I'm trying to show that $$ \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} $$
Evaluating $\mathbb{C}$omplex Fourier series of the following function $$ f(x) = x^3 - \pi^2 x, \quad x \in [-\pi, \pi] $$
I have to find the Fourier coefficients such that $$ f(x) = \frac{1}{\sqrt{2\pi}} \sum_{n=-\infty}^{+\infty} c_n e^{inx} $$
We know that $$ c_n = \frac{1}{T} \int_{-T/2}^{T/2} f(x) e^{-inx} dx $$
Computing $c_0$ $$ c_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} (x^3 - \pi^2 x) = \frac{1}{2\pi} \left[ \frac{x^4}{4} - \frac{\pi^2x^2}{2} \right]_{-\pi}^{\pi} = 0 $$
Computing $c_{n \neq 0}$, \begin{align*} c_{n \neq 0} = \frac{1}{2\pi} \int_{-\pi}^{\pi} (x^3 - \pi^2 x) e^{-inx} dx \end{align*}
We need to compute the two following integrals \begin{align*} \int_{-\pi}^{\pi} x^3 e^{-inx} dx &= -\frac{1}{in} 2 \pi^3 (-1)^n + \frac{1}{in^3}12 \pi (-1)^n \end{align*}
\begin{align*} \int_{-\pi}^{\pi} x e^{-inx} dx &= -\frac{1}{in} 2\pi (-1)^n \end{align*}
Therefore, $c_{n \neq 0}$, \begin{align*} c_n = c_{n \neq 0} &=\frac{1}{2\pi} \int_{-\pi}^{\pi} (x^3 - \pi^2 x) e^{-inx} dx \\ &= -\frac{1}{in} \pi^2 (-1)^n + \frac{1}{in^3} 6 (-1)^n + \frac{1}{in} \pi^2 (-1)^n \\ &= \frac{6}{in^3} (-1)^n \end{align*}
Therefore, can I write the following ? $$ Sf(x) = \frac{1}{\sqrt{2\pi}} \sum_{n=-\infty}^{+\infty} \frac{6}{in^3} (-1)^n e^{inx} $$
We can then use the fact that $f(0) = Sf(0)$, \begin{align*} Sf(0) &= f(0) \\ \frac{1}{\sqrt{2\pi}} \sum_{n=-\infty}^{+\infty} \frac{6}{in^3} (-1)^n &= 0 \\ \sum_{n=-\infty}^{+\infty} \frac{(-1)^n}{n^3} &= 0 \\ \end{align*}
But then I'm stuck...
Thanks !