P407练习2
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设
\begin\{aligned\} x\_1=\frac\{1\}\{2\}, x\_\{n+1\}=\frac\{1\}\{2\}-\frac\{x\_n^2\}\{2\}\left(n=1,2,\cdots\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 试证极限 $\displaystyle \lim\_\{n\to\infty\}x\_n=A$ 存在, 有限;
(2)、 证明级数 $\displaystyle \sum\_\{n=1\}^\infty (x\_n-A)$ 绝对收敛. (华中师范大学)
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(1)、 由数学归纳法知
\begin\{aligned\} &0<x\_n\leq\frac\{1\}\{2\}\Rightarrow |x\_n+x\_\{n+1\}|\leq 1\\\\ \Rightarrow&|x\_\{n+1\}-x\_n| =\frac\{1\}\{2\}|x\_n^2-x\_\{n-1\}^2| \leq \frac\{1\}\{2\}|x\_n-x\_\{n-1\}|\leq \cdots \leq\frac\{1\}\{2^\{n-2\}\}|x\_2-x\_1|\\\\ \Rightarrow&|x\_\{n+p\}-x\_n|\leq \sum\_\{k=n\}^\{n+p-1\}|x\_\{k+1\}-x\_k| \leq\sum\_\{k=n\}^\infty \frac\{1\}\{2^\{k-1\}\}|x\_2-x\_1| =\frac\{1\}\{2^\{n-2\}\}|x\_2-x\_1|\left(n\to\infty\right)\\\\ \Rightarrow&\lim\_\{n\to\infty\}x\_n=A\mbox\{存在\}\Rightarrow A=\frac\{1\}\{2\}-\frac\{A^2\}\{2\}\Rightarrow A=\sqrt\{2\}-1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 由
\begin\{aligned\} \frac\{|x\_\{n+1\}-A|\}\{|x\_n-A|\}=\frac\{\left|\frac\{1\}\{2\}-\frac\{x\_n^2\}\{2\}-\left(\frac\{1\}\{2\}-\frac\{A^2\}\{2\}\right)\right|\}\{|x\_n-A|\} =\frac\{|x\_n+A|\}\{2\}\xrightarrow\{n\to\infty\}|A|<1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及 D'Alembert 判别法知级数 $\displaystyle \sum\_\{n=1\}^\infty (x\_n-A)$ 绝对收敛.
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