<s>[INST] Convergence of $\sqrt{n}x_{n}$ where $x_{n+1} = \sin(x_{n})$ Consider the sequence defined as
$x_1 = 1$
$x_{n+1} = \sin x_n$
I think I was able to show that the sequence $\sqrt{n} x_{n}$ converges to $\sqrt{3}$ by a tedious elementary method which I wasn't too happy about.
(I think I did this by showing that $\sqrt{\frac{3}{n+1}} < x_{n} < \sqrt{\frac{3}{n}}$, don't remember exactly)
This looks like it should be a standard problem. 
Does anyone know a simple (and preferably elementary) proof for the fact that the sequence $\sqrt{n}x_{n}$ converges to $\sqrt{3}$?
 [/INST] If your sequence is of the form
$$x_{n+1} = f(x_n)$$
with
$$f(x) = x( 1 - c x^{\alpha} + \textrm{h. o. t} )$$
then notice ( e.g. ) that
$$\frac{1}{f(x)^{\alpha}} = \frac{1}{x^{\alpha}} +  \alpha c+ \textrm{h. o. t.}$$
and so
$$\frac{1}{x_{n+1}^{\alpha}} - \frac{1}{x_n^{\alpha}} \to  \alpha c$$
and with Cesaro-Stolz we get
$$\frac{1}{n x_n^{\alpha} } \to  \alpha c$$
In the particular case $f(x) = \sin x= x( 1 - \frac{x^2}{6} + \cdots)$, we get
$$\frac{1}{n x_n^2} \to 2 \cdot \frac{1}{6}  = \frac{1}{3}$$
$\bf{Added}$ We can get effective estimates for the sequence
$x_{n+1} = \sin x_n$ as follows
Consider the Taylor expansion
$$\frac{\frac{3}{\sin^2 x} - \frac{3}{x^2} - 1}{x^2} = \frac{1}{5} + \frac{2 x^2}{63} + \cdots$$
where all the remaining coefficients are positive. We conclude that for $0<|x|< \pi$ we get
$$\frac{\frac{3}{\sin^2 x} - \frac{3}{x^2} -1}{x^2} > \frac{1}{5}$$
and for $0 < |x|< 1$ we have
$$\frac{\frac{3}{\sin^2 x} - \frac{3}{x^2} -1}{x^2} <  
\frac{\frac{3}{\sin^2 x} - \frac{3}{x^2} -1}{x^2} \ _{x=1}=0.2368\ldots < \frac{1}{4}$$
Therefore we have for $0< |x|\le 1$
$$\frac{3}{x^2} + 1 + \frac{x^2}{5}<\frac{3}{\sin^2 x} < \frac{3}{x^2} + 1 + \frac{x^2}{4}$$
Now for the sequence $y_n= \frac{3}{x_n^2}$ we get the inequalities
$$y_n+ 1 + \frac{3}{5 y_n} < y_{n+1} < y_n + 1 + \frac{3}{4 y_n}$$
and we can use this to get  estimates for $y_n$ and so $x_n$.
</s>