$a_n$ is the Fourier coefficient of $$f(x) = \left(1 - \frac{|x|}{\pi}\right)^4$$
The answer is infinity, but can someone give an answer that doesn't require explicit computation of the $a_n$? I'm looking for an explanation that applies convergence and divergence results, and maybe even growth and decay conditions of the coefficients, of the Fourier series.
Thanks,
Edit: Two things to note -
1) $a_n$ for $f(x) = \left(1 - \frac{|x|}{\pi}\right)$ decays like $\frac{1}{n^2}$. Can I use this to say something about the decay of $a_n$ for $f(x) = \left(1 - \frac{|x|}{\pi}\right)^4$?
2) $f(x) = \left(1 - \frac{|x|}{\pi}\right)^4$ is not differentiable at zero.
The function: $$ f(x) = \left(1-\frac{|x|}{\pi}\right)^4 $$ vanishes in the endpoints of $[-\pi,\pi]$ together with its (weak) derivatives up to order $3$, hence the Fourier coefficient $$ c_n = \int_{-\pi}^{\pi}f(x)\, e^{-nix}\,dx $$ may be easily computed through integration by parts. That gives in advance that the rate of decay of the Fourier coefficients of $f(x)$ is the same as the rate of decay of the Fourier coefficients of $\left(1-\frac{|x|}{\pi}\right)$. For non-believers, $c_n$ defined as above is: $$ c_n = \frac{8(n^2\pi^2-6)}{\pi^3 n^4}=\Theta\left(\frac{1}{n^2}\right).$$