Find the value of $\sin(\tfrac{\pi}3) + \tfrac12\sin(\tfrac{2\pi}3) + \tfrac13\sin(\tfrac{3\pi}3)+\dots$ up to infinity 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.
 A: 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 $(\pi-x)/2$.
A: $$\sum^{\infty}_{j=1}\frac{1}{j}\sin \bigg(\frac{j\pi}{3}\bigg)$$
$$=\frac{\sqrt{3}}{2}\bigg[1+\frac{1}{2}-\frac{1}{4}-\frac{1}{7}+\frac{1}{8}-\frac{1}{10}-\frac{1}{11}+\cdots\bigg]$$
$$=\frac{\sqrt{3}}{2}\int^{1}_{0}\bigg(1+x-x^3-x^4+x^6+x^7-x^9-x^{10}\cdots\bigg)dx=\frac{\sqrt{3}}{2}\int^{1}_{0}\frac{1+x}{1+x^3}dx$$
So Sum is $$\frac{\sqrt{3}}{2}\int^{1}_{0}\frac{1}{1-x+x^2}dx=\frac{\pi}{3}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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.

