I want to show that $\sum_{k=0}^{\infty} \frac{1}{(2k + 1)^4} = \frac{\pi^4}{96}$ using Parseval's theorem and the function $f: \mathbb{R} \to \mathbb{R}, f(x) = |x|$. However, my calculations are always short of a factor of $1/4$ and I can't find the mistake.
Here are the relevant definitions used in my class: Let $L_{2}^{2\pi}$ be the space of $2\pi$-periodic, square-integrable (on the interval $[-\pi, \pi]$) functions $f: \mathbb{R} \to \mathbb{C}$. We define the inner product: $\langle f, g \rangle := \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \overline{g(x)} dx$. This induces a norm on the space and in particular we have: $$\| f \|_0^2 = \frac{1}{2\pi} \int_{-\pi}^{\pi} |f(x)|^2 dx.$$ Our statement of Parseval's theorem: For $f,g \in L_{2}^{2\pi}$ we have $$\langle f , g\rangle = \sum_{k = -\infty}^{\infty} \hat{f}(k)\overline{\hat{g}(k)}\quad\textrm{,in particular}\quad \| f \|_{0}^{2} = \sum_{k = -\infty}^{\infty} |\hat{f}(k)|^2. $$ where $\hat{f}(x), \hat{g}(x)$ denote the Fourier coefficients as usual. For a real-valued $f$ this reduces to: $$\| f \|_{0}^{2} = a_0^2 + 2\sum_{k=1}^{\infty} \left(a_k^2 + b_k^2\right), $$ where $$a_k := \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \cos{(kx)} dx \quad\textrm{and}\quad b_k := \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \sin{(kx)}dx. $$
So much for the definitions. Now my derivation: $f$ is even so all $b_k =0$ Now, $$a_k = \frac{1}{2\pi} \int_{-\pi}^{\pi} |x| \cos{(kx)} dx = \frac{1}{\pi k^2}\left((-1)^k - 1\right).$$ It follows that $a_0 = 0$ and using Parseval's theorem we have that $$\| f \|_{0}^2 = a_0^2 + 2\sum_{k=1}^{\infty} \left(a_k^2 + b_k^2\right) = 2\sum_{k=0}^{\infty} a_k^2 = 2\sum_{k=0}^{\infty} \frac{1}{\pi^2 k^4}\left((-1)^k - 1\right)^2 = \frac{8}{\pi^2} \sum_{k=0}^{\infty} \frac{1}{(2k + 1)^4}. $$ The last equality can be seen from the fact that the parenthesis vanishes for even $k$ and evaluates to $4$ for odd $k$. Using this we can now calculate (using the definition of the norm): $$\sum_{k=0}^{\infty} \frac{1}{(2k + 1)^4} = \frac{\pi^2}{8} \| f \|_{0}^2 = \frac{\pi^2}{8} \frac{1}{2\pi}\int_{-\pi}^{\pi} x^2 dx = \frac{\pi}{16} \frac{2\pi^3}{3} = \frac{\pi^4}{24}$$. So there's a $1/4$ missing in the result. Where's the mistake?