P407练习2

zhangzujin

手机查看请在浏览器中打开, 到了支付页面请截图, 并用支付宝或微信扫描之, 稍等后获得金钱, 即可购买. 偶偶因为网络问题充值不成功, 请与微信 pdezhang 联系, 发送论坛昵称与付款时间即可处理, 稍安勿躁. 购买后刷新网页才能正常显示数学公式.

# P407练习2 --- 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) [购书](https://mp.weixin.qq.com/s/9RUc7PhEAjT7vlZk4lqZSw) / [答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A) / [pdf1](https://mbd.pub/o/zhangzujin) / [pdf2](https://mbd.pub/o/gjsx) / (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. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw) / [公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow) / [资料目录](https://mbd.pub/o/bread/mbd-YpiZmJdv) / [视频](https://space.bilibili.com/507709073) / [微信群](https://mbd.pub/o/bread/mbd-YZmTkp1w)