Let $0<x_0<1$ if $x_{n+1} = sin(x_n)$ show that $\lim_{n\to\infty} \frac{x_n}{\sqrt{3}/n} = 1$
Asked
Active
Viewed 93 times
-1
-
thanks, I already solved the problem with Stolz–Cesàro theorem. . – Sergio MNZ Nov 18 '18 at 04:39
-
1The first appearance of this question seems to be MSE question "Convergence of $\sqrt{n}x_{n}$ where $x_{n+1} = \sin(x_{n})$" from 2010. – Somos Nov 18 '18 at 05:29
1 Answers
1
We have that:
$$\frac{x_n}{(\frac {\sqrt3}{n})}=\frac{nx_n}{\sqrt3}$$
Also, since for $0 <x <1$ we have that $x >\sin x$, $$\lim_{n\to \infty}{x_n}=0$$
You can do the rest.

Rhys Hughes
- 12,842