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?

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 $\pi$.
