# recursively defined trigonometric sequence and Cesàro means

So, we have a function

$$f(x) = \frac{3\sin(x)}{2 + \cos(x)}$$

and we fix a point $$x_0 \in \left(0, \frac{2\pi}{3}\right]$$ and define a sequence by setting $$x_{n+1} = f(x_n)$$

Now I need to show that $$\lim_{n\to\infty} n^{1/4}x_n = 45^{1/4}$$

I am pretty sure this has something to do with Cesàro means. In a previous part of the problem we proved that the sequence is strictly decreasing and has a limit of zero, and we also found the 5th order Taylor approximation of the function (centered at 0), which happens to be $$p_5(f, x) = x - \frac{x^5}{180}$$ but we have not proved very many theorems involving Taylor polynomials (and none that seem to apply), and the section in Rudin (our textbook) is quite brief. Any help or suggestions would be welcome.

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kind of a brutal homework problem :) – Zarrax Dec 15 '10 at 15:30

Check David Speyer's excellent answer here: Convergence of $\sqrt{n}x_{n}$ where $x_{n+1} = \sin(x_{n})$
Let $f$ be a function $0 \leq f(u) > \leq u$ on an interval $[0,c]$ then the sequence $x_n:=f(f(f(\cdots > f(c)\cdots)$ approaches $0$. If $f(u)=u-a u^{k+1} + O(u^{k+2})$ (with $a>0$) then $x_n \approx \alpha n^{-1/k}$ and you can prove that by the same methods here.