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?

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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}

$$\color{#66f}{\large%
\sum_{k = 1}^{\infty}{\sin\pars{k} \over k} = {\pi - 1 \over 2}}
$$
