# How Find the $f(x)$ such $\lim_{x\to 1^{-}}\frac{\sum_{n=0}^{\infty}x^{n^2}}{f(x)}=1$

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

• 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
• 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

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

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

• Numerical calculations do not confirm that: $$evalf(Sum(.999^{n^2}, n = 0 .. infinity))$$ produces $28.51794780$, but $$evalf(sqrt(Pi)/(-2*ln(.999)))$$ gives $885.7837378$. If we put $x=0.9999$, then $$evalf(Sum(.9999^{n^2}, n = 0 .. 2*10^5))$$ 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
• @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
• – math110 Sep 21 '13 at 15:23

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.