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 $\sqrt{3}$?
 A: 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 $x_n$.
A: This problem can be found in Kaczor, Nowak: Problems in Mathematical Analysis I, Real Numbers, Sequences and Series. I'll copy their solution here.

Problem 2.5.22, p.50, a solution is given on p.215.

Problem 2.5.22. The sequence $(a_n)$ is defined inductively as follows:
  $$0<a_1<\pi \qquad a_{n+1}=\sin a_n \text{ for }n\ge 1$$
  Prove that 
  $\lim\limits_{n\to\infty} \sqrt n a_n = \sqrt3$.

Solution:
It is easy to see that the sequence $(a_n)$ is monotonically
decreasing to zero. Moreover, an application of I'Hospital's rule gives
$$\lim\limits_{x\to 0}\frac{x^2-\sin^2x}{x^2\sin^2x}=\frac13.$$
Therefore
$$\lim\limits_{n\to\infty}\left(\frac1{a_{n+1}^2}-\frac1{a_n^2}\right)=\frac13$$
Now, by the result in Problem 2.3.14, $\lim\limits_{n\to\infty} na_n^2 = 3$.

Problem 2.3.14, 
p.38, a solution is given on p.184.

Problem 2.3.14. Prove that if $(a_n)$ is a sequence for which
  $$\lim\limits_{n\to\infty}(a_{n+1}-a_n)=a$$
  then
  $$\lim\limits_{n\to\infty}\frac{a_n}n=a.$$

Solution: In Stolz theorem we set $x_{n}=a_{n+1}$ and $y_n=n$.
Formulation of Stolz theorem in this book is the following

Let $(x_n)$, $(y_n)$ be two sequences that satisfy the conditions:  
  
  
*
  
*$(y_n)$ strictly increases to $+\infty$,
  
*$$\lim\limits_{n\to\infty} \frac{x_n-x_{n-1}}{y_n-y_{n-1}}=g.$$
  
  
  Then $$\lim\limits_{n\to\infty} \frac{x_n}{y_n}=g.$$

For Stolz-Cesaro theorem see also this question: Stolz-Cesàro Theorem 
Perhaps it is also worth mentioning that there are two equivalent forms of Stolz-Cesaro theorem: see e.g. this answer.
A: Before getting into the details, let me say: The ideas I'm talking about, including this exact example, can be found in chapter 8 of Asymptotic Methods in Analysis (second edition), by N. G. de Bruijn. This is a really superb book, and I recommend it to anyone who wants to learn how to approximate quantities in "calculus-like" settings. (If you want to do approximation in combinatorial settings, I recommend Chapter 9 of Concrete Mathematics.) 
Also, this isn't just about $\sin$. Let $f$ be a function with $f(0)=0$ and $0 \leq f(u) < u$ for $u$ in $(0,c]$ 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.
Having said that, the answer to your question. On $[0,1]$, we have
$$\sin x=x-x^3/6+O(x^5).$$
Setting $y_n=1/x_n^2$, we have
$$1/x_{n+1}^2 = x_n^{-2} \left(1-x_n^2/6+O(x_n^4) \right)^{-2} =  1/x_n^2 + 1/3 + O(x_n^2)$$
so
$$y_{n+1} = y_n + 1/3 + O(y_n^{-1}).$$
We see that
$$y_n = \frac{n}{3} + O\left( \sum_{k=1}^n y_k^{-1} \right)$$
and
$$\frac{1}{n}y_n = \frac{1}{3} + \frac{1}{n} O\left( \sum_{k=1}^n y_k^{-1} \right)$$
Since we already know that $x_n \to 0$, we know that $y_n^{-1} \to 0$, so the average goes to zero and we get $\lim_{n \to \infty} y_n/n=1/3$. Transforming back to $\sqrt{n} x_n$ now follows by the continuity of $1/\sqrt{t}$.  
