Limit of a Square Series

Calculus Level 5

$\large f(x) = \sum_{n=0}^{\infty} x^{n^2}, \quad \lim_{x \to 1^-} \sqrt{1-x} f(x) = \frac{\pi^a}{b}$

The above holds for rational numbers $a$ and $b$ . Evaluate $2ab$ .

The answer is 2.

1 solution

A Former Brilliant Member
Dec 18, 2018

$\sum _{n=0}^{\infty } x^{n^2}$ is $\frac{1}{2}(1+\vartheta _3(0,x))$ . $\underset{x\to 1^-}{\text{lim}}\frac{1}{2} \sqrt{1-x} \left(\vartheta _3(0,x)+1\right)$ is $\frac{\sqrt{\pi }}{2}$ . $2 \frac{1}{2} 2$ is $2$ .

First, I think in fact: f(x)=(1/2) (1+theta_3(0, ln(x)/i*pi )) ?

Florian Wechslberger - 2 years, 5 months ago

Limit of a Square Series

The answer is 2.

1 solution

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