This question popped up in a recent exam of mine to which I didn't know how to answer. The problem was as following
Given the function $$f(\theta) = \frac{1+\cos(\theta)}{3+2\cos(\theta)}$$ and it's Fourier coefficients $$f_n=\frac{1}{2\pi}\int_{-\pi}^\pi f(\theta)e^{in\theta}\,\mathrm d\theta\;\;\;\;\;\; n\in\mathbb Z$$
Prove that for $n\gg 1$, $\,f_n$ dacays to $0$ faster than $\;n^{-m}$ $\;\forall m > 0$
Find the limit $$\lim_{n\rightarrow +\infty}\frac{|f_{n+1}|}{|f_n|}$$
What I've done for the first question was to find the Laurent series of $f$ which turns out to be, setting $z=e^{i\theta}$ $$f(z) = \frac{1+{z\over 2}+\frac{1}{2z}}{3+z+{1\over z}} = \frac{1}{2}\frac{z^2+2z+1}{z^2+3z+1} = \frac{1}{2}\frac{(z+1)^2}{(z+1)^2+z} \\ = \frac{(z+1)^2}{2(z+1)^2}\frac{1}{1+\frac{z}{(z+1)^2}} = \frac{1}{2}\sum_{k=0}^\infty (-1)^k\frac{z^k}{(z+1)^{2k}}$$ I came back to the geometric series because clearly $\frac{z}{(z+1)^2}\lt 1$. Going back to the coefficients $$f_n = \frac{1}{2\pi}\oint\limits_{\partial D(0,1)}\frac{1}{2}\sum_{k=0}^\infty (-1)^n\frac{z^k}{(z+1)^{2k}}z^n\,\frac{\mathrm dz}{iz} \\ f_n = -\frac{i}{4\pi}\sum_{k=0}^\infty \oint\limits_{\partial D(0,1)}\frac{z^{n+k-1}}{(z+1)^{2k}}\,\mathrm dz$$
From here I got suck and didn't know were to go! I thought about using the residue theorem to evaluate the integral, but depending on the value of $k$, which is being summed over, the order of the pole $z=-1$ rises. So I'm here to ask you if my method was indeed good and how to proceed further.
As a disclamer: I'm a physics student and this was a question on my exam on mathematical methods for physicists. Although I'm quite interested in more advanced (and more detailed) math than what my course gave me, I want you to keep in mind that my knowledge of functional analysis is that of a undergraduate level physicist! If you want to give me a more insightful answer (mathematically speaking) I'll be quite happy to have it! But if there's a more direct and simpler solution, that would be better.
I think you know residue theorem.
Let $\displaystyle{f(\theta)= \frac{1}{2}\frac{z^2+2z+1}{z^2+3z+1}}$ where $z=e^{i\theta}$.
Let $z^2+3z+1=(z-c)(z-\overline{c})$ where $c=\frac{-3+\sqrt 5}{2}$.
Then, $$ \begin{align} f_n &=\frac{1}{2\pi}\int_{-\pi}^\pi f(\theta)e^{in\theta}\,\mathrm d\theta \\ &=\frac1{2\pi}\oint_{|z|=1}\frac{1}{2}\frac{z^2+2z+1}{z^2+3z+1}z^n\frac{dz}{iz} \qquad{\text{let }z=e^{i\theta}} \\ &=\frac1{2\pi i}\oint_{|z|=1}\frac{1}{2}\frac{z^2+2z+1}{z^2+3z+1}z^{n-1}dz \\ &=\frac1{2\pi i}\cdot2\pi i\operatorname*{Res}_{z=c}\frac{1}{2}\frac{z^2+2z+1}{z^2+3z+1}z^{n-1} \qquad{\text{for $n\ge 1$}} \\ &=\operatorname*{Res}_{z=c}\frac{1}{2}\frac{(z+1)^2}{(z-c)(z-\overline{c})}z^{n-1} \\ &=\frac{1}{2}\frac{(c+1)^2}{c-\overline{c}}c^{n-1} \\ &=\underbrace{\frac{(c+1)^2}{2\sqrt5}}_{=k}\cdot c^{n-1} \\ &=k\cdot c^{n-1} \end{align} $$
Clearly, since $|c|<1$, $|f_n|$ exhibits exponential decay as $n\to\infty$, which is faster than any $n^{-m}$ decay (I call it polynomial decay). This answers your first bullet point.
For the second bullet point, $$ \begin{align} \lim_{n\rightarrow +\infty}\frac{|f_{n+1}|}{|f_n|} &=\lim_{n\rightarrow +\infty}\frac{|kc^n|}{|kc^{n-1}|} \\ &=\lim_{n\rightarrow +\infty} |c| \\ &= |c|\\ &\color{red}{=\frac{3-\sqrt 5}{2}} \end{align} $$