## 张祖锦2023年数学专业真题分类70天之第04天
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70、 4、 设 $\displaystyle a > 1$, 求 $\displaystyle \lim\_\{n\to\infty\}\left(\frac\{1\}\{a\}+\frac\{2^2\}\{a^2\}+\cdots+\frac\{n^2\}\{a^n\}\right)$. (上海财经大学2023年数学分析考研试题) [数列极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle b=\frac\{1\}\{a\}\in (0,1)$, 则
\begin\{aligned\} \mbox\{原式\}=&\sum\_\{n=1\}^\infty n^2b^n =b^2\sum\_\{n=1\}^\infty n(n-1)b^\{n-2\}+b\sum\_\{n=1\}^\infty nb^\{n-1\}\\\\ =&b^2\left.\left(\sum\_\{n=1\}^\infty x^n\right)''\right|\_\{x=b\} +b\left.\left(\sum\_\{n=1\}^\infty x^n\right)'\right|\_\{x=b\} =b^2\frac\{2\}\{(1-b)^3\}+\frac\{b\}\{(1-b)^2\}\\\\ =&\frac\{b(b+1)\}\{(1-b)^3\}=\frac\{a(a+1)\}\{(a-1)^3\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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71、 1、 解答如下问题: (1)、 用 ‘$\varepsilon-N$ 语言叙述``极限 $\displaystyle \lim\_\{n\to\infty\}x\_n=a$‘; (上海大学2023年数学分析考研试题) [数列极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \forall\ \varepsilon > 0,\exists\ N,\mathrm\{ s.t.\} \forall\ n\geq N, |x\_n-a| < \varepsilon$.跟锦数学微信公众号. [在线资料](
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72、 2、 计算下列极限: (1)、 $\displaystyle \lim\_\{n\to\infty\}\left(\frac\{n^2+1\}\{n^2-2\}\right)^\{n^2+2n\}$; (上海大学2023年数学分析考研试题) [数列极限 ]
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https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}&=\mathrm\{exp\}\left\[\lim\_\{n\to\infty\}(n^2+2n)\ln\frac\{n^2+1\}\{n^2-2\}\right\]\\\\ &\xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \mathrm\{exp\}\left\[\lim\_\{n\to\infty\}(n^2+2n)\left(\frac\{n^2+1\}\{n^2-2\}-1\right)\right\]\\\\ &=\mathrm\{exp\}\left\[\lim\_\{n\to\infty\}(n^2+2n)\frac\{3\}\{n^2-2\}\right\]=\mathrm\{e\}^3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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73、 3、 (20 分) 设函数 $\displaystyle f(x)$ 在区间 $\displaystyle [0,1]$ 上可导, 其值域也是区间 $\displaystyle [0,1]$, 且存在 $\displaystyle 0 < \theta < 1$, 使得 $\displaystyle |f'(x)|\leq \theta$. 证明: $\displaystyle f(x)=x$ 在 $\displaystyle [0,1]$ 上存在唯一解. (上海交通大学2023年数学分析考研试题) [数列极限 ]
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https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 这就是压缩映射原理. (1)、 存在性. 任取 $\displaystyle x\_0\in [0,1]$, 定义 $\displaystyle x\_\{n+1\}=f(x\_n), n\geq 0$, 则由
\begin\{aligned\} |x\_\{n+1\}-x\_n|=&|f(x\_n)-f(x\_\{n-1\})| \leq \theta |x\_n-x\_\{n-1\}|\\\\ &\leq \cdots \leq \theta ^n|x\_1-x\_0| \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} |x\_\{n+p\}-x\_n|\leq&\sum\_\{k=n\}^\{n+p-1\}|x\_\{k+1\}-x\_k| \leq \sum\_\{k=n\}^\{n+p-1\}\theta ^k|x\_1-x\_0|\\\\ \leq& \sum\_\{k=n\}^\infty \theta ^k|x\_1-x\_0| =\frac\{\theta ^n\}\{1-\theta \}|x\_1-x\_0|\xrightarrow\{n\to\infty\}0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 Cauchy 收敛准则知 $\displaystyle \left\\{x\_n\right\\}$ 收敛. 设极限为 $\displaystyle x$, 则在递推式中令 $\displaystyle n\to\infty$, 并注意到 $\displaystyle f$ 的 Lipschitz 连续性知 $\displaystyle x=f(x)$. (2)、 唯一性. 设 $\displaystyle y$ 也满足 $\displaystyle y=f(y)$, 则
