Proof that $\sum\limits_{n=1}^{\infty}{(-1)^{n+1}\sin(n)\over{n}}={1\over2}$ While messing around with WolframAlpha, I came across this identity that
${\sin{1}\over{1}}-{\sin{2}\over{2}}+{\sin{3}\over{3}}-{\sin{4}\over{4}}+{\sin{5}\over{5}}\cdots={1\over{2}}$.
One would perhaps expect such a seemingly simple identity to have a clean / intuitive proof, however after I have had trouble finding one while looking around online.
What is the simplest/ most intuitive way to prove this?
 A: 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 $f(1)-f(2) = \frac{1}{2}$
A: $S$ 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)
A: 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
$${\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\sin( n )}{n}=\frac{1}{2} }$$
