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?

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 $$\arctan\left(\frac{\sin 1}{1-\cos 1}\right)=\arctan\left(\cot\frac12\right) =\frac{\pi}{2}-\frac{1}{2}$$
