Compute $\sum\limits_{n = 1}^{\infty} \frac{1}{n^4}$. 
Compute the Fourier series for $x^3$ and use it to compute the value of $\sum\limits_{n = 1}^{\infty} \frac{1}{n^4}$.

I determined the coefficients of the Fourier series, which are
$$a_0 = \dfrac{\pi^3}{2}; \qquad a_n = \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}$$
Then, I get
$$x^3 = \dfrac{\pi^3}{4} + \sum\limits_{n = 1}^{\infty} \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}\cos(nx)$$
If $x = \pi$, then
$$\begin{aligned}
\pi^3 &= \dfrac{\pi^3}{4} + \sum\limits_{n = 1}^{\infty} \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}\cos(n\pi)\\
\dfrac{3\pi^3}{4} &= \sum\limits_{n = 1}^{\infty} \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}(-1)^n
\end{aligned}$$
I'm stuck.  It's easy to compute $\sum\limits_{n = 1}^{\infty} \frac{1}{n^2}$, using the Fourier series, but for this type of problem I'm stuck.
Any comments or suggestions?  By the way, I know that
$$\sum\limits_{n = 1}^{\infty} \dfrac{1}{n^4} = \dfrac{\pi^4}{90}$$
I need to know how to get there.

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\begin{align}
\pi x\cot\pars{\pi x}&=1 + \sum_{n = 1}^{\infty}{2x^{2} \over x^{2} - n^{2}}
=1 - 2x^{2}\sum_{n = 1}^{\infty}{1 \over n^{2}}
- 2x^{4}\sum_{n = 1}^{\infty}{1 \over n^{4}}
- 2x^{6}\sum_{n = 1}^{\infty}{1 \over n^{6}} - \cdots
\end{align}

$$
\pi x^{1/2}\cot\pars{\pi x^{1/2}}=1 - 2x\sum_{n = 1}^{\infty}{1 \over n^{2}}
- 2x^{2}\sum_{n = 1}^{\infty}{1 \over n^{4}}
- 2x^{3}\sum_{n = 1}^{\infty}{1 \over n^{6}} - \cdots
$$
  For $\verts{z} \sim 0$:
  \begin{align}
z\cot\pars{z}&={z \over \tan\pars{z}} \sim {z \over z + z^{3}/3 + 2z^{5}/15}
={1 \over 1 + z^{2}/3 + 2z^{4}/15}
\\[3mm]&\sim 1 - \pars{{z^{2} \over 3} + {2z^{4} \over 15}}
+ \pars{{z^{2} \over 3} + {2z^{4} \over 15}}^{2}
\sim 1 - {z^{2} \over 3} - {2z^{4} \over 15} + {z^{4} \over 9}
=1 - {z^{2} \over 3} - {z^{4} \over 45}
\end{align}

$$
\pi x^{1/2}\cot\pars{\pi x^{1/2}}\sim
1 - {\pi^{2} \over 3}\,x - {\pi^{4} \over 45}\,x^{2}
$$

$$
\color{#00f}{\large\sum_{n = 1}^{\infty}{1 \over n^{4}}}
= -\,\half\,\pars{-\,{\pi^{4} \over 45}}
= \color{#00f}{\large{\pi^{4} \over 90}}
$$