\begin\{aligned\} &|x-y|=|f(x)-f(y)|\leq \theta |x-y|\\\\ \Rightarrow& (1-\theta )|x-y|\leq 0\Rightarrow |x-y|=0\Rightarrow x=y. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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74、 1、 (1)、 求极限 $\displaystyle \lim\_\{n\to\infty\}n^2\left(1-n\sin\frac\{1\}\{n\}\right)$. (首都师范大学2023年数学分析考研试题) [数列极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
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\begin\{aligned\} \mbox\{原式\}\xlongequal[\tiny\mbox\{原理\}]\{\tiny\mbox\{归结\}\}& \lim\_\{x\to 0\}\frac\{1-\frac\{\sin x\}\{x\}\}\{x^2\} =\lim\_\{x\to 0\}\frac\{x-\sin x\}\{x^3\}\\\\ \xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\}& \lim\_\{x\to 0\}\frac\{x-\left\[x-\frac\{x^3\}\{6\}+o(x^4)\right\]\}\{x^3\}=\frac\{1\}\{6\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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75、 1、 计算下列极限 (每小题 8 分, 共 32 分). (1)、 $\displaystyle \lim\_\{n\to\infty\}\frac\{1^p+2^p+\cdots+n^p\}\{n^\{p+1\}\}$ ($p > 0$). (四川大学2023年数学分析考研试题) [数列极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}&\xlongequal\{\tiny\mbox\{Stolz\}\} \lim\_\{n\to\infty\}\frac\{n^p\}\{n^\{p+1\}-(n-1)^\{p+1\}\} \xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} \lim\_\{n\to\infty\}\frac\{n^p\}\{(p+1)\xi\_n^p\}=\frac\{1\}\{p+1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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76、 (2)、 $\displaystyle \lim\_\{n\to\infty\}\left\[\frac\{1\}\{\ln (n+1)-\ln n\}-n\right\]$. (四川大学2023年数学分析考研试题) [数列极限 ]
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https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
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\begin\{aligned\} \mbox\{原式\}=&\lim\_\{n\to\infty\}\left\[\frac\{1\}\{\ln\left(1+\frac\{1\}\{n\}\right)\}-\frac\{1\}\{\frac\{1\}\{n\}\}\right\] \xlongequal[\tiny\mbox\{原理\}]\{\tiny\mbox\{归结\}\} \lim\_\{x\to 0\}\left\[\frac\{1\}\{\ln(1+x)\}-\frac\{1\}\{x\}\right\]\\\\ =&\lim\_\{x\to 0\}\frac\{x\ln(1+x)\}\{x\ln(1+x)\} \xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\} \lim\_\{x\to 0\}\frac\{x-\left\[x-\frac\{x^2\}\{2\}+o(x^2)\right\]\}\{x[x+o(x)]\}=\frac\{1\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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77、 1、 (20 分) 求下列极限: (1)、 $\displaystyle \lim\_\{n\to\infty\}\left(1+\frac\{1\}\{n^2\}\right)^n$; (苏州大学2023年数学分析考研试题) [数列极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}=\mathrm\{exp\}\left\[\lim\_\{n\to\infty\}n\ln\left(1+\frac\{1\}\{n^2\}\right)\right\] \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \mathrm\{exp\}\left\[\lim\_\{n\to\infty\}n\frac\{1\}\{n^2\}\right\]=\mathrm\{e\}^0=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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78、 1、 求极限 $\displaystyle \lim\_\{n\to\infty\}\left(1+\frac\{1\}\{n^2\}\right)\left(1+\frac\{2\}\{n^2\}\right)\cdots\left(1+\frac\{n\}\{n^2\}\right)$. (太原理工大学2023年数学分析考研试题) [数列极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 由
