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Suppose $$x_2 \in \left(0,\frac{\pi}{2}\right), x_{n+1}=\left(1-\frac{1}{n}\right)\sin x_n(n \ge 2).$$ Evaluate $\lim\limits_{n \to \infty} nx_n.$

Note that $$x_{n+1}=\left(1-\frac{1}{n}\right)\sin x_n\le \sin x_n\le x_n,$$ and $$|x_{n+1}|=\left|\left(1-\frac{1}{n}\right)\sin x_n\right|\le 1,$$ it follows that $\{x_n\}$ converges. Therefore, we can readily obtain $x_n \to 0$.

As for the limit wanted, we can consider apply Stolz theorem. But it's too complicated.

mengdie1982
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    @UmbertoP. Observe that the range of the sine function is $[0,1] \in [0,\frac\pi2]$, and the factor $1-\frac1n \in (0,1)$, so $x_n$ decreases strictly to some number between $0$ and $1$. But this limit can't be positive, as you can apply the same process to get a smaller number. – GNUSupporter 8964民主女神 地下教會 Dec 21 '20 at 13:11
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    I wonder if there's an easy closed form expression. I tried with Python: def x(n): return (1-1/n)*math.sin(x(n-1)) if n >= 3 else 0.5 and n*x(n) for $n = 100,500$ and $995$. I got 0.9064076753543608, 0.9054104297085305 and 0.9052871869587333 respectively. – GNUSupporter 8964民主女神 地下教會 Dec 21 '20 at 13:17
  • @GNUSupporter 8964民主女神 地下教會 Your argument doesn't make sense. "You can apply the same process to get a smaller number" doesn't mean the limit is $0$. However, it is clear that $ x_n < \frac{1}{n-1}$ for $n \geq 3$. which shows that $x_n \to 0.$ – Adam Rubinson Dec 21 '20 at 13:26
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    @GNUSupporter8964民主女神地下教會. In the same spirit, using $x_2=\frac \pi 2$, compute $n^2 x_n$,It is a straight line with a slope equal to $0.90406$ $(\sigma=0.00030)$. This linear behaviour is observed for several $x_2$. – Claude Leibovici Dec 21 '20 at 13:35
  • Can we prove at least that the limit $\lim n x_n$ exist? – Peanut Dec 21 '20 at 13:46
  • The existence of the limit is obvious, hopes for a closed form are... erm... very optimistic, so I'm not sure what this question is about. –  Dec 21 '20 at 14:02
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    Something to do with Contraction Mapping/ Banach Fixed-point theorem? Also note that $\lim_{n \to \infty}nx_n = \lim_{n \to \infty}(n-1)x_n,$ the latter being the limit of the nicer-looking sequence $\left(x_2,\ \sin x_2,\ 2\sin\left(\frac12 \sin x_2\right),\ 3 \sin\left( \frac23 \sin\left(\frac12 \sin x_2\right) \right), 4\sin\left(\frac34\sin\left( \frac23 \sin\left(\frac12 \sin x_2\right)\right) \right), ... \right).$ – Adam Rubinson Dec 21 '20 at 14:05

1 Answers1

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Work in progress. We can conclude that $(nx_n)_{n\in\mathbb{N}}$ converges ...

Using induction we can show that $x_n \geq 0$ and, thus, $nx_n\geq0$.

Function $\sin{x}\geq0$, for $x\in\left[0,\frac{\pi}{2}\right]$. As a result $\sin{x_2}>0$ and $$0\leq x_3=\frac{\sin{x_2}}{2}\leq 1<\frac{\pi}{2} \Rightarrow x_3\in\left[0,\frac{\pi}{2}\right]$$ And using the induction hypothesis for $x_n\in \left[0,\frac{\pi}{2}\right]$ $$0\leq x_{n+1}=\left(1-\frac{1}{n}\right)\sin{x_n}\leq 1<\frac{\pi}{2} \Rightarrow x_{n+1}\in\left[0,\frac{\pi}{2}\right]$$

From

$$|x_{n+1}|= \left|1-\frac{1}{n}\right||\sin{x_n}|\leq \left|1-\frac{1}{n}\right||x_n|\leq \\ \left|1-\frac{1}{n}\right|\left|1-\frac{1}{n-1}\right||\sin{x_{n-1}}|\leq ... \\ \leq \left|1-\frac{1}{n}\right|\left|1-\frac{1}{n-1}\right|\left|1-\frac{1}{n-2}\right|...\left|1-\frac{1}{2}\right||\sin{x_2}|=\frac{|\sin{x_2}|}{n}$$ Thus $$|(n+1)x_{n+1}|\leq \left(1+\frac{1}{n}\right)|\sin{x_2}|$$ and $(nx_n)_{n\in\mathbb{N}}$ is bounded.

Now:

$$0\leq \frac{(n+1)x_{n+1}}{nx_{n}}=\left(1+\frac{1}{n}\right)\left(1-\frac{1}{n}\right) \frac{\sin{x_n}}{x_n} \leq 1-\frac{1}{n^2}<1$$

and $(nx_n)_{n\in\mathbb{N}}$ is descending.

rtybase
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    Well done. Further the limit must be $\le 1$. – Peanut Dec 21 '20 at 14:13
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    I may have missed something, but the fact that $(nx_n)$ converges seems straightforward, since $$x_{n+1}=\left(1-\frac{1}{n}\right)\sin x_n \leq \left(1-\frac{1}{n}\right) x_n $$

    So $$nx_{n+1} \leq (n-1)x_n$$ so $(nx_{n+1})$ is decreasing and positive, so it converges.

     – TheSilverDoe Dec 21 '20 at 17:44
  • @TheSilverDoe plus bounded and positive ... 2 important details. And $nx_n$. – rtybase Dec 21 '20 at 17:52
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    @TheSilverDoe thank you for being a great human being! Merry Christmas! – rtybase Dec 21 '20 at 17:57
  • @rtybase Can you give more details ? You can prove in two ligns that $(nx_n)$ is convergent, or did I miss something ? Why do you need such a long proof ? And I don't understand your last comment... but merry christmas anyway... – TheSilverDoe Dec 21 '20 at 21:48
  • @TheSilverDoe the proof is not long, just wordy and it's a work in progress. When I joined MSE, almost 9 years ago, I was "instructed" not be lazy and provide as many details as possible (different people are reading the answers). If you look at the edit history, you'll see I started with $|nx_n|$ (which converges, but actual $nx_n$ can oscillate between the positive and negative values). Then, obviously, it is important to show that $nx_n$ is positive (1st section), then decreasing (3rd section, one line). – rtybase Dec 21 '20 at 22:19
  • The 2nd section is a leftover from the previous edits and another easy way to show the convergence of $x_n$ to zero. – rtybase Dec 21 '20 at 22:19
  • Also, if you look at the 3rd section, we can multiply $\frac{(n+1)x_{n+1}}{nx_n}$, terms will cancel each other and the result will be $\sim \prod\limits_{n}\left(1-\frac{1}{n^2}\right)$ which can be used to approximate the limit or even find it, since $\frac{\sin{x_n}}{x_n} \sim 1$ ... as I wrote, work in progress. – rtybase Dec 21 '20 at 22:25
  • Re "don't understand your last comment", well ... your comment and downvote came at the same time. So, it was a thank you. Of course you can prove me wrong by downvoting me ;) ... that will teach me a lesson. – rtybase Dec 21 '20 at 22:27