Question

$$\sum_{k\in\mathbb{Z}^+}\:\frac{\cos(2kx)}{k^2}$$

Serisini hesaplayın.

Answer 1

Serimiz :

$$\sum_{k\in\mathbb{Z}^+}\:\frac{\cos(2kx)}{k^2}$$

Buradaki serinin integralini alalım.

$$\int\sum_{k\in\mathbb{Z}^+}\:\frac{\sin(2kx)}{k}\,dx=\int\frac{\pi}{2}-x\,dx$$

$$\sum_{k\in\mathbb{Z}^+}\:\frac{\cos(2kx)}{k^2}\,dx=x^2-\pi{x}+C$$

$x$ yerine $0$ verirsek , $C$ ' yi $\zeta(2)=\frac{\pi^2}{6}$ olarak buluruz.

$$\large\color{#A00000}{\boxed{\sum_{k\in\mathbb{Z}^+}\:\frac{\cos(2kx)}{k^2}=x^2-\pi{x}+\frac{\pi^2}{6}\:\:\:,\:\:\:x\in(0,\pi)}}$$