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can someone help me to prove this please

$\lim_{x\rightarrow 1^{-}}\sqrt{1-x}\sum_{n=1}^{\infty }x^{n^{2}} = \frac{\sqrt{\pi }}{2}$

I try to use the expansion of 1/sqrt(1-x) but I don't get anything, thanks for your attention

ruka
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    My guess, which might lead nowhere, is that your are supposed to convert it into a Reimann sum, and then construct and evaluate the corresponding integration problem. – user2661923 Jan 06 '23 at 01:39
  • Looks like you should compare it to a gaussian function – David Raveh Jan 06 '23 at 02:02
  • Denoting the theta function $\theta(s)=\sum_{n=-\infty}^\infty e^{-n^2\pi s}$ and using the functional equation $\theta(s)=\frac{\theta(\frac{1}{s})}{\sqrt s}$(http://www.math.columbia.edu/~woit/fourier-analysis/theta-zeta.pdf , page 4) $$\sqrt{1-x}\sum_{n=0}^\infty e^{-n^2\ln\frac{1}{x}}=\sqrt{1-x}\Big(\frac{1}{2}\theta\Big(\frac{\ln\frac{1}{x}}{\pi}\Big)+\frac{1}{2}\Big)=\frac{\sqrt{1-x}}{2}\Big(\sqrt\frac{\pi}{\ln\frac{1}{x}}\theta\Big(\sqrt{\frac{\pi}{\ln\frac{1}{x}}}\Big)+1\Big)$$ – Svyatoslav Jan 06 '23 at 13:52
  • All the terms of $\theta\Big(\sqrt{\frac{\pi}{\ln\frac{1}{x}}}\Big)$ are exponentially small at $x\to1^-$, except for $n=0;\quad \theta\Big(\sqrt{\frac{\pi}{\ln\frac{1}{x}}}\Big)\to 1$. The limit follows. – Svyatoslav Jan 06 '23 at 13:52

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