What is the value of $\sum_{k=1}^{\infty}\frac{\cos\left(\frac{2\pi k}{3}\right)}{k^2}$?

Question

I've come across the following trigonometric series:

$$\sum_{k=1}^{\infty}\frac{\cos\left(\frac{2\pi k}{3}\right)}{k^2}$$

for which WolframAlpha gives the answer $-\dfrac{\pi^2}{18}$.

How do you evaluate it? (I'm guessing it's using Fourier analysis which I'm not very familiar with.)

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Also, the similar sum $$\sum_{k=1}^{\infty}\frac{\sin\left(\frac{2\pi k}{3}\right)}{k^2}$$

does not give a "nice" answer on WolframAplha. Could you evaluate it in a similar way?

Answer 1

I will provide an elementary proof that doesn't require Fourier analisys. Consider the sequence in the numerator: $$\cos\left(\frac{2\pi k}{3}\right)_{k\ge 1}$$ notice that the arguments of the cosines of the sequence start with: $$\frac{2\pi}{3},\ \frac{4\pi}{3},\ \frac{6\pi}{3}$$ and then repeat with $$ \frac{8\pi}{3}=2\pi+\frac{2\pi}{3}$$ $$ \frac{10\pi}{3}=2\pi+\frac{4\pi}{3}$$ $$ \frac{12\pi}{3}=2\pi+\frac{6\pi}{3}$$ and so on. So it is clear that, because of the periodicity of the cosine function, the sequence is just: $$\cos\left(\frac{2\pi k}{3}\right)_{k\ge 1}=\cos\left(\frac{2\pi}{3}\right),\ \cos\left(\frac{4\pi}{3}\right),\ \cos\left(\frac{6\pi}{3}\right),\ \cos\left(\frac{2\pi}{3}\right),\ \cos\left(\frac{4\pi}{3}\right),\dots$$ Now, we know that $$\cos\left(\frac{2\pi}{3}\right)=-\frac 12$$ $$\cos\left(\frac{4\pi}{3}\right)=-\frac 12$$ $$\cos\left(\frac{6\pi}{3}\right)=\cos(2\pi)=1$$ Therefore our series is just: $$\sum_{k=1}^{\infty}\frac{\cos(\frac{2\pi k}{3})}{k^2}=$$ $$=\sum_{k=1}^{\infty}\frac{1}{(3k)^2}-\frac12\left(\sum_{k=1}^{\infty}\frac{1}{{k}^2}-\sum_{k=1}^{\infty}\frac{1}{(3k)^2}\right) $$ where the first sum is counting all the $n=3k$, where the numerator is $1$, while in brackets there is the the sum over all $n$ of the form $3k+1,\ 3k+2$ (with $-\frac12$ as numerator), which is just the sum over all $n$ minus the sum over the $n=3k$. We continue as follows: $$=\sum_{k=1}^{\infty}\frac{1}{(3k)^2}-\frac12\left(\sum_{k=1}^{\infty}\frac{1}{{k}^2}-\sum_{k=1}^{\infty}\frac{1}{(3k)^2}\right)= $$ $$=\frac19\sum_{k=1}^{\infty}\frac{1}{k^2}-\frac12\sum_{k=1}^{\infty}\frac{1}{k^2}+\frac{1}{18}\sum_{k=1}^{\infty}\frac{1}{k^2}=$$ $$=\left(\frac19-\frac12+\frac{1}{18}\right)\sum_{k=1}^{\infty}\frac{1}{k^2}=$$ $$=\left(\frac19-\frac12+\frac{1}{18}\right)\left(\frac{\pi^2}{6}\right)=$$ $$=-\frac{\pi^2}{18}$$

Answer 2

Note that for $k=1,2,3$

$$\cos\left(\frac{2\pi k}{3}\right)= -\frac1 2, -\frac1 2,1$$

and it is periodic therefore

$$S_N=\sum_{k=1}^{3N}\frac{\cos(\frac{2\pi k}{3})}{k^2}=\sum_{k=0}^{N-1}\frac{\cos(\frac{2\pi (3k+1)}{3})}{(3k+1)^2}+\sum_{k=0}^{N-1}\frac{\cos(\frac{2\pi (3k+2)}{3})}{(3k+2)^2}+\sum_{k=0}^{N-1}\frac{\cos(\frac{2\pi (3k+1)}{3})}{(3k+3)^2}=$$

$$=-\frac1 2\sum_{k=0}^{N-1}\frac{1}{(3k+1)^2}-\frac 12\sum_{k=0}^{N-1}\frac{1}{(3k+2)^2}+\sum_{k=0}^{N-1}\frac{1}{(3k+3)^2}$$

then we can use the following idea to obtain

$$=-\frac1 2\sum_{k=0}^{N-1}\frac{1}{(3k+1)^2}-\frac 12\sum_{k=0}^{N-1}\frac{1}{(3k+2)^2}-\frac12\sum_{k=0}^{N-1}\frac{1}{(3k+3)^2}+\frac{3}{2}\sum_{k=0}^{N-1}\frac{1}{(3k+3)^2}=$$

$$S_N=-\frac1 2\sum_{k=0}^{N-1}\frac{1}{(k+1)^2}+\frac{1}{6}\sum_{k=0}^{N-1}\frac{1}{(k+1)^2}=-\frac{1}{3}\sum_{k=0}^{N-1}\frac{1}{(k+1)^2}\to -\frac{\pi^2}{18}$$

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For the other one, in a similar way we obtain

$$S_N=\sum_{k=1}^{3N}\frac{\sin\left(\frac{2\pi k}{3}\right)}{k^2}=\frac{\sqrt 3} 2\sum_{k=0}^{N-1}\frac{1}{(3k+1)^2}-\frac{\sqrt 3} 2\sum_{k=0}^{N-1}\frac{1}{(3k+2)^2}$$

which in the limit gives by Polygamma function

$$S_N\to \frac{\sqrt 3} {18} \left(\psi'\left(\frac13\right)-\psi'\left(\frac23\right)\right)\approx \frac23$$

Answer 3

Using the complex represntation of trigonometric fuctions and making the problem more general, the partial sum $$S_n=\sum_{n=1}^p \frac {\cos(ak)}{k^b}$$ is given in terms of the Lerch transcendent function and $$S=\sum_{n=1}^\infty \frac {\cos(ak)}{k^b}=\frac{1}{2} \left(\text{Li}_b\left(e^{-i a}\right)+\text{Li}_b\left(e^{i a}\right)\right)$$ $$T=\sum_{n=1}^\infty \frac {\sin(ak)}{k^b}=\frac{1}{2} i \left(\text{Li}_b\left(e^{-i a}\right)-\text{Li}_b\left(e^{i a}\right)\right)$$

If $a=\frac{2\pi}3$ and $b=2$ $$T=\frac{1}{6 \sqrt{3}}\left(\psi ^{(1)}\left(\frac{1}{3}\right)-\psi^{(1)}\left(\frac{2}{3}\right)\right)$$ already given by @user.