Krdytkyu

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

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

Hints: Using elementary calculus, you can show that $displaystylelim_{x to 0}dfrac{1}{sin^2 x} - dfrac{1}{x^2} = L$ where $L$ is some constant (which you can explicitly work out).

Then, you can use the fact that $displaystylelim_{n to infty}x_n = 0$ to show that $displaystylelim_{n to infty}dfrac{1}{x_{n+1}^2} - dfrac{1}{x_n^2} = lim_{n to infty}dfrac{1}{sin^2(x_n)} - dfrac{1}{x_n^2} = lim_{x to 0}dfrac{1}{sin^2 x} - dfrac{1}{x^2} = L$.

Finally, you can use the result $displaystylelim_{n to infty}dfrac{1}{x_{n+1}^2} - dfrac{1}{x_n^2} = L$, 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
$$u_{n+1}-u_ntofrac13$$
as $ntoinfty$.

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,
$$L=lim_{ntoinfty}frac{1/x_n^2}{n}=lim_{ntoinfty}left( frac{1}{x^2_{n+1}} - frac{1}{x^2_{n}}right)=lim_{ntoinfty}left( frac{1}{sin^2(x_n)} - frac{1}{x^2_{n}}right).$$
Finally, in order to find $L$, show that $x_nto 0$ and use the Taylor expansion of $sin(x)$ at $x=0$.