Why does $\sum_{k=1}^{\infty}\frac{{\sin(k)}}{k}={\frac{\pi-1}{2}}$? Inspired by this question (and far more straightforward, I am guessing), Mathematica tells us that $$\sum_{k=1}^{\infty}\dfrac{{\sin(k)}}{k}$$ converges to $\dfrac{\pi-1}{2}$. 
Presumably, this can be derived from the similarity of the Leibniz expansion of $\pi$ $$4\sum_{k=1}^{\infty}\dfrac{(-1)^{k+1}}{2k-1}$$to the expansion of $\sin(x)$ as $$\sum_{k=0}^{\infty}\dfrac{(-1)^k}{(2k+1)!}x^{2k+1},$$ but I can't see how...
Could someone please explain how $\dfrac{\pi-1}{2}$ is arrived at?
 A: As a generalization, if $a$ is not an integer,
$$ \sum_{k=-\infty}^{\infty} \frac{\sin k}{k+a} = \pi \Big(\cos (a) -\sin (a) \cot (\pi a) \Big).$$
This result can be derived using contour integration.
Using the kernel function $\pi \cot(\pi z)$ won't work in the sense that $$\int_{|z|=N+\frac{1}{2}} \frac{\pi \cot(\pi z) e^{iz}}{z+a} \, dz$$ won't vanish as $N \to \infty$ through the positive integers.
So instead we will use the kernel $\pi e^{-i \pi z} \csc(\pi z)$, which also has simple poles at the integers with residue $1$.
To see that the integral vanishes, notice that as $\text{Im}(z) \to + \infty$, $\left|e^{-i \pi z} \csc(\pi z) e^{iz} \right|$ decays like  $2 e^{-\text{Im}(z)}$. And as $\text{Im}(z) \to - \infty$,  $\left|e^{-i \pi z} \csc(\pi z) e^{iz} \right|$ decays like $2 e^{(2 \pi -1) \, \text{Im}(z)}$.
So summing the residues, we get
$$2 \pi i \sum_{k=-\infty}^{\infty} \text{Res} \left[\frac{\pi e^{- i \pi z} \csc(\pi z)e^{iz}}{z+a}, k \right] + 2 \pi i \ \text{Res} \left[\frac{\pi e^{-i \pi z} \csc(\pi z)e^{iz}}{z+a},-a \right] = 0,$$
which implies
$$ \begin{align} \sum_{k=-\infty}^{\infty}\frac{e^{ik}}{k+a} &= - \text{Res} \left[\frac{\pi e^{-i \pi z}\csc(\pi z) e^{iz}}{z+a},-a \right] \\ &= - \lim_{z \to -a} \pi e^{-i \pi z} \csc (\pi z) e^{iz}   \\ &= \pi e^{i \pi a} \csc(\pi a)e^{-ia}   \\ &= \pi \Big(\cos (\pi a) + i \sin (\pi a) \Big) \csc (\pi a)\Big( \cos (a) - i \sin (a)\Big)   \\ &= \pi \ \frac{\cos ( a) \cos (\pi a) + \sin (a) \sin (\pi a)}{\sin (\pi a)} + i \pi  \ \frac{\cos(a) \sin (\pi a) - \sin(a) \cos (\pi a)}{\sin (\pi a)}. \end{align} $$
Then equating the imaginary parts on both sides of the equation, we get
$$ \sum_{k=-\infty}^{\infty} \frac{\sin k}{k+a} = \pi \Big(\cos (a) - \sin (a) \cot (\pi a) \Big).$$
Notice that nothing changes if we remove $k=0$ from the summation.
