sum of the series $\cos \frac{\pi}{3}+ \frac{1}{2}\cos\frac{2\pi}{3}+\frac{1}{3}\cos \frac{3\pi}{3}+........$ is

Question

The sum of the series $\displaystyle \cos \frac{\pi}{3}+\frac{1}{2}\cos\frac{2\pi}{3}+\frac{1}{3}\cos \frac{3\pi}{3}+........$ is

solution i try

$\displaystyle \frac{1}{2}\left[1-\frac{1}{2}-\frac{2}{3}-\frac{1}{4}+\frac{1}{5}....\right]$

$\displaystyle \frac{1}{2} \int^{1}_{0}\left(1-x-x^2-x^3+x^4+.........\right)dx$

I am trying to convert into some series form but i do not understand

How i write it

Answer 1

Before converting your series into an integral, you have to check it is convergent. Indeed, that follows from Dirichlet's test, since the partial sums of $\cos\frac{\pi n}{3}$ are bounded and $n$ is increasing towards $+\infty$. By periodicity we have $$ \sum_{n\geq 1}\frac{1}{n}\cos\frac{\pi n}{3}=\sum_{k\geq 0}\left[\frac{\frac{1}{2}}{6k+1}+\frac{-\frac{1}{2}}{6k+2}+\frac{-1}{6k+3}+\frac{-\frac{1}{2}}{6k+4}+\frac{\frac{1}{2}}{6k+5}+\frac{1}{6k+6}\right]$$ and the RHS can be written as $$ \frac{1}{2}\sum_{k\geq 0}\int_{0}^{1}x^{6k}\left(1-x-2x^2-x^3+x^4+2x^5\right)\,dx=\frac{1}{2}\int_{0}^{1}\frac{1-2x}{1-x(1-x)}\,dx $$ where the last integral equals $\color{red}{0}$ since the integrand function is odd with respect to $x=\frac{1}{2}$.

Answer 2

Let $\omega = e^{\pi i/3}$. What you want can be rewritten as $Re(\sum_{n=0}^\infty\frac1n\omega^n)$.

Let $A(z) = \sum_{n=0}^\infty\frac1n z^n$. You need to figure out how to evaluate this series (at least I suppose you know how to do if $z$ is real). It converges when $|z| \leq 1$ except when $z = 1$. In particular, it converges when $z = \omega$. Now you take the real part of it. The answer is 0.

Another thing you might need is $Re(\log z) = \log |z|$.

Answer 3

Hint: $\operatorname{Re}\left(-\log\left(1-e^{\pi i/3}\right)\right)=0$ since $\left|\,1-e^{\pi i/3}\,\right|=1$.