# Sum of infinite series $\cos \theta - \frac{\cos 5\theta}{5} + \frac{\cos 7\theta}{7} - \frac{\cos 11\theta}{11} …, \theta \in (-\pi/3, \pi/3)$

I'm stumped at the following exercise on series summation:

$$\cos \theta - \frac{\cos 5\theta}{5} + \frac{\cos 7\theta}{7} - \frac{\cos 11\theta}{11} ..., \theta \in (-\pi/3, \pi/3)$$

The range gave me a hint that I should (probably?) split this into two summations: $$\cos\theta+\frac{\cos7\theta}{7}+\frac{\cos13\theta}{13}+...$$ and $$\frac{\cos5\theta}{5}+\frac{\cos11\theta}{11}+\frac{\cos17\theta}{17}...$$

However I'm stumped at this point. The only hint I was given was that

$$r\cos\theta-r^2\frac{\cos2\theta}{2}+r^3\frac{\cos3\theta}{3}-+...=\frac{1}{2}ln(1+2r\cos\theta+r^2)$$

and that

$$r\sin\theta-r^2\frac{\sin2\theta}{2}+r^3\frac{\sin3\theta}{3}-+...=\arctan\frac{r\sin\theta}{1+r\cos\theta}$$

(This can be obtained by the Taylor expansion of $\ln(1-z)$ and setting $z=re^{i\theta}$, then solving for the real and imaginary parts separately.)

Can anyone give me a hint here please? Thanks!

• That is not a trivial exercise, a bit of complex analysis is involved. – Jack D'Aurizio Oct 20 '16 at 17:52

## 2 Answers

For any $x$ such that $|x|<1$ we have $$S(x)=\sum_{n\geq 0}\left(\frac{x^{6n+1}}{6n+1}-\frac{x^{6n+5}}{6n+5}\right) = \int_{0}^{x}\sum_{n\geq 0}\left(t^{6n}-t^{6n+4}\right)\,dx = \int_{0}^{x}\frac{1-t^4}{1-t^6}\,dt$$ hence $$\begin{eqnarray*} S(x) = \int_{0}^{x}\frac{1+t^2}{1+t^2+t^4}\,dt &=& \frac{1}{\sqrt{3}}\left(\arctan\frac{2x-1}{\sqrt{3}}+\arctan\frac{2x+1}{\sqrt{3}}\right)\\&=&\frac{1}{\sqrt{3}}\,\arctan\left(\frac{x\sqrt{3}}{1-x^2}\right)\end{eqnarray*}$$ and $\lim_{x\to 1^-}S(x)=\frac{\pi}{2\sqrt{3}}$. If $|\theta|<\frac{\pi}{3}$, the function $f(t)=\frac{1+t^2}{1+t^2+t^4}$ is holomorphic in a neighbourhood of the circle sector delimited by the angles $0$ and $\theta$ in the unit disk, hence $$S(e^{i\theta}) = \frac{\pi}{2\sqrt{3}}+\int_{0}^{\theta}\frac{1-e^{4it}}{1-e^{6it}} ie^{it}\,dt = \frac{\pi}{2\sqrt{3}}+2i\int_{0}^{\theta}\frac{\cos t}{1+2\cos(2t)}\,dt$$ where the last integral is a purely imaginary number. It follows that by considering the real parts of both sides of the last identity, $$\forall\theta\in\left(-\frac{\pi}{3},\frac{\pi}{3}\right),\qquad \sum_{n\geq 0}\left(\frac{\cos((6n+1)\theta)}{6n+1}-\frac{\cos((6n+5)\theta)}{6n+5}\right)=\color{red}{\frac{\pi}{2\sqrt{3}}}$$ holds.