\begin\{aligned\} &x-\frac\{x^2\}\{2\} < \ln (1+x)\left(x > 0\right)\\\\ \Leftrightarrow&f(x)=\ln (1+x)-x+\frac\{x^2\}\{2\} > 0\left(x > -1\right)\\\\ \Leftarrow& f(0)=0, f'(x)=\frac\{x^2\}\{1+x\} > 0\left(x > -1\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及
\begin\{aligned\} &\ln (1+x) < x\left(x > 0\right)\\\\ \Leftrightarrow&f(x)=\ln (1+x)-x < 0\left(x > 0\right)\\\\ \Leftarrow& f(0)=0, f'(x)=\frac\{-x\}\{1+x\} < 0\left(x > 0\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle x-\frac\{x^2\}\{2\} < \ln (1+x) < x, x > 0$. (2)、 由
\begin\{aligned\} \sum\_\{k=1\}^n \ln \left(1+\frac\{k\}\{n^2\}\right)& < \sum\_\{k=1\}^n \frac\{k\}\{n^2\}=\frac\{1\}\{n^2\}\frac\{n(n+1)\}\{2\}\to \frac\{1\}\{2\}\left(n\to\infty\right),\\\\ \sum\_\{k=1\}^n \ln \left(1+\frac\{k\}\{n^2\}\right)& > \sum\_\{k=1\}^n \left(\frac\{k\}\{n^2\}-\frac\{k^2\}\{2n^4\}\right)\\\\ &=\frac\{1\}\{n^2\}\cdot\frac\{n(n+1)\}\{2\}-\frac\{1\}\{2n^4\}\frac\{n(n+1)(2n+1)\}\{6\}\\\\ &\to \frac\{1\}\{2\}\left(n\to\infty\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及迫敛性知 $\displaystyle \lim\_\{n\to\infty\}\sum\_\{k=1\}^n \ln \left(1+\frac\{k\}\{n^2\}\right)=\frac\{1\}\{2\}$, 而原式 $\displaystyle =\sqrt\{\mathrm\{e\}\}$.跟锦数学微信公众号. [在线资料](
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79、 8、 设数列 $\displaystyle \left\\{x\_n\right\\}$ 满足
\begin\{aligned\} x\_1 > 0, x\_\{n+1\}=\ln(1+x\_n), n=1,2,\cdots. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: $\displaystyle \left\\{x\_n\right\\}$ 收敛, 并求 $\displaystyle \lim\_\{n\to\infty\}nx\_n$. (太原理工大学2023年数学分析考研试题) [数列极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由数学归纳法知 $\displaystyle x\_n > 0$. 再由不等式
\begin\{aligned\} \ln(1+x) < x\left(x > -1, x\neq 0\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} x\_\{n+1\}=\ln(1+x\_n) < x\_n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \left\\{x\_n\right\\}$ 递减有下界. 据单调有界定理及极限的保不等式性知
\begin\{aligned\} l=\lim\_\{n\to\infty\}x\_n\geq 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
存在. 在 $\displaystyle x\_\{n+1\}=\ln (1+x\_n)$ 中令 $\displaystyle n\to\infty$ 得
\begin\{aligned\} l=\ln (1+l)\Rightarrow l=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
进一步,
\begin\{aligned\} \lim\_\{n\to\infty\}nx\_n=&\lim\_\{n\to\infty\}\frac\{n\}\{\frac\{1\}\{x\_n\}\}\xlongequal\{\tiny\mbox\{Stolz\}\} \lim\_\{n\to\infty\}\frac\{1\}\{\frac\{1\}\{x\_n\}-\frac\{1\}\{x\_\{n-1\}\}\}\\\\ =&\lim\_\{n\to\infty\}\frac\{x\_nx\_\{n-1\}\}\{x\_\{n-1\}-x\_n\}=\lim\_\{n\to\infty\}\frac\{x\_\{n-1\}\ln(1+x\_\{n-1\})\}\{x\_\{n-1\}-\ln(1+x\_\{n-1\})\}\\\\ \xlongequal[\tiny\mbox\{原理\}]\{\tiny\mbox\{归结\}\}&\lim\_\{x\to 0\}\frac\{x\ln(1+x)\}\{x-\ln(1+x)\} \xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\} \lim\_\{x\to 0\}\frac\{x\left\[x+o(x)\right\]\}\{x-\left\[x-\frac\{x^2\}\{2\}+o(x^2)\right\]\}=2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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80、 1、 计算题. (1)、 求极限 $\displaystyle \lim\_\{n\to\infty\}\frac\{\displaystyle\int\_0^\{n\pi\}x|\sin x|\mathrm\{ d\} x\}\{n(n+1)\}$. (武汉大学2023年数学分析考研试题) [数列极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
