299. A class of the recurrent sequencies.

by Virgil Nicula, Jul 18, 2011, 2:14 AM

Theorem. Let $f:(0,b)\rightarrow (0,1)$ be a continue function for which $\lim_{x\to 0}f(x)=1$ and exists $r>0$ so that $\lim_{x\to 0}\frac {1-f(x)}{x^r}=l>0$ .

Then the recurrent sequence $x_1\in (0,b)$ and $x_{n+1}=x_nf\left(x_n\right)$ for any $n\in\mathbb N^*$ has the properties $x_n\rightarrow 0$ and $nx_n^r\rightarrow \frac {1}{rl}$ .


Proof. Let $g:[0,b)\rightarrow [0,b)$ be a function for which $g(0)=0$ and $g(x)=xf(x)$ for any $x\in (0,b)$ . Observe that the function $g$ is continue and for any

$x\in (0,b)$ we have $g(x)\in (0,x)\subset (0,b)$ ($g$ is a dynamic function). Thus, the sequence $\left(x_n\right)_n$ is defined correctly : $x_1\in (0,b)$ and $x_{n+1}=g\left(x_n\right)$ ,

where $n\in\mathbb N^*$ . For any $n\in\mathbb N^*$ we have $0<x_{n+1}=g\left(x_n\right)<x_n$ , what means the sequence $x_n\searrow x\in [0,b)$ (strict decreasing). For $n\rightarrow\infty$

in the recurrent relation obtain that $x=g(x)$ , i.e. $x=0$ . Indeed, if $x\ne 0$ , then $f(x)=1$ , absurd. In conclusion, $\boxed {\ x_n\searrow 0\ }$ . Observe that

$\lim_{n\to\infty}\frac {x_{n+1}}{x_n}=\lim_{n\to\infty}\frac {g\left(x_n\right)}{x_n}=\lim_{n\to\infty}f(x_n)\stackrel{(\mathrm{Heine})}{=}\lim_{x\searrow 0}f(x)=1$ and $\lim_{n\to\infty}\left(\frac {1}{x_{n+1}^r}-\frac {1}{x_n^r}\right)=$ $\lim_{n\to\infty}\frac {1}{x_n^r}\cdot \frac {\left(\frac {x_n}{x_{n+1}}\right)^r-1}{\frac {x_n}{x_{n+1}}-1}\cdot \left(\frac {x_n}{x_{n+1}}-1\right)=$

$r\cdot \lim_{n\to\infty}\frac {1}{x_n^r}\cdot\left(\frac {x_n}{x_{n+1}}-1\right)=$ $r\cdot \lim_{n\to\infty}\frac {x_n-x_{n+1}}{x_n^{r+1}}\cdot\frac {x_n}{x_{n+1}}=$ $r\cdot\lim_{n\to\infty}\frac {x_n-x_{n+1}}{x_n^{r+1}}=$ $r\cdot\lim_{n\to\infty}\frac {x_n-x_nf\left(x_n\right)}{x_n^{r+1}}=$ $r\cdot\lim_{n\to\infty}\frac {1-f\left(x_n\right)}{x_n^{r}}\stackrel{\mathrm{Heine}}{=}$

$r\cdot\lim_{t\searrow 0}\frac {1-f(t)}{t^r}=rl\implies$ $\lim_{n\to\infty}\left(\frac {1}{x_{n+1}^r}-\frac {1}{x_n^r}\right)=rl$ . From the Stolz's theorem
obtain that $\lim_{n\to\infty}\frac {1}{nx_n^r}=\lim_{n\to\infty}\frac {\frac {1}{x_n^r}}{n}=rl$ , i.e. $\boxed{\ \lim_{n\to\infty}nx_n^r=\frac {1}{rl}\ }$ .

