I am working on the following exercise:
Find the Fourier series of the function $f(x) = (x^2-\pi^2)^2$ and $-\pi < x < \pi$. Use it to calculate the sum of $1-\frac{1}{2^4}+\frac{1}{3^4}-\frac{1}{4^4} \ldots$
The Fourier series is defined as: $$ \frac{a_0}{2} + \sum_{n=1}^\infty a_n cos(nt) + b_nsin(nt) \ dt$$
Where $$a_n = \frac{1}{\pi}\int_{-\pi}^\pi f(t) cos(nt) \ dt $$ for $n = 0,1,2, \ldots$ .
And $$b_n = \frac{1}{\pi} \int_{-\pi}^\pi f(t) sin(nt) \ dt $$ for $n = 1,2,3 \ldots$ .
Using the computer I found that $a_0 = \dfrac{16{\pi}^4}{15}$ and for $n>0$: $$a_n = -\dfrac{\left(16{\pi}^2n^2-48\right)\sin\left({\pi}n\right)+48{\pi}n\cos\left({\pi}n\right)}{\pi n^5} = \pm \frac{48}{n^4}$$ and $$b_n = 0.$$
But how can we use this to calculate $1-\frac{1}{2^4}+\frac{1}{3^4}-\frac{1}{4^4} \ldots$ ?
EDIT: If someone else should ever need this, here is the answer: In total we have that $$(t^2-\pi^2)^2 = \dfrac{8{\pi}^4}{15} + 48 \sum_{n=1}^\infty \frac{(-1)^n}{n^4} cos(nt)$$
We set $t= 0$ and we are done.
Note that $\sin(n\pi)=0$ and $\cos(n\pi)=(-1)^n$. Can you go on from this?