Use a fórmula para provar a igualdade:
Fórmula:
\( \sum_{n=1}^{\infty} \frac{cos(nx)}{n^2} = \frac{x^2}{4} - \frac{\pi x}{2} + \frac{\pi^2}{6} \:\:\:\: se \:\:\:\: 0\le x\le2\pi\)
Igualdade:
\( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(2n - 1)^3} = \frac{\pi^3}{32}\)
Utilizando-se da fórmula :
\( \sum_{n=1}^{\infty} \frac{cos(nx)}{n^2} = \frac{x^2}{4} - \frac{\pi x}{2} + \frac{\pi^2}{6}\)
\( \int\sum_{n=1}^{\infty} \frac{cos(nx)}{n^2} dx = \int(\frac{x^2}{4} - \frac{\pi x}{2} + \frac{\pi^2}{6})dx\)
\(\sum_{n=1}^{\infty} \frac{sin(nx)}{n^3} = \frac{x^3}{12} - \frac{\pi x^2}{4} + \frac{\pi^2 x}{6}\)
Usando o valor de \(x = \frac{\pi}{2} \) para a integral, então:
\(\sum_{n=1}^{\infty} \frac{sin(\frac{\pi}{2} . n)}{n^3} = \frac{\frac{\pi^3}{8}}{12} - \frac{\frac{\pi^3}{4}}{4} + \frac{\frac{\pi^3}{2}}{6}\)
\( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(2n - 1)^3} = \frac{\pi^3}{32}\)