Proof that $\sum\limits_{n=1}^{\infty}{(-1)^{n+1}\sin(n)\over{n}}={1\over2}$

If you see https://math.stackexchange.com/a/13494/706414 the author (in an elementary way) shows that $$f(x) = \lim_{n\to\infty}\ \sum_{k=1}^{n}\frac{\sin kx}{k}=\frac{\pi-x}{2},\qquad x\in(0,2\pi).$$ Note that $$f(2) = 2\sum_{n=1}^\infty \frac{\sin(2k)}{2k}$$ Your desired sum then becomes $f(1)-f(2) = \frac{1}{2}$

$S$ is the imaginary part of $$-\sum_{r=1}^\infty\dfrac{(-1)^re^{ir}}r$$

which is $$=\ln(1-(-1)e^i)=\ln(e^{i/2}+e^{-i/2})+\ln(e^{i/2})=\ln\left(2\cos\dfrac12\right)+\dfrac i2$$(considering the principal branch)

Let

$$S=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin(2 \pi n x)$$

$$=\frac{1}{2 i}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} \Big\{ e^{2 \pi i n x}-e^{-2 \pi i n x} \Big \}$$

$$=\frac{1}{2 i}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}e^{2 \pi i n x}}{n} -\frac{1}{2 i}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}e^{-2 \pi i n x} }{n}$$

$$=\frac{1}{2 i}\Big\{ \log(1+e^{2 \pi i x})-\log(1+e^{-2 \pi i x}) \Big \}$$

$$=\frac{1}{2 i}\Big\{ \log(1+e^{2 \pi i x})-\log(1+\frac{1}{e^{2 \pi i x}}) \Big \}$$

$$=\frac{1}{2 i}\Big\{ \log(1+e^{2 \pi i x})-\log(1+e^{2 \pi i x}) + 2 \pi i x \Big \}$$

$$=\frac{1}{2 i}\Big\{ \log \bigg(\frac{1+e^{2 \pi i x}}{1+e^{2 \pi i x}} \bigg) + 2 \pi i x \Big \}$$

$$=\frac{1}{2 i}\Big\{ \log(1) + 2 \pi i x \Big \}$$

$$=\frac{1}{2 i} 2 \pi i x $$

$$\boxed{\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin(2 \pi n x)=\pi x}$$

If we let $$x=\frac{1}{2 \pi}$$ we get

$${\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\sin( n )}{n}=\frac{1}{2} }$$