2 \left(t+t^4+t^9\right)\right) (x-1)^2+O\left((x-1)^3\right)$$ where $$t=\exp\Bigg[\pi^2\left(\frac{1}{x-1}+\frac{1}{2}-\frac{x-1}{12}+\frac{1}{24} (x-1)^2-\frac{19}{720} (x-1)^3 \right) \Bigg]$$ is very small; for $x=\frac 12$, $t=\exp\Big[ -\frac{8321 }{5760}\pi ^2\Big]=6.43\times 10^{-7}$.
So, using $t=0$, we have as another approximation $$h(x)=-\frac{1}{x-1}-\frac{1}{4}+\frac{7 }{96}(x-1)-\frac{5}{128} (x-1)^2+O\left((x-1)^3\right)$$
Using $t=0$, we could extend the expansion and have $$h(x)=-\frac{1}{x-1}-\frac{1}{4}+\frac{7 }{96}(x-1)-\frac{5}{128} (x-1)^2+\frac{309 }{10240}(x-1)^3-$$ $$\frac{763 }{40960}(x-1)^4+\frac{893209 }{61931520}(x-1)^5+O\left((x-1)^6\right)$$
Using the last $h(x)$, consider the norm $$\Phi=\int_{\frac 12}^{1} \Big[f(x)-h(x)\Big]^2\,dx=9.80 \times 10^{-9}$$
Edit
I have been in touch with my former PhD student who continued working in this area. His latest results are $$\operatorname{\vartheta}_3\left(0,x\right)=\sqrt{\frac \pi t}\Bigg[1-\frac 14\sum_{n=1}^\infty a_n\,t^n\Bigg]\qquad \text{with}\qquad t=1-x$$ where the $a_n$ form the sequence $$\left\{1,\frac{7}{24},\frac{5}{32},\frac{787}{7680},\frac{763}{10240},\frac{893209}{15482880},\frac{2885597}{61931520},\frac{1153151299}{29727129600},\frac{261937547}{7927234560},\frac{3997632829}{139519328256},\frac{30141297349}{1195879956480},\frac{4101 190700056349}{182826127746662400},\cdots\right\}$$ This makes
$$f(x)=\frac{\operatorname{\vartheta}_3\left(0,x\right)}{\sqrt{\pi\!\left(1-x\right)}}=\frac 1{1-x} -\frac 14\sum_{n=1}^\infty a_n\,(1-x)^{n-1}$$
Using the above coefficients $$\Phi=\int_{\frac 12}^{1} \Big[f(x)-h(x)\Big]^2\,dx=7.83 \times 10^{-15}$$