# Limit (5)

• July 29th 2009, 09:06 PM
NonCommAlg
Limit (5)
Define the sequence $\{x_n \}$ by $x_1=\frac{\pi}{2}$ and $x_{n+1}=\sin x_n, \ n \geq 1.$ Evaluate $\lim_{n\to\infty}\sqrt{n} x_n.$
• July 30th 2009, 05:41 AM
DeMath
Non-rigorous proff (solution)
Quote:

Originally Posted by NonCommAlg
Define the sequence $\{x_n \}$ by $x_1=\frac{\pi}{2}$ and $x_{n+1}=\sin x_n, \ n \geq 1.$ Evaluate $\lim_{n\to\infty}\sqrt{n} x_n.$

Assume that the limit exists and is equal to $A$. Substitute $A$ in the recurrence relation $\mathop{\lim }\limits_{n \to \infty } \frac{A}{{\sqrt {n + 1} }} = \mathop {\lim }\limits_{n \to \infty } \frac{{\sin A}}{{\sqrt n }}$. Then use the Taylor's formula for sinus.
So we have

${A^2} = 6\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{{\sqrt n }} - \frac{1}{{\sqrt {n + 1} }}} \right)\sqrt {{n^3}} = \ldots = 3.$

$\mathop {\lim }\limits_{n \to \infty } \sqrt n {x_n} = \sqrt 3 .$
• August 6th 2009, 06:32 PM
luobo
Define $y_n=\sqrt{n} x_n$, then $y_{n+1}=\sqrt{n+1} \sin \frac {y_n}{\sqrt{n}}$

If
$\lim_{n\to\infty}y_n$ exists, it would indicate:
(1)
$\lim_{n\to\infty} \frac {y_n}{\sqrt{n}} = 0$
(2) Series $z_1=y_1$, $z_{n+1}=y_{n+1}-y_n$ converges, since $\sum_{i=1}^{n+1} z_i=$ $y_{n+1}$.

$z_{n+1}=\sqrt{n+1} \sin \frac{y_n}{\sqrt{n}}-y_n$
$= \sqrt{n+1} \{ \sin \frac{y_n}{\sqrt{n}} - \frac{y_n}{\sqrt{n}} + \frac{y_n^3}{6n\sqrt{n}} \}+\frac{y_n}{n} \{\sqrt{n^2+n}-n-\frac{y_n^2}{6}\sqrt{\frac{n+1}{n}} \}$

The first term converges since it is $O(\frac{1}{n^2})$. The second term is $\frac{y_n(3-y_n^2)}{6}\frac{1}{n}+O(\frac{1}{n^2})$.

For the second term to converge, it requires $\lim_{n\to\infty}y_n(3-y_n^2)=0$ (limit comparison test against the p-series), so $\lim_{n\to\infty}y_n=\sqrt{3}$, noting $\lim_{n\to\infty}y_n>0$. This also confirms $\lim_{n\to\infty}y_n$ does exist.

This method also applies to limit (1).

• August 8th 2009, 06:05 AM
halbard
It is clear that $x_n>0$ and that $x_{n+1}=\sin x_n for all $n\geq1$. Therefore $x_n$ is decreasing, bounded below and converges to $L$, say.

Since $L=\sin L$ follows from the recurrence we must have $L=0$. Thus $x_n\to 0$.

Now $x_n\to 0$ implies that $\frac{x_{n+1}}{x_n}=\frac{\sin x_n}{x_n}\to 1$.

Now consider $\frac1{x_{n+1}^2}-\frac1{x_n^2}=\frac{x_n^2-\sin^2 x_n}{x_n^2x_{n+1}^2}$. By Taylor's theorem, $\sin^2 x=x^2-\frac{x^4}3+o(x^5)$ and so

$
\frac1{x_{n+1}^2}-\frac1{x_n^2}=\frac{x_n^4/3+o(x_n^5)}{x_n^2x_{n+1}^2}=\frac{x_n^2}{x_{n+1}^2 }\left(\frac13+o(x_n)\right)\to\frac13$
.

Thus $\frac{(n+1)-n}{\frac1{x_{n+1}^2}-\frac1{x_n^2}}\to 3$. By the Stolz-Cesaro theorem, $\frac n{\frac1{x_n^2}}\to 3$, i.e. $nx_n^2\to 3$. Thus $\lim_{n\to\infty}\surd n x_n=\surd 3$.

