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}

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

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

A: From your identity
\begin{equation*}
\frac{3\pi ^{3}}{4}=\sum_{n=1}^{\infty }\frac{6(\pi ^{2}n^{2}-2)(-1)^{n}+12}{
\pi n^{4}}(-1)^{n}
\end{equation*}
expanding the right hand side and using the result $\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi^2}{6}$, we get
\begin{eqnarray*}
\frac{3\pi ^{4}}{4} &=&\sum_{n=1}^{\infty }\frac{6(\pi
^{2}n^{2}-2)(-1)^{n}+12}{n^{4}}(-1)^{n} \\
&=&6\pi ^{2}\sum_{n=1}^{\infty }\frac{1}{n^{2}}-12\sum_{n=1}^{\infty }\frac{1
}{n^{4}}+12\sum_{n=1}^{\infty }\frac{(-1)^{n}}{n^{4}} \\
&=&\pi ^{4}-12\sum_{n=1}^{\infty }\frac{1}{n^{4}}-12\sum_{n=1}^{\infty }
\frac{(-1)^{n-1}}{n^{4}}.
\end{eqnarray*}
Now we need to express the alternating series $\sum_{n=1}^{\infty }\frac{
(-1)^{n-1}}{n^{4}}$ in terms of $\sum_{n=1}^{\infty }\frac{1}{n^{4}}$, e.g.
as follows
\begin{eqnarray*}
\sum_{n=1}^{\infty }\frac{\left( -1\right) ^{n-1}}{n^{4}} &=&\sum_{n=1}^{
\infty }\frac{1}{n^{4}}-2\sum_{n=1}^{\infty }\frac{1}{(2n)^{4}}
=\sum_{n=1}^{\infty }\frac{1}{n^{4}}-\frac{1}{2^{3}}\sum_{n=1}^{\infty }
\frac{1}{n^{4}}=\frac{7}{8}\sum_{n=1}^{\infty }\frac{1}{n^{4}}.
\end{eqnarray*}
Then
\begin{eqnarray*}
\frac{3\pi ^{4}}{4} &=&\pi ^{4}-12\sum_{n=1}^{\infty }\frac{1}{n^{4}}-\frac{
21}{2}\sum_{n=1}^{\infty }\frac{1}{n^{4}} \\
&=&\pi ^{4}-\frac{45}{2}\sum_{n=1}^{\infty }\frac{1}{n^{4}}.
\end{eqnarray*}
Solving for $\sum_{n=1}^{\infty }\frac{1}{n^{4}}$ we finally obtain
\begin{equation*}
\sum_{n=1}^{\infty }\frac{1}{n^{4}}=\frac{2}{45}\left( \pi ^{4}-\frac{3\pi
^{4}}{4}\right) =\frac{\pi ^{4}}{90}.
\end{equation*}
