Sequence – Limit

Bài toán : Cho dãy số $(x_n)$ như sau :

$$\{\begin{array}{c}{x}_1\in \left(1,2\right)\\ x_{n+1}=1+x_n-\frac{x_n^2}{2},\forall n=1,2,...\end{array}$$

Tính $\underset{n\to +\infty }{\lim }{x}_n$.

Lời giải :

Ta có :

$x_{n+1}$$=\frac{-1}{2}(x_n-1{)}^2+\frac{3}{2}$$\le \frac{3}{2}$$\Rightarrow x_n$$\le \frac{3}{2},\forall n\in {\mathbb{N}}^{\ast }$

Từ đó :

$$x_{n+1}=1+\frac{x_n(2-x_n)}{2}>1\Rightarrow x_n>1,\forall n\in {\mathbb{N}}^{\ast }$$

Đặt $\sqrt{2}=a$ ta có $a=1+a-\frac{a^2}{2}$. Ta xét :

$\left|x_{n+1}-a\right|$$=\left|\left(1+x_n-\frac{x_n^2}{2}\right)-\left(1+a-\frac{a^2}{2}\right)\right|$$=\left|(x_n-a)-\frac{(x_n-a)(x_n+a)}{2}\right|$$=\frac{1}{2}\left|(x_n-a)\right|\left|2-a-x_n\right|$

Chú ý vì $1<x_n\le \frac{3}{2},\forall n\in {\mathbb{N}}^{\ast }$ nên :

$$\left|2-a-x_n\right|<1$$

Từ đó mà :

$0$$\le \left|x_{n+1}-a\right|$$<\frac{1}{2}\left|x_n-a\right|$$<...$$<\frac{1}{2^n}\left|x_1-a\right|$

Lại có $\underset{n\to +\infty }{\lim }\frac{1}{2^n}|x_1-a|=0$ nên theo nguyên lí kẹp ta được :

$$\underset{n\to +\infty }{\lim }{x}_n=\sqrt{2}$$