By the way, the limit $\lim_{n\to\infty}\left(\frac1{\sin^2 x_n}-\frac1{x_n^2}\right)$ can be obtained from l'Hopitals rule. Since $\lim_{x\to 0}\frac{\sin x}x=1$ we only need to find $\lim_{x\to 0}\left(\frac{x^2-\sin^2 x}{x^4}\right)$. Thus

$\lim_{x\to 0}\left(\frac1{\sin^2 x}-\frac1{x^2}\right)=\lim_{x\to 0}\frac{x^2-\sin^2 x}{x^4}=\lim_{x\to 0}\frac{2x-\sin 2x}{4x^3}$ $=\lim_{x\to 0}\frac{2-2\cos 2x}{12x^2}=\lim_{x\to 0}\frac{4\sin 2x}{24x}=\lim_{x\to 0}\frac{8\cos 2x}{24}=\frac13$.
• August 12th 2009, 05:24 AM
luobo
Quote:

Originally Posted by halbard
It is clear that $x_n>0$ and that $x_{n+1}=\sin x_n for all $n\geq1$. Therefore $x_n$ is decreasing, bounded below and converges to $L$, say.

Since $L=\sin L$ follows from the recurrence we must have $L=0$. Thus $x_n\to 0$.

Now $x_n\to 0$ implies that $\frac{x_{n+1}}{x_n}=\frac{\sin x_n}{x_n}\to 1$.

Now consider $\frac1{x_{n+1}^2}-\frac1{x_n^2}=\frac{x_n^2-\sin^2 x_n}{x_n^2x_{n+1}^2}$. By Taylor's theorem, $\sin^2 x=x^2-\frac{x^4}3+o(x^5)$ and so

$
\frac1{x_{n+1}^2}-\frac1{x_n^2}=\frac{x_n^4/3+o(x_n^5)}{x_n^2x_{n+1}^2}=\frac{x_n^2}{x_{n+1}^2 }\left(\frac13+o(x_n)\right)\to\frac13$
.

Thus $\frac{(n+1)-n}{\frac1{x_{n+1}^2}-\frac1{x_n^2}}\to 3$. By the Stolz-Cesaro theorem, $\frac n{\frac1{x_n^2}}\to 3$, i.e. $nx_n^2\to 3$. Thus $\lim_{n\to\infty}\surd n x_n=\surd 3$.

By the way, the limit $\lim_{n\to\infty}\left(\frac1{\sin^2 x_n}-\frac1{x_n^2}\right)$ can be obtained from l'Hopitals rule. Since $\lim_{x\to 0}\frac{\sin x}x=1$ we only need to find $\lim_{x\to 0}\left(\frac{x^2-\sin^2 x}{x^4}\right)$. Thus

$\lim_{x\to 0}\left(\frac1{\sin^2 x}-\frac1{x^2}\right)=\lim_{x\to 0}\frac{x^2-\sin^2 x}{x^4}=\lim_{x\to 0}\frac{2x-\sin 2x}{4x^3}$ $=\lim_{x\to 0}\frac{2-2\cos 2x}{12x^2}=\lim_{x\to 0}\frac{4\sin 2x}{24x}=\lim_{x\to 0}\frac{8\cos 2x}{24}=\frac13$.

Great. Limit (5) and Limit (1) are problems of the same type. So can be solved in the same way.

Can Stolz-Cesaro theorem be used for complex numbers? What happens if $x$ is in complex domain?
• August 12th 2009, 06:43 AM
luobo
Quote:

Originally Posted by NonCommAlg
Define the sequence $\{x_n \}$ by $x_1=\frac{\pi}{2}$ and $x_{n+1}=\sin x_n, \ n \geq 1.$ Evaluate $\lim_{n\to\infty}\sqrt{n} x_n.$

It would be interesting to investigate when $x$ is defined in complex domain, i.e.

Define the sequence $\{z_n \}$ by $z_1=1+i$ and $z_{n+1}=\sin z_n, \ n \geq 1.$ Assuming the limit exists, evaluate $\lim_{n\to\infty}\sqrt{n} z_n$, and find how the initial value of $z_1$ influences of the value of the limit.

Actually it can be found that the limit also exists except for some initial values of $z_1$. If the limit does exist, then it is either $\sqrt{3}$ or $-\sqrt{3}$ for $z_1 \neq 0$, depending on the initial value of $z_1$, even when $z_n$ is a complex number.