sum_02

Similar solution with @Hassan Abdulla's

$\begin{aligned} S & = \sum_{k=1}^\infty \frac {\sin k}k & \small \color{#3D99F6} \text{By Euler's formula } e^{i \theta} = \cos \theta + i \sin \theta \\ & = \sum_{k=1}^\infty \frac {\Im (e^{ik})}k & \small \color{#3D99F6} \text{where } \Im (z) \text{ is the imaginary part of }z. \\ & = \Im \left(\sum_{k=1}^\infty \frac {e^{ik}}k \right) & \small \color{#3D99F6} \text{By Maclaurin series } \ln (1-x) = - \sum_{n=1}^\infty \frac {x^n}n \\ & = \Im \left(- \ln (1-e^i) \right) & \small \color{#3D99F6} \text{By Euler's formula} \\ & = \Im \left(- \ln (1-\cos 1 - i\sin 1) \right) \\ & = \Im \left(- \ln \left(\sqrt{(1-\cos 1)^2+\sin^2 1}\left(\frac {1-\cos 1 - i\sin 1}{\sqrt{(1-\cos 1)^2+\sin^2 1}} \right) \right) \right) \\ & = \Im \left(- \ln \left(\sqrt{2-2\cos 1}\left(\frac {1-\cos 1}{\sqrt{2-2\cos 1}} + i \frac {-\sin 1}{\sqrt{2-2\cos 1}} \right) \right) \right) \\ & = \Im \left(- \ln \left(\sqrt{2-2\cos 1}\ {\color{#3D99F6}\exp} \left(i \tan^{-1} \frac {-\sin 1}{1-\cos 1} \right) \right) \right) & \small \color{#3D99F6} \text{where }\exp(x) = e^x \\ & = \Im \left(- \ln \left(\sqrt{2-2\cos 1}\right) - \ln \left(\exp \left(i \tan^{-1} \frac {\sin 1}{1-\cos 1} \right) \right) \right) \\ & = \Im \left(- \ln \left(\sqrt{2-2\cos 1}\right) + i \tan^{-1} \frac {\sin 1}{1-\cos 1} \right) \\ & = \tan^{-1} \frac {\sin 1}{1-\cos 1} & \small \color{#3D99F6} \text{Using half-angle tangent substitution} \\ & = \tan^{-1} \frac {\frac {2t}{1+t^2}}{1-\frac {1-t^2}{1+t^2}} = \tan^{-1} \frac 1t = \tan^{-1} \frac 1{\tan \frac 12} & \small \color{#3D99F6} \text{and let }t = \tan \frac 12 \\ & = \tan^{-1} \cot \frac 12 = \frac \pi 2 - \tan^{-1} \tan \frac 12 \\ & = \boxed{\dfrac {\pi - 1}2} \end{aligned}$

$\implies a+b+c = 1+1+2 = \boxed{4}$

It is not hard to observe that the series given is actually the Fourier series expansion of the function $f(x)=\frac{\pi-x}{2}$ for $x\in(0,2\pi)$ in disguise, i.e. $\sum^{\infty}_{k=1}\frac{\sin kx}{k}=\frac{\pi-x}{2}.$ Substituting $x=1$ yields the final result.

let I= $\sum _{ k=1 }^{ \infty }{ \frac { \sin { \left( k \right) } }{ k } }$

I= $\sum _{ k=1 }^{ \infty }{ \frac { Im\left\{ { e }^{ ik } \right\} }{ k } }$

I= $Im\left\{ \sum _{ k=1 }^{ \infty }{ \frac { { e }^{ ik } }{ k } } \right\} =Im\left\{ \sum _{ k=1 }^{ \infty }{ \frac { { \left( { e }^{ i } \right) }^{ k } }{ k } } \right\}$

I= $Im\left\{ -\ln { \left( 1-{ e }^{ i } \right) } \right\}$

let z= $1-{ e }^{ i }=\left( 1-\cos { \left( 1 \right) } \right) +i\left( -\sin { \left( 1 \right) } \right)$

I= $m\left\{ -\ln { \left( z \right) } \right\}$

I= $Im\left\{ -\left( \ln { \left( \left| z \right| \right) +i } \theta \right) \right\}$ * \\ definition of logarithm in complex *

I= $-\theta =-\tan ^{ -1 }{ \left( \frac { -\sin { \left( 1 \right) } }{ 1-\cos { \left( 1 \right) } } \right) } =\tan ^{ -1 }{ \left( \frac { \sin { \left( 1 \right) } }{ 1-\cos { \left( 1 \right) } } \right) }$

Now

$1-\cos { \left( 1 \right) } =2\cdot \frac { 1-\cos { \left( 2\cdot \frac { 1 }{ 2 } \right) } }{ 2 } =2\sin ^{ 2 }{ \left( \frac { 1 }{ 2 } \right) }$ * \\ since $\sin ^{ 2 }{ \left( x \right) } =\frac { 1-\cos { \left( 2x \right) } }{ 2 }$ *

and

$\sin { \left( 1 \right) } =\sin { \left( 2\cdot \frac { 1 }{ 2 } \right) } =2\sin { \left( \frac { 1 }{ 2 } \right) } \cos { \left( \frac { 1 }{ 2 } \right) }$

so I= $\tan ^{ -1 }{ \left( \frac { 2\sin { \left( \frac { 1 }{ 2 } \right) } \cos { \left( \frac { 1 }{ 2 } \right) } }{ 2\sin ^{ 2 }{ \left( \frac { 1 }{ 2 } \right) } } \right) }$

I= $\tan ^{ -1 }{ \left( \cot { \left( \frac { 1 }{ 2 } \right) } \right) }$

I= $\tan ^{ -1 }{ \left( \tan { \left( \frac { \pi }{ 2 } -\frac { 1 }{ 2 } \right) } \right) }$

I= $\frac { \pi ^{ 1 }-1 }{ 2 }$

a=1,b=1 and c=2

a+b+c=4