Serimiz :
$$\sum_{k\in\mathbb{Z}^+}\:\frac{\sin(2kx)}{k}$$
Sinüsü açalım.
$$\sum_{k\in\mathbb{Z}^+}\:\frac{e^{2ikx}-e^{-2ikx}}{2ik}$$
Sadeleştirelim.
$$\frac{1}{2i}\Bigg[\sum_{k\in\mathbb{Z}^+}\:\frac{e^{2ikx}}{k}-\sum_{k\in\mathbb{Z}^+}\:\frac{e^{-2ikx}}{k}\Bigg]$$
Artık serileri hesaplayabiliriz.
$$\frac{1}{2i}\bigg[-\ln(1-e^{2ix})+\ln(1-e^{-2ix})\bigg]$$
Sadeleştirelim.
$$\frac{1}{2i}\ln\bigg(\frac{1-e^{-2ix}}{1-e^{2ix}}\bigg)$$
$$\frac{1}{2i}\ln\big(-e^{-2ix}\big)$$
$$\large\color{#A00000}{\boxed{\sum_{k\in\mathbb{Z}^+}\:\frac{\sin(2kx)}{k}=\frac{\pi}{2}-x\:\:\:,\:\:\:x\in(0,\pi)}}$$