Proving $\sqrt{n}(x_n)$ converges when $x_n = \sin(x_{n-1}), x_1=1$ [duplicate]

This is a problem that showed up on a qual exam that I have been stuck on for a while.

Let $$$$x_n = \sin(x_{n-1}), x_1 = 1$$$$ Prove $$\lim_{n \rightarrow \infty} \sqrt{n} x_n$$ exists and compute its value. The problem gave the following hint: Show that $$$$\frac{1}{x^2_{n+1}} - \frac{1}{x^2_{n}}$$$$ converges to a constant.

I have shown that $$x_n$$ converges to $$0$$, but I am unsure on how to begin bounding $$\sqrt{n}x_n$$ or how the hint is helpful in this problem. I have tried using the MVT to show $$\sin(x_n) \rightarrow 0$$ faster than $$\sqrt{n} \rightarrow \infty$$, but I didn't get far. Any help will be appreciated.

marked as duplicate by rtybase, Community♦Dec 30 '18 at 0:16

• Also, covered here – rtybase Dec 29 '18 at 22:45

Hints: Using elementary calculus, you can show that $$\displaystyle\lim_{x \to 0}\dfrac{1}{\sin^2 x} - \dfrac{1}{x^2} = L$$ where $$L$$ is some constant (which you can explicitly work out).
Then, you can use the fact that $$\displaystyle\lim_{n \to \infty}x_n = 0$$ to show that $$\displaystyle\lim_{n \to \infty}\dfrac{1}{x_{n+1}^2} - \dfrac{1}{x_n^2} = \lim_{n \to \infty}\dfrac{1}{\sin^2(x_n)} - \dfrac{1}{x_n^2} = \lim_{x \to 0}\dfrac{1}{\sin^2 x} - \dfrac{1}{x^2} = L$$.
Finally, you can use the result $$\displaystyle\lim_{n \to \infty}\dfrac{1}{x_{n+1}^2} - \dfrac{1}{x_n^2} = L$$, to show that $$\dfrac{1}{x_n^2} \approx Ln + \text{const}$$ for large $$n$$.
Let $$u_n=\frac1{x_n^2}.$$ Then $$u_{n+1}=\frac1{\sin^2 (1/\sqrt{u_n})}=\frac{u_n}{1-1/(3u_n)+O(u_n^{-2})} =u_n+\frac13+O(u_n^{-1}).$$ Then $$u_{n+1}-u_n\to\frac13$$ as $$n\to\infty$$.
From this you can deduce that $$u_n=n/3+o(n)$$ etc.
Why the given hint is useful. It suffices to find the limit $$L$$ of $$\frac{1/x_n^2}{n}$$, then $$\sqrt{n}x_n$$ is convergent to $$1/\sqrt{L}$$. Now, by Stolz-Cesaro Theorem, $$L=\lim_{n\to\infty}\frac{1/x_n^2}{n}=\lim_{n\to\infty}\left( \frac{1}{x^2_{n+1}} - \frac{1}{x^2_{n}}\right)=\lim_{n\to\infty}\left( \frac{1}{\sin^2(x_n)} - \frac{1}{x^2_{n}}\right).$$ Finally, in order to find $$L$$, show that $$x_n\to 0$$ and use the Taylor expansion of $$\sin(x)$$ at $$x=0$$.