5th April

If we consider the function which is periodic of period $2\pi$ and defined on $[-\pi,\pi]$ by the formula $F(\theta) \; = \; \theta^2 \hspace{2cm} |\theta| \le \pi$ then we define the Fourier coefficients $a_n \; = \; \tfrac{1}{2\pi}\int_{-\pi}^\pi \theta^2 e^{-in\theta}\,d\theta \; = \; \left\{ \begin{array}{lll} \frac{2(-1)^n}{n^2} & \hspace{1cm} & n \neq 0 \\ \frac13\pi^2 & & n = 0 \end{array}\right.$ This means that we have the identity $\tfrac13\pi^2 + 4\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\cos n\theta \; = \; F(\theta) \hspace{2cm} \theta \in \mathbb{R}$ For this problem, we want the value of $F\big(\tfrac{1}{\sqrt{3}}\pi\big) = \tfrac13\pi^2$ , which makes the answer $\boxed{0}$ .

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Here is the detailed proof of the convergence. Using the periodicity of $F$ , $\begin{aligned} F_N(\theta) = \sum_{n=-N}^N a_n e^{in\theta} & = \; \tfrac{1}{2\pi} \sum_{n=-N}^N\int_{-\pi}^\pi F(\phi)e^{in(\theta-\phi)}\,d\phi \; =\; \tfrac{1}{2\pi} \sum_{n=-N}^N\int_{-\pi}^\pi F(\theta + \phi)e^{-in\phi}\,d\phi \\ & = \; \frac{1}{2\pi}\int_{-\pi}^\pi F(\theta+\phi) \frac{\sin(N+\frac12)\pi}{\sin\frac12\pi}\,d\phi \; =\; \frac{1}{\pi}\int_{-\frac12\pi}^{\frac12\pi} F(\theta+2\phi)\frac{\sin(2N+1)\phi}{\sin\phi}\,d\phi \\ &=\; \frac{1}{\pi}\int_0^{\frac12\pi} \big[F(\theta+2\phi) + F(\theta-2\phi)\big]\frac{\sin(2N+1)\phi}{\sin\phi}\,d\phi \end{aligned}$ for any $N \in \mathbb{N}$ . The above calculation is valid for any continuous periodic function, and not just $F$ , so we deduce that $\frac{2}{\pi}\int_0^{\frac12\pi} \frac{\sin(2N+1)\phi}{\sin\phi}\,d\phi \; = \; 1$ for any $N \in \mathbb{N}$ , and hence $F_N(\theta) - F(\theta) \; = \; \frac{1}{\pi}\int_0^{\frac12\pi}\big[F(\theta+2\phi) + F(\theta-2\phi) - 2F(\theta)\big] \frac{\sin(2N+1)\phi}{\sin\phi}\,d\phi$ for any $N \in \mathbb{N}$ . Since $F$ is continuous on $\mathbb{R}$ , and since the left and right-hand derivatives $F'(\theta-)$ and $F'(\theta+)$ exist for all $\theta \in \mathbb{R}$ , we deduce that, for any $-\pi \le \theta \le \pi$ , we can find $A_\theta > 0$ such that $\big|F(\theta + 2\phi) + F(\theta - 2\phi) - 2F(\theta)\big| \; \le \; A_\theta \phi \hspace{2cm} 0 \le \phi \le \tfrac12\pi$ and this implies that the function $\phi \; \mapsto \; \frac{F(\theta + 2\phi) + F(\theta - 2\phi) - 2F(\theta)}{\sin\phi}$ is continuous on $(0,\tfrac12\pi]$ and bounded as $\phi \to 0+$ , and hence belongs to $L^1(0,\tfrac12\pi)$ . The Riemann-Lebesgue Lemma tells us that $\lim_{N \to \infty} F_N(\theta) \; =\; F(\theta)$ for all $-\pi \le \theta \le \pi$ . Thus we deduce the identity $\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\cos n\theta \; = \; \tfrac14\theta^2 - \tfrac{1}{12}\pi^2 \hspace{2cm} -\pi \le \theta \le \pi$ with the matching identity for $F(\theta)$ valid for all real $\theta$ .

Here the sum is $\displaystyle \sum_{n=1}^∞ \dfrac{\cos\left(1+\sqrt{\dfrac{1}{3}}\right)nπ}{n^2}$

In previous I have generalised that previous problem

$\displaystyle\sum_{n=1}^∞ \dfrac{\sin(n\theta)}{n} = \dfrac{π-\theta}{2}$ $(0<\theta≤π)$

Now taking integral in both sides

$\displaystyle\sum_{n=1}^∞ \int_0^{\theta} \dfrac{\sin(n\theta)}{n} d\theta = \int_0^{\theta} \dfrac{π-\theta}{2} d\theta$

$\implies\displaystyle\sum_{n=1}^∞ \dfrac{\cos(n\theta)}{n^2} -\sum_{n=1}^∞ \dfrac{1}{n^2} = \dfrac{\theta^2}{4} -\dfrac{π\theta}{4}$ So

$\displaystyle\sum_{n=1}^∞ \dfrac{\cos(n\theta)}{n^2} = \dfrac{π^2}{6} +\dfrac{\theta^2}{4} -\dfrac{π\theta}{2}$ $\left(0≤\theta≤π\right)$

Here $\theta=\left( 1+\sqrt{\dfrac{1}{3}} \right)π$

put that and you will get $\displaystyle \sum_{n=1}^∞ \dfrac{\cos\left(1+\sqrt{\dfrac{1}{3}}\right)nπ}{n^2} = 0$