Let $\hat{f}$ be of compact support $[-M,M]$ and $f(z)=\sum_{n=0}^{\infty}a_nz^n$ then I want to prove that:
$$\limsup_{n-> \infty}(n!|a_n|)^{1/n} \le 2 \pi M$$
given that I know that:
$$a_n=\frac{(2 \pi i)^n}{n!}\int_{-M}^{M}\hat{f}(\zeta)\zeta^{n}d\zeta$$
I was trying to use Cauchy's inequalities then I will get the following
$$|a_n| \le \frac{n!}{R^n}||f||_{C}$$
for an appropiate election of the radius of the circle, so I want to pick the circle with center in zero and of radius $M$ but then How can I bound the integral? and Am I right in my reasoning? (I don't know if Paley- Weiner theorem can be useful here)
Thanks a lot in advance.
Another idea is the following:
$$|a_n|\le \frac{(2 \pi )^n}{n!}\int_{-M}^{M}|\hat{f}(\zeta)||\zeta^{n}|d\zeta$$
but then I don't know how to ensure that the integral is less than $M^n$