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\begin\{aligned\} \mbox\{原式\}&\xlongequal\{\tiny\mbox\{Stolz\}\} \lim\_\{n\to\infty\}\frac\{\int\_\{n\pi\}^\{(n+1)\pi\}x|\sin x|\mathrm\{ d\} x\}\{(n+1)(n+2)-n(n+1)\} =\frac\{1\}\{2\}\lim\_\{n\to\infty\}\frac\{\int\_\{n\pi\}^\{(n+1)\pi\}x|\sin x|\mathrm\{ d\} x\}\{n+1\}\\\\ &\xlongequal\{\tiny\mbox\{Stolz\}\} \frac\{1\}\{2\}\lim\_\{n\to\infty\}\left\[\int\_\{(n+1)\pi\}^\{(n+2)\pi\}x|\sin x|\mathrm\{ d\} x -\int\_\{n\pi\}^\{(n+1)\pi\}x|\sin x|\mathrm\{ d\} x\right\]\\\\ &\stackrel\{x-\pi=t\}\{=\}\frac\{1\}\{2\}\lim\_\{n\to\infty\}\left\[\int\_\{n\pi\}^\{(n+1)\pi\}(\pi+t)|\sin t|\mathrm\{ d\} t -\int\_\{n\pi\}^\{(n+1)\pi\}x|\sin x|\mathrm\{ d\} x\right\]\\\\ &=\frac\{\pi\}\{2\}\lim\_\{n\to\infty\}\int\_\{n\pi\}^\{(n+1)\pi\} |\sin t|\mathrm\{ d\} t =\frac\{\pi\}\{2\} \sin t\mathrm\{ d\} t=\pi. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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81、 3、 证明题. (1)、 已知 $\displaystyle x\_0 > 0$, 当 $\displaystyle n\geq 1$ 时, $\displaystyle x\_n=\arctan x\_\{n-1\}$, 证明: $\displaystyle \lim\_\{n\to\infty\}x\_n=0$. (武汉理工大学2023年数学分析考研试题) [数列极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由
\begin\{aligned\} \tan x > x, 0 < x < \frac\{\pi\}\{2\}\Rightarrow x > \arctan x, x > 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle x\_n\searrow, \geq 0$. 由单调有界定理知 $\displaystyle \lim\_\{n\to\infty\}x\_n=l$ 存在, 且
\begin\{aligned\} l=\arctan l, 0\leq l\leq 1\Rightarrow l=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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---
82、 1、 填空题. (1)、 $\displaystyle \lim\_\{n\to\infty\}\left(\frac\{1\}\{2\}\cdot\frac\{3\}\{4\}\cdots \frac\{2n-1\}\{2n\}\right)=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (西安交通大学2023年数学分析考研试题) [数列极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle \frac\{2k-1\}\{2k\} < \frac\{2k\}\{2k+1\}$ 知
\begin\{aligned\} 0 < &\left(\frac\{1\}\{2\}\cdot\frac\{3\}\{4\}\cdots \frac\{2n-1\}\{2n\}\right)^2\\\\ < &\left(\frac\{1\}\{2\}\cdot\frac\{3\}\{4\}\cdots \frac\{2n-1\}\{2n\}\right)\left(\frac\{2\}\{3\}\cdot\frac\{4\}\{5\}\cdots \frac\{2n\}\{2n+1\}\right) =\frac\{1\}\{2n+1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由迫敛性知原式 $\displaystyle =0$.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](
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---
83、 2、 证明题. (1)、 已知数列 $\displaystyle \left\\{x\_n\right\\}$ 满足 $\displaystyle x\_1=0, x\_\{n+1\}=\cos x\_n\ (n=1,2,\cdots)$. 证明: (1-1)、 $\displaystyle \left\\{x\_\{2n\}\right\\}, \left\\{x\_\{2n-1\}\right\\}$ 均单调; (1-2)、 $\displaystyle \left\\{x\_n\right\\}$ 收敛. (西安交通大学2023年数学分析考研试题) [数列极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle x\_0=0$,
\begin\{aligned\} 0\leq x\_k\leq 1\Rightarrow 0\leq \cos 1\leq x\_\{k+1\}=\cos x\_k\leq \cos 0=1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及数学归纳法知 $\displaystyle 0\leq x\_n\leq 1, \forall\ n\geq 1$. 设 $\displaystyle f(x)=\cos \cos x, 0\leq x\leq 1$, 则
