Someone could compute the sign of the following series ?
\begin{equation} \underset{k > 0}{\sum} \frac{\sin (kx)}{k} \end{equation}
I expect that is the same as the first term $\sin x$ because of the pseudo terms-alternate property of the serie, but it's not clear. If you have an idea, it would be nice.
Thanx for helping.
Your series is the Fourier-Series of the $2\pi$-peridodic extension of the function $$ f(x):=\frac{\pi-x}{2} $$ defined on $[0,2\pi]$, hence the sign will depend on $x$… EDIT: Thus your statement $$\operatorname{sgn}\left(\sum\limits_{k=1}^\infty\frac{\sin (kx)}{k}\right)=\operatorname{sgn}(\sin x)$$ is correct.
Note that $$\sum_{k=1}^\infty\frac{\sin kx}{k}=\Im\sum_{k=1}^\infty\frac{e^{ikx}}{k}=-\Im\ln\left(1-e^{ix}\right)$$ where we use Taylor series of $\ln(1-x)$. Now, we use the principal value of complex logarithm. We get $$\ln\left(1-e^{ix}\right)=\ln\left(1-\cos x-i\sin x\right)=\ln\sqrt{(1-\cos x)^2+\sin^2x}-i\arctan\left(\frac{\sin x}{1-\cos x}\right)$$ Hence $$\sum_{k=1}^\infty\frac{\sin kx}{k}=-\Im\ln\left(1-e^{ix}\right)=\arctan\left(\frac{\sin x}{1-\cos x}\right)=\arctan\left(\cot\left(\frac{x}{2}\right)\right)$$ where we use identities $$\sin x=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$ and $$\cos x=2\cos^2\left(\frac{x}{2}\right)-1$$ Using identity $$\arctan\left(x\right)+\arctan\left(\frac{1}{x}\right)=\frac{\pi}{2}$$ we get $$\sum_{k=1}^\infty\frac{\sin kx}{k}=\frac{\pi}{2}-\arctan\left(\tan\left(\frac{x}{2}\right)\right)=\frac{\pi-x}{2}$$
