The answer is 0.

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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 $\displaystyle \sum_{n=1}^∞ \dfrac{\cos\left(1+\sqrt{\dfrac{1}{3}}\right)nπ}{n^2} = 0$

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1 Interesting
2 Brilliant
1 Confused

Very nice - I wonder if this leads to nice ways of evaluating the Riemann zeta function.

One question - why isn't this result periodic? If we call the function $f(\theta)$ , we'd expect $f(\theta+2\pi)=f(\theta)$ , but this clearly isn't the case in your formula. Also, since it's not, which set of $\theta$ values is it correct for?

Chris Lewis
- 2 months, 1 week ago

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@Chris Lewis $0≤\theta≤π$

Yes sir! I have found a way to find the values of Riemann zeta functions . I will write an discussion about that

Thanks sir

Dwaipayan Shikari
- 2 months, 1 week ago

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@Dwaipayan Shikari . The function satisfies Dirichlet condition in $[0,\pi]$ .

Arghyadeep Chatterjee
- 2 months, 1 week ago

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@Arghyadeep Chatterjee – Yes - that's exactly it, but perhaps this should be mentioned/explained in the solution. In fact the definition of the function means it's correct on $[0,2\pi]$ , since it's symmetric in the line $\theta=\pi$ .

Chris Lewis
- 2 months, 1 week ago

The convergence results from the function sarisfying dirichlet's condition in the given interval. So the fourier series converges. It does not guarantee periodicty. Ofcourse the function itself is not periodic but can be redefined and extended so that it forms a periodic function. @Chris Lewis

Nicholas Flammel
- 2 months, 1 week ago

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Just a clarification. Nicholas Flammel and me are the same. One of the accounts are linked to my google account in my phone and the other one on my pc.

Arghyadeep Chatterjee
- 2 months, 1 week ago

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@Arghyadeep Chatterjee – "Nicholas Flammel"seems familiar . May be from Harry Potter . Alchemist may be

Dwaipayan Shikari
- 2 months ago

Dwaipayan Shikari
- 2 months, 1 week ago

Although your formula is correct, your justification is wrong. You establish your formula for $0 < \theta < \pi$ , and then try to plug in a value of $\theta$ that is greater than $\pi$ . It is much easier to separate out the sign changes... My proof goes into all the details of the Fourier analysis. @Chris Lewis @Nicholas Flammel @Dwaipayan Shikari

Mark Hennings
- 2 months, 1 week ago

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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 $\boxed{0}$ .

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 $\theta$ .