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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?

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    $\begingroup$ It follows from the Fourier series of the fractional part but that is not particularly elementary $\endgroup$ – Conrad Sep 19 '19 at 0:35
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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 $f(1)-f(2) = \frac{1}{2}$

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

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