Particular case. For the function $f(x)=\frac {\sin x}{x}\ :\left(0,\frac {\pi}{2}\right)\rightarrow (0,1)$ we have $\lim_{x\to 0}=1$ and $\lim_{x\to 0}\frac {1-f(x)}{x^2}=\lim_{x\to 0}\frac {x-\sin x}{x^3}=\frac 16$ , i.e. $r=2$ and $l=\frac 16$ . In conclusion, the sequence $x_{n+1}=x_n\cdot\frac {\sin x_n}{x_n}$ ,

i.e. $\boxed{x_{n+1}=\sin x_n}$ has the properties $x_n\rightarrow 0$ and $nx_n^2\rightarrow \frac {1}{2\cdot\frac 16}=3$ , i.e. $\boxed{x_n\sqrt n\rightarrow\sqrt 3}$ , Vezi
articolul meu din G.M.B. - O clasa de siruri recurente.


$2.2.\blacktriangleright$ Examples. Ascertain the limits of the following strings $x_n,\ y_n,\ z_n$, where $n\in N^*$ :

1. $x_1\in (0,1),\ x_{n+1}=x_n(1-x_n),\ y_n=nx_n,\ z_n=\sum_{k=1}^n x^2_k\ .$

2. $x_1>0,\ x_{n+1}=\ln (1+x_n),\ y_n=nx_n\ .$

3. $x_1\in (0,\pi),\ x_{n+1}=\sin {x_n},\ y_n=x_n\sqrt{n}\ .$

4. $x_1\in R,\ x_{n+1}=\ln (1+\arctan x_n),\ y_n=nx_n\ .$

5. $x_1\ne 0,\ x_{n+1}=x_n+e^{-x_n},\ y_n=\frac{x_n}{\ln n},\ z_n=\frac{x_n}{n}\ .$

6. $x_1>0,\ x_{n+1}=e^{x^n-1}\ .$

7. $x_1>0,\ x_{n+1}+x^3_{n+1}=x_n,\ y_n=x_n\sqrt n\ .$

8. $x_1>0,\ x_{n+1}=\arctan x_n,\ y_n=x_n\sqrt n\ .$

9. $x_1>0,\ x_{n+1}=\left(1+\frac{x_n}{a}\right)^a-1, 0<a<1,\ y_n=nx_n$ $\left(a:=\frac{1}{p},\ p\in N^*\right)\ .$

10. $x_1\in (0,1),\ x_{n+1}=x_n(1-x_n)^a,\ a>0,\ y_n=nx_n\ .$

11. $x_1\in (0,1),\ x_{n+1}=x_n-x^2_n+x^3_n-x^4_n,\ y_n=nx_n\ .$

!2. $x_1>0,\ x_{n+1}=|x_n-n|,\ y_n=\frac{x_n}{n}\ .$

13. $x_1\in \left(0,\frac{\pi}{2}\right),\ x_{n+1}=x_n\cos x_n,\ y_n=x_n\sqrt n\ .$

14. $x_1>0,\ x_{n+1}=x_n a^{x_n},\ a\in (0,1),\ y_n=nx_n\ .$

15. $x_1=1,\ e^{1+x_{n+1}}=1+\ln (2+x_n)\ .$

16. $x_n\in (0,\pi),\sin {x_n}+\cos {x_n}=1\ .$

17. $x_1=1,\ x_{n+1}=\sqrt {1+nx_n},\ y_n=\frac{x_n}{n},\ z_n=n-x_n\ .$

18. $x_1=0,\ x_{n+1}=\sqrt {n-x_n},\ y_n=\frac{x_n}{\sqrt {n}},\ z_n=x_n-\sqrt n\ .$

19. $x_1>1,\ x_{n+1}=\sqrt [3] {3x_n-2},\ y_n=n(x_n-1)\ .$

20. $x_1\in (0,1),\ x_{n+1}=\frac{\sqrt {x_n}}{\sqrt {x_n}+\sqrt {1-x_n}},\ y_n=\frac{x_n}{1-x_n}\ .$

21. $x_1=1,\ x_{n+1}=x_n+\frac{1}{x_n},\ y_n=\frac{x_n}{\sqrt {2n}},\ z_n=\frac{1}{n\sqrt {n}}\cdot \sum_{k=1}^n x_k$ .


See here and here
This post has been edited 21 times. Last edited by Virgil Nicula, Nov 21, 2015, 7:48 AM

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