📝 题目
1.3.2 $\displaystyle{x}_{n + 1} = 1 - \sqrt{1 - {x}_{n}}\overset{\text{ 写成 }}{ = }\frac{{x}_{n}}{1 + \sqrt{1 - {x}_{n}}} \Rightarrow 0 < {x}_{n} \downarrow \Rightarrow \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = 0}$ ,
$\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{{x}_{n + 1}}{{x}_{n}} = \frac{1}{2}.}$
💡 答案与解析
1.3.2 $\displaystyle{x}_{n + 1} = 1 - \sqrt{1 - {x}_{n}}\overset{\text{ 写成 }}{ = }\frac{{x}_{n}}{1 + \sqrt{1 - {x}_{n}}} \Rightarrow 0 < {x}_{n} \downarrow \Rightarrow \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = 0}$ ,
$\displaystyle{\mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{{x}_{n + 1}}{{x}_{n}} = \frac{1}{2}.}$