[Resolução] 11.7.20.a
Posted: 10 Oct 2022 11:27
Segundo a fórmula dada,
\( \sum _{n=1}^{\infty} \frac{cos(nx)}{n^2}=\frac{x^4}{4}-\frac{\pi x}{2} + \frac{\pi}{6} \)
Aplicando ela para \(x=0\), temos
\( \sum _{n=1}^{\infty} \frac{cos(0)}{n^2}=\frac{0}{4}-\frac{0}{2} + \frac{\pi}{6} \)
\( \sum _{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi}{6} \)
\( \sum _{n=1}^{\infty} \frac{cos(nx)}{n^2}=\frac{x^4}{4}-\frac{\pi x}{2} + \frac{\pi}{6} \)
Aplicando ela para \(x=0\), temos
\( \sum _{n=1}^{\infty} \frac{cos(0)}{n^2}=\frac{0}{4}-\frac{0}{2} + \frac{\pi}{6} \)
\( \sum _{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi}{6} \)