Note that \begin{align} &\left.\sum_{n = 0}^{\infty}\pars{% {x^{6n + 1} \over 6n + 1} - {x^{6n + 5} \over 6n + 5}} \right\vert_{\ R\ =\ 1_{-}} = \sum_{n = 0}^{\infty}\bracks{\pars{% {x^{n + 1} \over n + 1} - {x^{n + 5} \over n + 5}}{1 \over 6} \sum_{r\ \in\ \Omega}r^{n}}_{\ R\ =\ 1^{-}}\label{2}\tag{2} \end{align} where $\ds{\braces{r}}$ are $\ds{\pars{~z^{6} = 1~}}$-roots such that $\ds{r \in \Omega \equiv \braces{\exp\pars{k\,{\pi\ic \over 3}}\ {\Large \mid}\ k = 0,1,\ldots,5}}$. For notation simplicity; we switch back and forth, along the result derivation, from $\ds{\sum_{r\ \in\ \Omega}}$ to $\ds{\sum_{k = 0}^{5}}$.
From identity \eqref{2}: \begin{align} &\left.\sum_{n = 0}^{\infty}\pars{% {x^{6n + 1} \over 6n + 1} - {x^{6n + 5} \over 6n + 5}} \right\vert_{\ R\ =\ 1^{-}} = \bracks{{1 \over 6}\sum_{r\ \in\ \Omega}{1 \over r} \sum_{n = 1}^{\infty}{\pars{rx}^{n} \over n} - {1 \over 6}\sum_{r\ \in\ \Omega}{1 \over r^{5}} \sum_{n = 5}^{\infty}{\pars{rx}^{n} \over n}}_{\ R\ =\ 1^{-}} \\[1cm] = &\ \bracks{-\,{1 \over 6}\sum_{r\ \in\ \Omega}r^{5}\ln\pars{1 - xr} + {1 \over 6}\sum_{r\ \in\ \Omega}r\ln\pars{1 - xr}}_{\ R\ =\ 1^{-}} \\[5mm] &\phantom{=}+ {1 \over 6} \underbrace{\sum_{r\ \in\ \Omega}\pars{rx + {r^{2}x^{2} \over 2} + {r^{3}x^{3} \over 3} + {r^{4}x^{4} \over 4}}}_{\ds{=\ 0}} = \left.-\,{1 \over 6}\sum_{r\ \in\ \Omega}\pars{r^{5} - r}\ln\pars{1 - xr} \,\right\vert_{\ R\ =\ 1^{-}} \end{align} Note that, in the limit $\ds{R \to 1^{-}}$, there isn't any contribution from $\ds{r = 1}$ to the above sum such that we can set $\ds{R = 1}$ in the RHS of the following expressions: \begin{align} &\left.\sum_{n = 0}^{\infty}\pars{% {x^{6n + 1} \over 6n + 1} - {x^{6n + 5} \over 6n + 5}} \right\vert_{\ R\ =\ 1^{-}} = -\,{1 \over 6}\sum_{r\ \in\ \Omega}r^{3}\pars{r^{2} - r^{-2}}\ln\pars{1 - xr} \\[5mm] = &\ -\,{1 \over 6}\sum_{k = 0}^{5}\exp\pars{3\,{k\pi \over 3}\ic} \bracks{2\ic\,\Im\pars{\exp\pars{2\,{k\pi \over 3}\ic}}} \ln\pars{1 - \exp\pars{\bracks{\theta + {k\pi \over 3}}\ic}} \\[5mm] = &\ -\,{1 \over 3}\,\ic\sum_{k = 0}^{5}\pars{-1}^{k}\sin\pars{2k\pi \over 3} \ln\pars{1 - \exp\pars{\ic\phi_{k}}}\qquad\mbox{where}\qquad \phi_{k} \equiv \theta + {1 \over 3}\,k\pi\label{3}\tag{3} \end{align}
\begin{align} &\mbox{Then,}\quad\left.\sum_{n = 0}^{\infty}\pars{% {x^{6n + 1} \over 6n + 1} - {x^{6n + 5} \over 6n + 5}} \right\vert_{\ R\ =\ 1^{-}} \\[5mm] = &\ -\,{1 \over 3}\,\ic\sum_{k = 0}^{5}\pars{-1}^{k}\sin\pars{2k\pi \over 3} \ln\pars{\vphantom{\Huge A}\exp\pars{{1 \over 2}\,\phi_{k}\ic}\bracks{% \exp\pars{-\,{1 \over 2}\,\phi_{k}\ic} - \exp\pars{{1 \over 2}\,\phi_{k}\ic}}} \\[5mm] = &\ -\,{1 \over 3}\,\ic\sum_{k = 0}^{5}\pars{-1}^{k}\sin\pars{2k\pi \over 3} \ln\pars{\vphantom{\Huge A}\exp\pars{{\phi_{k} + \pi \over 2}\,\ic} \bracks{2\sin\pars{-\,{\phi_{k} \over 2}}}} \end{align}
With the last expression and identity \eqref{1} it's clear that $\pars{~\vphantom{\large A}\mbox{see}\ \phi_{k}\ \mbox{definition in}\ \eqref{3}~}$: \begin{align} &\sum_{n = 0}^{\infty}\braces{% {\cos\pars{\bracks{6n + 1}\theta} \over 6n + 1} - {\cos\pars{\bracks{6n + 5}\theta} \over 6n + 5}} = {1 \over 6}\sum_{k = 0}^{5}\pars{-1}^{k}\sin\pars{2k\pi \over 3} \pars{\phi_{k} + \pi} \\[5mm] = &\ {1 \over 18}\,\pi\ \underbrace{\sum_{k = 0}^{5}\pars{-1}^{k}\sin\pars{2k\pi \over 3}k} _{\ds{=\ 3\root{3}}}\ +\ {1 \over 6}\pars{\theta + \pi}\ \underbrace{\sum_{k = 0}^{5}\pars{-1}^{k}\sin\pars{2k\pi \over 3}} _{\ds{=\ 0}}\ =\ \bbox[10px,#ffe,border:1px groove navy]{\ds{{\root{3} \over 6}\,\pi}} \end{align}