\begin\{aligned\} f'(x)=-\sin \cos x(-\sin x)=\sin \cos x\cdot \sin x > 0, 0 < x < 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle x\_\{2n\}=f(x\_\{2(n-1)\}), x\_\{2n+1\}=f(x\_\{2(n-1)+1\})$ 即知 $\displaystyle \left\\{x\_\{2n\}\right\\}, \left\\{x\_\{2n-1\}\right\\}$ 均递增. 由单调有界定理知 $\displaystyle \lim\_\{n\to\infty\}x\_\{2n\}=a, \lim\_\{n\to\infty\}x\_\{2n-1\}=b$ 都存在. 进而
\begin\{aligned\} &x\_\{2n\}=\cos x\_\{2n-1\}, x\_\{2n+1\}=\cos x\_\{2n\}\\\\ \Rightarrow& a=\cos b, b=\cos a \Rightarrow a=f(a), b=f(b). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle g(x)=f(x)-x,0\leq x\leq 1$, 则 $\displaystyle g(0)=\cos 1 > 0, g(1)=\cos \cos 1-1 < 0, g'(x)=f'(x)-1 < 0$. 从而 $\displaystyle f$ 的不动点是唯一的, 即 $\displaystyle a=b\Rightarrow \lim\_\{n\to\infty\}x\_n=a$.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](
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---
84、 (2)、 $\displaystyle \lim\_\{n\to\infty\}\frac\{\sqrt[n]\{(n+1)(n+2)\cdots(n+n)\}\}\{n\}$. (西北大学2023年数学分析考研试题) [数列极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}=\mathrm\{exp\}\left\[\lim\_\{n\to\infty\}\frac\{1\}\{n\}\sum\_\{k=1\}^n \ln \left(1+\frac\{1\}\{k\}\right)\right\] =\mathrm\{exp\}\left\[\int\_0^1 \ln (1+x)\mathrm\{ d\} x\right\] \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\}\frac\{4\}\{\mathrm\{e\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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)/
---
85、 2、 (20 分) 设 $\displaystyle 0 < x\_0 < \frac\{\pi\}\{2\}$, 作迭代序列 $\displaystyle x\_n=\sin x\_\{n-1\}\ (n=1,2,\cdots)$. (1)、 证明: $\displaystyle \lim\_\{n\to\infty\}x\_n=0$; (2)、 证明: $\displaystyle \left\\{nx\_n^2\right\\}$ 收敛, 并求其极限. (西北大学2023年数学分析考研试题) [数列极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 易知 $\displaystyle x\_n\searrow, > 0$. 由单调有界定理知 $\displaystyle \lim\_\{n\to\infty\}x\_n=\ell$ 存在, 且 $\displaystyle \ell=\sin \ell, \ell\in\left\[0,\frac\{\pi\}\{2\}\right)\Rightarrow \ell=0$. (2)、
\begin\{aligned\} \mbox\{原式\}=&\lim\_\{n\to\infty\}\frac\{n\}\{\frac\{1\}\{x\_n^2\}\} \xlongequal\{\tiny\mbox\{Stolz\}\}\lim\_\{n\to\infty\}\frac\{1\}\{\frac\{1\}\{x\_\{n+1\}^2\}-\frac\{1\}\{x\_n^2\}\} =\lim\_\{n\to\infty\}\frac\{x\_n^2x\_\{n+1\}^2\}\{x\_n^2-x\_\{n+1\}^2\}\\\\ =&\lim\_\{n\to\infty\}\frac\{x\_n^2\sin^2x\_n\}\{x\_n^2-\sin^2x\_n\} \xlongequal[\tiny\mbox\{原理\}]\{\tiny\mbox\{归结\}\}\lim\_\{x\to 0\}\frac\{x^2\sin^2x\}\{x^2-\sin^2x\}\\\\ \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\}&\lim\_\{x\to 0\}\frac\{x^4\}\{(x-\sin x)(x+\sin x)\} =\frac\{1\}\{\frac\{1\}\{3!\}\cdot 2\} =3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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---
86、 1、 计算题. (1)、 (10 分) 求极限 $\displaystyle \lim\_\{n\to\infty\}n^2\left(1-n\sin\frac\{1\}\{n\}\right)$. (西南大学2023年数学分析考研试题) [数列极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}\xlongequal[\tiny\mbox\{原理\}]\{\tiny\mbox\{归结\}\}&\lim\_\{x\to 0\}\frac\{1-\frac\{1\}\{x\}\sin x\}\{x^2\} =\lim\_\{x\to 0\}\frac\{x-\sin x\}\{x^3\}\xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\} \frac\{1\}\{6\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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](
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---
