Question

$$\sum_{k\in\mathbb{Z}}\:\frac{\sin^2k}{k^2}$$

Serisini hesaplayın.

Answer 1

Serimiz :

$$\sum_{k\in\mathbb{Z}}\:\frac{\sin^2k}{k^2}$$

Seriyi şöyle de yazabiliriz :

$$1+2\sum_{k\in\mathbb{Z}^+}\:\frac{\sin^2k}{k^2}$$

$$1+\sum_{k\in\mathbb{Z}^+}\:\frac{1-\cos(2k)}{k^2}$$

$$1+\sum_{k\in\mathbb{Z}^+}\:\frac{1}{k^2}-\sum_{k\in\mathbb{Z}^+}\:\frac{\cos(2k)}{k^2}$$

$$1+\frac{\pi^2}{6}-\sum_{k\in\mathbb{Z}^+}\:\frac{\cos(2k)}{k^2}$$

Sondaki seriyi daha önce burada çözmüştük.

$$1+\frac{\pi^2}{6}-1+\pi-\frac{\pi^2}{6}$$

$$\large\color{#A00000}{\boxed{\sum_{k\in\mathbb{Z}}\:\frac{\sin^2k}{k^2}=\pi}}$$