Find the value of $\sin(\tfrac{\pi}3) + \tfrac12\sin(\tfrac{2\pi}3) + \tfrac13\sin(\tfrac{3\pi}3)+\dots$ up to infinity

Question

Initially, I thought of doing this by first evaluating the function f(x) = cos(x) + cos(2x) + cos(3x) +.....
and then integrating it. However, I cant seem to find a proper integratable function (if that's a word) to replace f(x) with. Am I on the right track, or is the solution on a completely different path? Anyways, I would appreciate solutions of 10+2 level, if it exists.

Answer 1

It's well-known that $$\sum_{n=1}^\infty\frac{\sin nx}n=\frac{\pi-x}2$$ for $0<x<2\pi$. One should see the sum as the real part of $$\sum_{n=1}^\infty\frac{\exp(inx)}n=\sum_{n=1}^\infty\frac{z^n}n$$ where $z=\exp(ix)$. For $|z|<1$, $$\sum_{n=1}^\infty\frac{z^n}n=\ln\frac1{1-z}$$ (principal branch) and Abel's theorem implies that this extends continuously to points on the unit circle where the series converges, that is when $z\ne1$. For $0<x<2\pi$, $$\ln\frac1{1-\exp(i x)} =\frac{\pi i} 2-\frac {xi}2+\ln\frac{i}{\exp(ix/2)-\exp(-ix/2)} =\frac{(\pi -x)i}2+\ln\frac{1}{2\sin(x/2)}$$ which has imaginary part $(\pi-x)/2$.

Answer 2

$$\sum^{\infty}_{j=1}\frac{1}{j}\sin \bigg(\frac{j\pi}{3}\bigg)$$

$$=\frac{\sqrt{3}}{2}\bigg[1+\frac{1}{2}-\frac{1}{4}-\frac{1}{7}+\frac{1}{8}-\frac{1}{10}-\frac{1}{11}+\cdots\bigg]$$

$$=\frac{\sqrt{3}}{2}\int^{1}_{0}\bigg(1+x-x^3-x^4+x^6+x^7-x^9-x^{10}\cdots\bigg)dx=\frac{\sqrt{3}}{2}\int^{1}_{0}\frac{1+x}{1+x^3}dx$$

So Sum is $$\frac{\sqrt{3}}{2}\int^{1}_{0}\frac{1}{1-x+x^2}dx=\frac{\pi}{3}.$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With Abel-Plana Formula:

\begin{align} \sum_{j = 1}^{\infty}{\sin\pars{j\pi/3} \over j} & = {\pi \over 3}\sum_{j = 1}^{\infty}\mrm{sinc}\pars{{\pi \over 3}\, j} = {\pi \over 3} \bracks{-1 + \sum_{j = 0}^{\infty}\mrm{sinc}\pars{{\pi \over 3}\, j}} \\[5mm] & = {\pi \over 3} \bracks{-1\ +\ \underbrace{\int_{0}^{\infty}\mrm{sinc}\pars{{\pi \over 3}\, j}\dd j + {1 \over 2}\,\mrm{sinc}\pars{{\pi \over 3}\, 0}}_{\ds{Abel\!-\!Plana\ Formula}}} \\[5mm] & = {\pi \over 3} \pars{-1 + {3 \over 2} + {1 \over 2}} = \bbx{\pi \over 3} \end{align}

Why can we use the Abel-Plana Formula ?: It's discussed in this link.