Given the function $f(x) = 1 - \dfrac{|x|}{\pi}$, I had computed its Fourier coefficients, using integration by parts and got:
$$ a_n = \begin{cases} 0, & \text{for $n$ even}, \\[6pt] \dfrac{2}{n^2\pi^2}, & \text{for $n$ odd}, \\[6pt] \dfrac{1}{2}, & \text{for }n = 0. \end{cases} $$
So, I know the coefficients decay like $\dfrac{1}{n^2}$.
What if I now took $f(x)$ to be $\left(1 - \dfrac{|x|}{\pi}\right)^4$.
What is $\lim\limits_{n\to\infty}$ $n^3a_n$?
This is part 2 of a question that I am working on, and the question does not ask for a detailed proof but rather just the answer and the reasoning behind it.
Thanks,
Edit: Of course, if we iterate the integration by parts enough times, we can certainly compute the Fourier coefficients for the new function. But this is quite a bit of computation and I'm sure is not the point of the question - part 1 of the question already asked for explicit computation of the $a_n$'s. I'm not sure how to proceed, without direct computation. Perhaps there are decay / growth conditions of the $a_n$'s and convergence properties of the Fourier series that I am not aware of...
If $f$ has Fourier series $\sum a_n \cos nt+b_n\sin nt$, then $f'$ has Fourier series $ \sum na_n \cos nt-na_n\sin nt$. (I treat these as formal series, making no claims about their convergence). So, whenever you see Fourier coefficients multiplied by powers of $n$, think of derivatives.
So the natural idea is to consider $f'''$, but the differentiability of $f$ isn't that great... which brings to mind the second idea.
Suppose $n^3 a_n$ is bounded. Then the series for $f'$ has coefficients bounded by $C/n^2$. (In your case, all $b_n$ vanish since the function is even.) It follows $f'$ is represented by a uniformly convergent Fourier series, and is therefore continuous.
However, for $f(x)=(1-|x|/\pi)^4$, the derivative $f'$ is not continuous at $0$. Hence, $\lim n^3 a_n$ does not exist.