Sequence – Limit

Bài toán : Cho dãy số $(x_n)$ xác định bởi :

$$\{\begin{array}{c}{x}_1=1\\ x_{n+1}=x_n+x_n^2,\forall n\in {\mathbb{N}}^{\ast }\end{array}$$

Tính $\underset{n\to +\infty }{\lim }\left(\frac{x_1}{x_2}+\frac{x_2}{x_3}+...+\frac{x_n}{x_{n+1}}\right)$

Lời giải :

Dễ dàng chứng minh được $\underset{n\to +\infty }{\lim }{x}_n=+\infty$

Từ công thức xác định dãy số, ta suy ra :

$x_{n+1}-x_n$$=x_n^2$$\Rightarrow \frac{x_{n+1}-x_n}{x_n{x}_{n+1}}$$=\frac{x_n}{x_{n+1}}$$\Rightarrow \frac{x_n}{x_{n+1}}$$=\frac{1}{x_n}-\frac{1}{x_{n+1}}$$\Rightarrow \frac{x_1}{x_2}+\frac{x_2}{x_3}+...+\frac{x_x}{x_{n+1}}$$=\left(\frac{1}{x_1}-\frac{1}{x_2}\right)+\left(\frac{1}{x_2}-\frac{1}{x_3}\right)+...+\left(\frac{1}{x_n}-\frac{1}{x_{n+1}}\right)$$=1-\frac{1}{x_{n+1}}$

Do vậy mà :

$\underset{n\to +\infty }{\lim }\left(\frac{x_1}{x_2}+\frac{x_2}{x_3}+...+\frac{x_n}{x_{n+1}}\right)$$=\underset{n\to +\infty }{\lim }\left(1-\frac{1}{x_{n+1}}\right)$$=1$