# Convergence of \sqrt{n}x_{n}\sqrt{n}x_{n} where x_{n+1} = \sin(x_{n})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 $\sqrt{3}$?

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

Setting $y_n=1/x_n^2$, we have

so

We see that

and

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 $1/\sqrt{t}$.