# Proving that $\sum_{n=1}^\infty \frac{\sin^2 n}{n^2}=\sum_{n=1}^\infty \frac{\sin n}{n}$.

Proving that $$\sum_{n=1}^\infty \frac{\sin^2 n}{n^2}=\frac{\pi -1}{2}$$

I've known a similar conclusion $$\sum_{n=1}^\infty \frac{\sin nx}{n}= \begin{cases} \dfrac{\pi - x}{2} & x \in (0, 2\pi),\\ \quad 0 & x = 0, \\ f(x+2\pi) & x \in \Bbb{R}. \end{cases}$$ And one of my classmates found the equation mentioned above by mathematica.

I was amazed by the equation

$$\sum_{n=1}^\infty \frac{\sin^2 n}{n^2}=\sum_{n=1}^\infty \frac{\sin n}{n}$$

My attempt

\begin{align} \sum_{n=1}^\infty \frac{\sin^2 n}{n^2} & = \sum_{n=1}^\infty \frac{1-\cos 2n}{2n^2} \\ & = \sum_{n=1}^\infty \int_0^1 \frac{\sin 2n\theta}{n} d\theta \\ & = \int_0^1 \sum_{n=1}^\infty \frac{\sin 2n\theta}{n} d\theta \\ & = \int_0^1 \frac{\pi}{2}-\theta \,d\theta \\ & = \frac{\pi-1}{2} \end{align}

Oh. Actually I hadn't solved it before I edited this question, but I seemed to have worked it out.

So is there any other method to solve this problem? And deeper insights?

I've heard that it can be worked out via complex analysis and fourier analysis.

Thanks in advance!

Added

Thanks for your comments!

Here is another possible generalization

$$\sum_{n \in \mathbb{Z} } \left[\frac{\sin (n \alpha + \theta) }{ n \alpha + \theta} \right]^2 = \frac{\pi}{\alpha} \,\, \forall \alpha , \theta \in \mathbb{R}$$

I got stuck on it. For $$\theta=0$$ we can use the following equation $$\sum_{n \in \mathbb{Z} } \frac{\cos n\theta}{n^2} = \frac{\pi^2}{6}-\frac{\pi \theta}{2} + \frac{\theta^2}{4} \,\, \theta \in [0,2\pi]$$ But how to deal with the situation that $$\theta \ne 0$$?

Can you give me some hints? Thanks in advance!

• This is related. In the end, stuff usually boils down to the fact that the Fourier transform of $\sin x/x$ is a simple box function. So, maybe Poisson summation formula might help? – Fabian Apr 11 at 9:54
• I can't find it right now, but Math.SE has previously given some good answers to the related problem of showing $\int_{-\infty}^\infty\frac{\sin^2 xdx}{x^2}=\int_{-\infty}^\infty\frac{\sin xdx}{x}$ without evaluating either integral. – J.G. Apr 11 at 10:42

## 2 Answers

Too long for a comment.

You can make amazing equations with $$S_k=\sum_{n=1}^\infty \frac{\sin^k (n)}{n^k}$$ One of my former students worked them and showed that, for $$k \leq 6$$, they simply write as $$S_k=-\frac 12 +a_k\, \pi$$ where $$a_k$$ is a rational number.

These coefficients are $$\left\{\frac{1}{2},\frac{1}{2},\frac{3}{8},\frac{1}{3},\frac{115}{384},\frac{11}{40 }\right\}$$

$$\newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}{\mathcal{#1}} \newcommand{\mrm}{\mathrm{#1}} \newcommand{\pars}{\left(\,{#1}\,\right)} \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}{\left\vert\,{#1}\,\right\vert}$$ \begin{align} &\bbox[10px,#ffd]{\sum_{n = 1}^{\infty}{\sin^{k}\pars{n} \over n^{k}}} = \sum_{n = 1}^{\infty}\mrm{sinc}^{k}\pars{n} = -1 + \sum_{n = 0}^{\infty}\mrm{sinc}^{k}\pars{n} \\[5mm] = &\ -1 + \int_{0}^{\infty}\mrm{sinc}^{k}\pars{x}\dd x + {1 \over 2}\,\mrm{sinc}^{k}\pars{0}\qquad\pars{~Abel\mbox{-}Plana\ Formula~} \\[5mm] = &\ \bbx{\int_{0}^{\infty}{\sin^{k}\pars{x} \over x^{k}}\,\dd x - {1 \over 2}} \\[5mm] & \mbox{with}\quad \int_{0}^{\infty}{\sin^{k}\pars{x} \over x^{k}}\,\dd x = \left\{\begin{array}{ccrcl} \ds{\pi \over 2} & \mbox{if} & \ds{k} & \ds{\in} & \ds{\braces{1,2}} \\[2mm] \ds{3\pi \over 8} & \mbox{if} & \ds{k} & \ds{=} & \ds{3} \\[2mm] \ds{\pi \over 3} & \mbox{if} & \ds{k} & \ds{=} & \ds{4} \\[2mm] \ds{115\pi \over 384} & \mbox{if} & \ds{k} & \ds{=} & \ds{5} \\[2mm] \ds{11\pi \over 40} & \mbox{if} & \ds{k} & \ds{=} & \ds{6} \end{array}\right. \end{align}

In using Abel-Plana Formula,

$$\ds{\left.{\sin^{k}\pars{z} \over z^{k}}\expo{-2\pi\,\verts{\Im\pars{z}}} \right\vert_{\ z\ \in\ \mathbb{C}} \,\,\,\stackrel{\mrm{as}\ \verts{\Im\pars{z}}\ \to\ \infty}{\sim}\,\,\, \pm\,{\ic^{-k}\expo{\ic\,\Re\pars{z}} \over \bracks{\Im\pars{z}}^{k}} \exp\pars{\bracks{k - 2\pi}\verts{\Im\pars{z}}} \,\,\,\stackrel{\mrm{as}\ \verts{\Im\pars{z}}\ \to\ \infty}{\LARGE\to}\,\,\,\color{red}{\large 0}\implies \bbx{k \leq 6}}$$.

• What happens for k=7? – uhhhhidk Apr 15 at 23:21
• @uhhhhidk When $\displaystyle k = 7$ the Abel-Plana Formula is not valid for the above calculation. However, $\texttt{Mathematica}$ ( or any other method ) can perform that evaluation: It involves a linear combination of powers of $\displaystyle\left\{\pi^{0},\pi^{1},\pi^{2},\pi^{3},\pi^{4},\pi^{5},\pi^{6},\pi^{7}\right\}$. – Felix Marin Apr 16 at 23:57
• why is it not valid? Are growth conditions not met? It seems weird how it just stops working – uhhhhidk Apr 19 at 18:38
• @uhhhhidk See the Abel-Plana Formula link. – Felix Marin Apr 22 at 17:36