# 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?

• It follows from the Fourier series of the fractional part but that is not particularly elementary – Conrad Sep 19 '19 at 0:35

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 $$f(1)-f(2) = \frac{1}{2}$$
• (+1) You might want to explain (in your answer) why the desired sum is $f(1)-f(2)$. – robjohn Sep 19 '19 at 2:57
$$S$$ 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)