Fourier series proving $\sum \frac{1}{n^4} = \frac{\pi^4}{90}$ I have calculated the Fourier series of $x^4$ in $(-4,4)$  as
$$f(x) = \frac{256}{5} + \sum_{n=1}^\infty \left(\frac{2048}{\pi^2n^2} \cos(\pi n ) -\frac{12288}{\pi^4n^4} \cos(\pi n )\right)\cos (\frac{\pi n x}{4})$$
I have been asked to use
$$ \sum_{n=1}^∞ \frac{1}{n^2} = \frac{\pi^2}{6} $$
to show that 
$$ \sum_{n=1}^∞ \frac{1}{n^4} = \frac{\pi^4}{90} $$
Not sure how to go about this any pointers would help. 

Since $x^4$ is $C^1$ and has equal values at the endpoints of $[-4,4]$ we have
$$
x^4= \frac{256}{5} + \sum_{n=1}^\infty \left(\frac{2048}{\pi^2n^2} \cos(\pi n ) -\frac{12288}{\pi^4n^4} \cos(\pi n )\right)\cos \left(\frac{\pi n x}{4}\right)
$$
for $x\in[-4,4]$. Hence, taking $x=4$ we get
$$
\begin{split}
256
&= \frac{256}{5} + \sum_{n=1}^\infty \left(\frac{2048}{\pi^2n^2} \cos(\pi n ) -\frac{12288}{\pi^4n^4} \cos(\pi n )\right)\cos (\pi n )\\
&= \frac{256}{5} + \sum_{n=1}^\infty \left(\frac{2048}{\pi^2n^2} -\frac{12288}{\pi^4n^4}\right)\\
&= \frac{256}{5} +  \frac{2048}{\pi^2}\cdot\frac{\pi^2}{6} -\frac{12288}{\pi^4}\sum_{n=1}^\infty\frac{1}{n^4}.
\end{split}
$$
And after some simple computations we get
$$\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}.
$$
