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?
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:
$$\frac{\pi}{2}-\frac{1}{2}$$
As a generalization, if $a$ is not an integer, then
$$ \sum_{k=-\infty}^{\infty} \frac{\sin k}{k+a} = \pi \Big(\cos (a) -\sin (a) \cot (\pi a) \Big).$$
This result can be derived using contour integration.
Let $C_{N}$ be a rectangular contour with vertices at $\pm (N+1/2) \pm i (N+1/2)$.
Using the function $\pi \cot(\pi z)$ won't work in the sense that $$\int_{C_{N}} \frac{\pi \cot(\pi z) e^{iz}}{z+a} \, dz$$ won't vanish as $N \to \infty$ through the positive integers.
The issue is that the magnitude of $e^{iz}$ grows exponentially as $\Im(z) \to - \infty$.
To address this issue, we replace $\cot(\pi z)$ with $$\color{red}{\cot(\pi z)-i =\csc(\pi z) e^{-i \pi z}}. $$
Like $\pi \cot(\pi z)$, the function $\pi\left(\cot(\pi z)-i\right)$ has simple poles at the integers with residue $1$.
But an important difference is that $\cot(\pi z)-i \to 0$ uniformly as $\Im(z) \to - \infty$.
More importantly, as $\Im(z) \to -\infty$, $\left| \cot(\pi z)-i \right| \sim 2e^{2 \pi \Im(z)}$.
Now we have a situation where the magnitude of the numerator decays exponentially to zero as $\Im(z) \to \pm \infty$.
Combining this with the $\frac{1}{z+a}$, what we have are conditions that essentially satisfy Jordan's lemma.
(If $\frac{1}{z}$ were replaced with $\frac{1}{z^{2}}$, then the exponential decay is not important. In that situation what's important is that the magnitude of the numerator stays bounded on the contour.)
So since the integral now vanishes, we have
$$2 \pi i \sum_{k=-\infty}^{\infty} \text{Res} \left[\frac{\pi e^{- i \pi z} \csc(\pi z)e^{iz}}{z+a}, k \right] + 2 \pi i \ \text{Res} \left[\frac{\pi e^{-i \pi z} \csc(\pi z)e^{iz}}{z+a},-a \right] = 0,$$
from which we get
$$ \begin{align} \sum_{k=-\infty}^{\infty}\frac{e^{ik}}{k+a} &= - \text{Res} \left[\frac{\pi e^{-i \pi z}\csc(\pi z) e^{iz}}{z+a},-a \right] \\ &= - \lim_{z \to -a} \pi e^{-i \pi z} \csc (\pi z) e^{iz} \\ &= \pi e^{i \pi a} \csc(\pi a)e^{-ia} \\ &= \pi \Big(\cos (\pi a) + i \sin (\pi a) \Big) \csc (\pi a)\Big( \cos (a) - i \sin (a)\Big) \\ &= \pi \ \frac{\cos ( a) \cos (\pi a) + \sin (a) \sin (\pi a)}{\sin (\pi a)} + i \pi \ \frac{\cos(a) \sin (\pi a) - \sin(a) \cos (\pi a)}{\sin (\pi a)}. \end{align} $$
Equating the imaginary parts on both sides of the equation, we have
$$ \sum_{k=-\infty}^{\infty} \frac{\sin k}{k+a} = \pi \Big(\cos (a) - \sin (a) \cot (\pi a) \Big).$$
Notice that nothing changes if we remove $k=0$ from the summation.
Therefore, $2 \sum_{n=1}^{\infty} \frac{\sin k}{k}$ is the limiting case as $a \to 0$.
For some reason Wolfram Alpha gives a better approximation for different values of $a$ if you ask for more digits.
$$\sum_{n=1}^{+\infty}\frac{\sin n}{n}=\Im\sum_{n=1}^{+\infty}\frac{e^{in}}{n}=\Im\left(-\log(1-e^i)\right)=\pi-\arg(e^i-1)=\frac{\pi-1}{2}.$$
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \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}
$$\color{#66f}{\large% \sum_{k = 1}^{\infty}{\sin\pars{k} \over k} = {\pi - 1 \over 2}} $$
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 $\pi$.
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 $$\arctan\left(\frac{\sin 1}{1-\cos 1}\right)=\arctan\left(\cot\frac12\right) =\frac{\pi}{2}-\frac{1}{2}$$
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 $$\frac{\pi-1}2=\sum_{k=1}^{\infty }\frac{\sin k}k$$
