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find the value $f(x)$ such

$$\lim_{x\to 1^{-}}\dfrac{\displaystyle\sum_{n=0}^{\infty}x^{n^2}}{f(x)}=1$$

This problem is china (2009College students' mathematical contest comption) I have consider sometimes, and we know we can't find this sum $$\sum_{n=0}^{\infty}x^{n^2}$$ Thank you someone have nice methods

Antonio Vargas
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math110
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    I really enjoy your questions here, math110 :) – Antonio Vargas Sep 21 '13 at 12:52
  • Thank you very much @AntonioVargas – math110 Sep 21 '13 at 12:54
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    If I interpret this paper correctly, $\displaystyle;;\sum_{n=0}^{\infty} x^{n^2};;$ has following asymptotic expansion as $x \to 1^{-}$: $$\frac12\sqrt{\frac{\pi}{t}} + \frac{e^{-\frac{t^2}{4}}}{2} \left[\frac{\sinh\sqrt{t}}{\sqrt{t}} +\sum_{n=0}^{N-1}\frac{(2^{2n+1}-1)B_{2n+2}(\sqrt{t})^{2n+1}}{2^{2n}(2n+2)!}H_{2n+1}\left(\frac{\sqrt{t}}{2}\right) \right] + O(t^{N+1/2}) $$ where $t = -\log x$, $B_n$ is the $n^{th}$ Bernoulli number and $H_n$ is the $n^{th}$ Hermite polynomial. – achille hui Sep 21 '13 at 14:41

2 Answers2

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By the integral comparison test we have

$$ \int_0^\infty x^{t^2}\,dt \leq \sum_{n=0}^\infty x^{n^2} \leq 1 + \int_0^\infty x^{t^2}\,dt. $$

Now

$$ \begin{align} \int_0^\infty x^{t^2}\,dt &= \int_0^\infty \exp\left[-\left(t\sqrt{-\log x}\right)^2\right]\,dt \\ &= \frac{1}{\sqrt{-\log x}} \int_0^\infty e^{-u^2}\,du \\ &= \frac{1}{2} \sqrt{\frac{\pi}{-\log x}}, \end{align} $$

so

$$ \lim_{x \to 1^-} 2 \sqrt{\frac{-\log x}{\pi}} \sum_{n=0}^\infty x^{n^2} = 1. $$

To simplify this a little we could use the fact that

$$ \lim_{x\to 1} \frac{\log x}{x-1} = 1 $$

to get

$$ \lim_{x \to 1^-} 2 \sqrt{\frac{1-x}{\pi}} \sum_{n=0}^\infty x^{n^2} = 1. $$

Antonio Vargas
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  • Numerical calculations do not confirm that: $$evalf(Sum(.999^{n^2}, n = 0 .. infinity)) $$ produces $28.51794780 $, but $$evalf(sqrt(Pi)/(-2ln(.999))) $$ gives $ 885.7837378$. If we put $x=0.9999$, then $$evalf(Sum(.9999^{n^2}, n = 0 .. 210^5)) $$ gives $ 89.12047691$ and $$evalf(sqrt(Pi)/(-2*ln(.9999))) $$ outputs $ 8861.826138$. – user64494 Sep 21 '13 at 14:17
  • @user64494 if you have a question about the argument, feel free to ask. – Antonio Vargas Sep 21 '13 at 14:28
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    @user64494, numerical calculations would confirm my answer if you had entered the right expressions into your calculator. – Antonio Vargas Sep 21 '13 at 14:31
  • Numerical calculations confirm that: $$evalf(sqrt(Pi)/(2*sqrt(-ln(.999)))) $$ outputs $28.01794780 $. – user64494 Sep 21 '13 at 14:36
  • Nice,This http://sms.math.ecnu.edu.cn/contest/university01/nonmathclass09_answer.pdf – math110 Sep 21 '13 at 15:23
  • @math110 ah, it looks like I found the same answer they did. – Antonio Vargas Sep 21 '13 at 15:37
  • Pari/GP with 400 digits float precision gives $\small \begin{matrix} x & \sum (...) & 2 \sqrt{(...)}\cdot \sum (...) \ 0.95000000 & 4.4130432 & 1.1134693 \ 0.97500000 & 6.0697011 & 1.0829100 \ 0.98750000 & 8.4017929 & 1.0599418 \ 0.99375000 & 11.692435 & 1.0430377 \ 0.99687500 & 16.340912 & 1.0307572 \ 0.99843750 & 22.911203 & 1.0219107 \ 0.99921875 & 32.200424 & 1.0155742 \ 0.99960938 & 45.335550 & 1.0110531 \ 0.99980469 & 63.910140 & 1.0078360 \ 0.999902&90.1776& 1.00555 \end{matrix}$ by Mat(vectorv(10,r,[x=1-0.1/2^r, s=sumalt(k=0,x^k^2),s2sqrt((1-x)/Pi)])) – Gottfried Helms Sep 21 '13 at 21:19
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Maple does not compute the limit $$ \lim_{x \to 1^-} \sqrt{\frac{-\log x}{\pi}} (\vartheta_3(0,x)+1) $$ but does show numerically the value $1.0000000000$ for it.

added

"Code" was requested. Is that allowed in a math group?

CODE MAPLE 17

GEdgar
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