By comparing even numbers to odd numbers, we find that
\begin{align}
\sum_{n \in \mathbb{N}} \frac 1 {n^4} &= \sum_{n \text{ even}} \frac 1 {n^4} + \sum_{n \text{ odd}} \frac 1 {n^4} \\
&= \frac 1 {2^4} \sum_{k \in \mathbb{N}} \frac 1 {k^4} + \sum_{k \in \mathbb{N}} \frac{1}{(2k - 1)^4}
\end{align}
so that
$$\sum_{k \in \mathbb{N}} \frac 1 {(2k - 1)^4} = \frac {15}{16} \cdot \frac{\pi^4}{90} = \frac {\pi^4}{96}.$$
I can't check your calculations since you haven't included them, but it is clear that the Fourier series you found is not the Fourier series of $f$. Your function is not even, so it cannot have a Fourier cosine series. For example,
$$ b_1 = \frac{1}{\pi} \int_0^{2\pi} f(x) \sin(x) \, dx = \frac{1}{\pi} \int_0^{\pi} \sin^2(x) \, dx = \frac{1}{2}. $$
My guess is that you haven't been careful in checking the special case when integrating the complex form (that is, $\int e^{ikx} \, dx = \frac{e^{ikx}}{ik} + C$ only when $k \neq 0$). In fact, the Fourier series of $f$ is given by
$$ \sum_{k = 1}^{\infty} \frac{2}{\pi(1-4k^2)} \cos(2kx) + \frac{1}{\pi} + \frac{1}{2} \sin(x) $$
and you'll get your missing factor from the extra $\sin$ term.
The complex coefficient $c_1$ is given by
$$ c_1 = \frac{1}{2\pi} \int_0^{2\pi} f(x) e^{-ix} \, dx = \frac{1}{2\pi} \int_0^{\pi} \frac{e^{ix} - e^{-ix}}{2i} e^{-ix} \, dx = \frac{1}{4\pi i} \int_0^{\pi} (1 - e^{-2ix}) \ dx = \frac{1}{4\pi i} \left[x - \frac{e^{-2ix}}{-2i} \right]_{x = 0}^{x = \pi} = -\frac{1}{4}i. $$
Similarly,
$$ c_{-1} = \frac{1}{2\pi} \int_0^{2\pi} f(x) e^{ix} \, dx = \frac{1}{2\pi} \int_0^{\pi} \frac{e^{ix} - e^{-ix}}{2i} e^{ix} \, dx = \frac{1}{4\pi i} \int_0^{\pi} (e^{2ix} - 1) \ dx = \frac{1}{4\pi i} \left[\frac{e^{2ix}}{2i} - x \right]_{x = 0}^{x = \pi} = \frac{1}{4}i. $$
Hence,
$$ b_1 = i(c_1 - c_{-1}) = i(-\frac{1}{4}i - \frac{1}{4}i) = \frac{1}{2}. $$
Best Answer
Let $f(x)=x^2$ for $x\in(-\pi,\pi)$. Computing the Fourier coefficients gives
$$a_n=\frac{1}{2\pi}\int_{-\pi}^\pi x^2 e^{i n x} dx=\frac{2 \cos(\pi n)}{n^2}=2\frac{(-1)^n}{n^2}$$
for $n\in\mathbb{Z}$, $n\not=0$, and $a_0=\frac{1}{2\pi}\int_{-\pi}^\pi x^2 dx=\frac{\pi^2}{3}$.
Therefore $|a_n|^2=\frac{4}{n^4}$ for $n\in\mathbb{Z}$, $n\not=0$ and $|a_0|^2=\frac{\pi^4}{9}$.
By Plancherel/Parseval's theorem,
$$\frac{\pi^4}{9}+8\sum_{n=1}^\infty \frac{1}{n^4}=\sum_{n=-\infty}^\infty |a_n|^2=\frac{1}{2\pi}\int_{-\pi}^\pi x^4 dx=\frac{\pi^4}{5}$$
Simplifying, this gives
$$\sum_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{8}\left(\frac{1}{5}-\frac{1}{9}\right)=\frac{\pi^4}{90}$$