So if we let $a \to 0$ on both sides of the equation, we get
$$2 \sum_{k=1}^{\infty} \frac{\sin k}{k} =  \pi - 1.$$
A: $$\sum_{n=1}^{+\infty}\frac{\sin n}{n}=\Im\sum_{n=1}^{+\infty}\frac{e^{in}}{n}=\Im\left(-\log(1-e^i)\right)=\pi-\arg(e^i-1)=\frac{\pi-1}{2}.$$
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
\sum_{k = 1}^{\infty}{\sin\pars{k} \over k}&=
-1 + \sum_{k = 0}^{\infty}{\sin\pars{k} \over k}
\end{align}

With Abel-Plana Formula:
  \begin{align}
\sum_{k = 1}^{\infty}{\sin\pars{k} \over k}&=
\color{#c00000}{\Large -1} +\
\overbrace{\int_{0}^{\infty}{\sin\pars{x} \over x}\,\dd x}
^{\ds{=\ \color{#c00000}{\Large{\pi \over 2}}}}\
+\
\color{#c00000}{\Large\half}\ \overbrace{\lim_{x \to 0}{\sin\pars{x} \over x}}^{\ds{=\ 1}}
\\[3mm]&\phantom{=}+ \ic\ \underbrace{\int_{0}^{\infty}\bracks{%
{\sin\pars{\ic x} \over \ic x} - {\sin\pars{-\ic x} \over -\ic x}}
{\dd x \over \expo{2\pi x} - 1}}_{\ds{=\ 0}}
\end{align}

$$\color{#66f}{\large%
\sum_{k = 1}^{\infty}{\sin\pars{k} \over k} = {\pi - 1 \over 2}}
$$
A: Using the power series
$$
-\log(1-z)=\sum_{k=1}^\infty\frac{z^k}{k}
$$
we get
$$
\begin{align}
\sum_{k=1}^\infty\frac{\sin(k)}{k}
&=\frac1{2i}\sum_{k=1}^\infty\frac{e^{ik}-e^{-ik}}{k}\\
&=\frac1{2i}\left[-\log(1-e^i)+\log(1-e^{-i})\right]\\
&=\frac1{2i}\log(-e^{-i})\\
&=\frac{\pi-1}{2}
\end{align}
$$
That is, since $1-e^{-i}$ is in the first quadrant and $1-e^i$ is in the fourth, the imaginary part of $-\log(1-e^i)+\log(1-e^{-i})$ is between $0$ and $\pi$.
A: It follows from the formula $-log(1-z)=\sum_{j=1}^\infty \frac{z^j}{j}$ for $|z|\leq 1$ and $z\neq 1$. Then letting $z_0=e^{i}=\cos 1 + i\sin 1$. Then the sum you want is the imaginary part of $-\log(1-z_0)$.
Since the imaginary part of $\log$ is the angle, then your sum is $$\arctan\left(\frac{\sin 1}{1-\cos 1}\right)=\arctan\left(\cot\frac12\right) =\frac{\pi}{2}-\frac{1}{2}$$
A: You have a function $f:[0,2\pi]\to\Bbb R$ given by $x\mapsto \dfrac{\pi-x}2$ which you can extend periodically to get an odd function $f:\Bbb R\to\Bbb R$. Since $f$ is odd, the even Fourier coefficients vanish, and the odd coefficients of $f$ are given by $$\frac{1}{\pi}\int_0^{2\pi}\sin kx\frac{ \pi-x}2dx=-\frac 1 {2\pi}\int_0^{2\pi}x \sin kxdx$$
Integrating by parts gives $$\int_0^{2\pi}x \sin kxdx=-\frac{2\pi}{k}$$ so that the coefficients are $\dfrac 1 k$, i.e. we have the Fourier expansion $$ \sum_{k=1}^{\infty}\frac{\sin kx}k$$
Since $f$ is differentiable at $x=1$ we can plug in $x=1$ to get $$\frac{\pi-1}2=\sum_{k=1}^{\infty }\frac{\sin k}k$$
A: Here is one way, but it does not use the series you mention so much. I hope that's OK. 
The series is:  
$$\sin(1)+\frac{\sin(2)}{2}+\frac{\sin(3)}{3}+\cdot\cdot\cdot $$
$$\Im\left[e^{i}+\frac{e^{2i}}{2}+\frac{e^{3i}}{3}+\cdot\cdot\cdot \right]$$
Let $\displaystyle x=e^{i}$.
$$\Im\left[x+\frac{x^2}{2}+\frac{x^3}{3}+\cdot\cdot\cdot \right]$$
differentiate:
$$\Im \left[1+x+x^{2}+x^{3}+\cdot\cdot\cdot \right]$$
This is a geometric series, $\displaystyle \frac{1}{1-x}$ 
$$\Im [\frac{1}{1-x}]$$
Integrate:
$$-\Im[\ln(x-1)]=-\Im [\ln(e^{i}-1)]$$
Now, suppose $$\ln(e^{i}-1)=a+bi$$,
$$e^{i}-1=e^{a}e^{bi}$$
$$\cos(1)-1+i\sin(1)=e^{a}\left[\cos(b)+i\sin(b)\right]$$
Equate real and imaginary parts:
$$\cos(1)-1=e^{a}\cos(b)\\ \sin(1)=e^{a}\sin(b)$$
divide both:
$$\frac{\cos(1)-1}{\sin(1)}=\frac{e^{a}\sin(b)}{e^{a}\cos(b)}$$
$$-\cot(1/2)=\tan(b)$$
$$b=\tan^{-1}(\cot(1/2))=\frac{1}{2}-\frac{\pi}{2}$$
But we need the negative of this, so finally:
$$\frac{\pi}{2}-\frac{1}{2}$$
