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.
$$1+\frac{\pi^2}{6}-1+\pi-\frac{\pi^2}{6}$$
$$\large\color{#A00000}{\boxed{\sum_{k\in\mathbb{Z}}\:\frac{\sin^2k}{k^2}=\pi}}$$