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After playing with some series in a numerical math website, it seems to me like the following identity holds:

$$\sum_{n=-\infty}^{\infty}\frac{1-\cos(an)}{(an)^2}=\frac{\pi}{a}$$

It seems a little bit surprising to me, and I was wondering if there is an elementary way to see it. Convergence is trivial due to comparison with $\frac{1}{n^2}$, but the specific value is interesting. I would think that some Fourier analysis might be applicable, mainly because $\pi$ appeared here, but couldn't make it work.

P.S: Even a way to see the behavior $\sum_{k=-\infty}^{\infty}\frac{1-\cos(an)}{(an)^2}\propto \frac{1}{a}$ is interesting to me, and presumably simpler.

P.S2: It seems that for large $a$ (possibly just $a>2\pi$) the claim is incorrect, see comment. Still, it is interesting to calculate, even if only for $|a|<2\pi$.

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  • $\begingroup$ Is $n$ allowed to be $0$? $\endgroup$ – xbh Jun 9 '19 at 10:50
  • $\begingroup$ @xbh Thanks for asking. At $n=0$, using Taylor, $\frac{|1-\cos(an)|}{(an)^2}=\frac{1}{2}$ is reasonable. Yet, this shows that my hypothesis isn't true for $a>2\pi$. Indeed, going back to my numerical website, I see that for large $a$ the answer is different. For instance, for $a=100$, it is $\frac{\pi}{500}(155-24\pi)$. Still interesting to calculate this. It seems like the answer relates to the polylogarithm function. $\endgroup$ – The way of life Jun 9 '19 at 10:59
  • $\begingroup$ Well… according to WolframAlpha, the $\sum_1^\infty \cos (an)/(an)^2$ doesn't have some nice form, like this. $\endgroup$ – xbh Jun 9 '19 at 11:27
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Define

$$f(x)=\sum_{n=-\infty}^{\infty}\frac{1-\cos(nx)}{(nx)^2}$$

for $x\in (0,2\pi]$. We can differentiate this function to get

$$f'(x)=\frac{d}{dx}\left(\sum_{n=-\infty}^{\infty}\frac{1-\cos(nx)}{(nx)^2}\right)=\sum_{n=-\infty}^{\infty}\frac{d}{dx}\left(\frac{1-\cos(nx)}{(nx)^2}\right)$$

$$=\sum_{n=-\infty}^{\infty}\left(\frac{\sin (n x)}{n x^2}-\frac{2 (1-\cos (n x))}{n^2 x^3}\right)=\frac{1}{x^2}\sum_{n=-\infty}^{\infty}\frac{\sin(nx)}{n}-\frac{2}{x}f(x).$$

From the answer to this question, we know that for $x\in (0,2\pi)$

$$\sum_{n=1}^\infty \frac{\sin(nx)}{n}=\frac{\pi-x}{2}.$$

Since $\lim_{n\to 0}{\frac{\sin(nx)}{n}}=x$ and $\frac{\sin(nx)}{n}$ is even, this implies

$$\frac{1}{x^2}\sum_{n=-\infty}^\infty \frac{\sin(nx)}{n}=\frac{1}{x^2}\left(x+2\sum_{n=1}^\infty \frac{\sin(nx)}{n}\right)=\frac{1}{x^2}(x+\pi-x)=\frac{\pi}{x^2}.$$

This then gives us the ODE

$$f'(x)=\frac{\pi}{x^2}-\frac{2}{x}f(x).$$

Solving this, we get that

$$f(x)=\frac{\pi}{x}+\frac{C}{x^2}$$

for some constant $C$. We can find this by noting that

$$f(\pi)=\sum_{n=-\infty}^{\infty}\frac{1-\cos(nx)}{(nx)^2}=\frac{1}{2}+\frac{2}{\pi^2}\sum_{n=1}^\infty \frac{2}{(2n-1)^2}=\frac{1}{2}+\frac{2}{\pi^2}\frac{\pi^2}{4}=1.$$

Then $C=0$, and we get $f(x)=\frac{\pi}{x}$ for $x\in (0,2\pi)$. To finish the proof, note that

$$f(2\pi)=\sum_{n=-\infty}^{\infty}\frac{1-\cos(n2\pi)}{(n2\pi)^2}=...+0+0+\frac{1}{2}+0+0+...=\frac{1}{2}$$

as expected.

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  • $\begingroup$ Well… the differentiation is suspicious… $f(x) $ does not converge uniformly, since the general term does not converge to $0$ uniformly. $\endgroup$ – xbh Jun 9 '19 at 14:54
  • $\begingroup$ True, but it converges uniformly for any set $[\epsilon,2\pi]$, so for any $x$ just take $\epsilon$ to be $x/2$. $\endgroup$ – QC_QAOA Jun 9 '19 at 14:56
  • $\begingroup$ Thanks! I forgot this! +1 for this answer. $\endgroup$ – xbh Jun 9 '19 at 15:02

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