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This question already has an answer here:

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.

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

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  • $\begingroup$ Also, covered here $\endgroup$ – rtybase Dec 29 '18 at 22:45
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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 $n$.

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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.

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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 $x=0$.

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