87、 1、 (10 分) 若 $\displaystyle \lim\_\{n\to\infty\}a\_n=a$, 证明: $\displaystyle \lim\_\{n\to\infty\}\frac\{a\_1+\cdots+a\_n\}\{n\}=a$. (西南交通大学2023年数学分析考研试题) [数列极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle \lim\_\{n\to\infty\}a\_n=a$ 知
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ N\_1,\mathrm\{ s.t.\} \forall\ n\geq N\_1, |a\_n-a| < \frac\{\varepsilon\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由 $\displaystyle \lim\_\{n\to\infty\}\frac\{1\}\{n\}\sum\_\{k=1\}^\{N\_1\} |a\_k-a|=0$ 知
\begin\{aligned\} \exists\ N\_2,\mathrm\{ s.t.\} \forall\ n\geq N\_2, \frac\{1\}\{n\}\sum\_\{k=1\}^\{N\_1\}|a\_k-a| < \frac\{\varepsilon\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是当 $\displaystyle n\geq N=\max\left\\{N\_1,N\_2\right\\}$ 时,
\begin\{aligned\} &\left|\frac\{a\_1+\cdots+a\_n\}\{n\}-a\right|\leq \frac\{|a\_1-a|+\cdots+|a\_n-a|\}\{n\}\\\\ \leq& \frac\{1\}\{n\}\sum\_\{k=1\}^\{N\_1\}|a\_k-a| +\frac\{1\}\{n\}\sum\_\{k=N\_1+1\}^n |a\_k-a| < \frac\{\varepsilon\}\{2\}+\frac\{\varepsilon\}\{2\}=\varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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](
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---
88、 (3)、 $\displaystyle \lim\_\{n\to\infty\}\left(1+\frac\{1\}\{2\}+\cdots+\frac\{1\}\{n\}\right)^\frac\{1\}\{n\}$. (湘潭大学2023年数学分析考研试题) [数列极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}&=\mathrm\{exp\}\left\[\lim\_\{n\to\infty\}\frac\{1\}\{n\}\ln \left(1+\frac\{1\}\{2\}+\cdots+\frac\{1\}\{n\}\right)\right\]\\\\ &\xlongequal\{\tiny\mbox\{Stolz\}\} \mathrm\{exp\}\left\\{\lim\_\{n\to\infty\}\left\[\ln \left(1+\frac\{1\}\{2\}+\cdots+\frac\{1\}\{n+1\}\right) -\ln \left(1+\frac\{1\}\{2\}+\cdots+\frac\{1\}\{n\}\right)\right\]\right\\}\\\\ &=\mathrm\{exp\}\left\[\lim\_\{n\to\infty\}\ln \left(1+\frac\{\frac\{1\}\{n+1\}\}\{1+\frac\{1\}\{2\}+\cdots+\frac\{1\}\{n\}\}\right)\right\] =\mathrm\{exp\}\left\[\ln 1\right\]=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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](
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---
89、 4、 证明题. (1)、 (14 分) 解答如下问题. (1-1)、 设 $\displaystyle a\_n > 0, n\in\mathbb\{N\}\_+$, 且 $\displaystyle \lim\_\{n\to\infty\}a\_n=+\infty$, 证明: $\displaystyle \lim\_\{n\to\infty\}\sqrt[n]\{a\_1\cdots a\_n\}=+\infty$. (长安大学2023年数学分析考研试题) [数列极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle b\_n=\frac\{1\}\{a\_n\} > 0$, 则只要证 $\displaystyle \lim\_\{n\to\infty\}b\_n=0\Rightarrow \lim\_\{n\to\infty\}\sqrt[n]\{b\_1\cdots b\_n\}=0$. 由均值不等式及迫敛性知仅需验证 $\displaystyle \lim\_\{n\to\infty\}\frac\{b\_1+\cdots+b\_n\}\{n\}=0$. 一般的, 我们有如下结论. 若 $\displaystyle \lim\_\{n\to\infty\}b\_n=b$, 则$\lim\_\{n\to\infty\}\frac\{b\_1+\cdots+b\_n\}\{n\}=b$. 事实上, 由 $\displaystyle \lim\_\{n\to\infty\}b\_n=b$ 知
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ N\_1,\mathrm\{ s.t.\} \forall\ n\geq N\_1, |b\_n-b| < \frac\{\varepsilon\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由 $\displaystyle \lim\_\{n\to\infty\}\frac\{1\}\{n\}\sum\_\{k=1\}^\{N\_1\} |b\_k-b|=0$ 知
\begin\{aligned\} \exists\ N\_2,\mathrm\{ s.t.\} \forall\ n\geq N\_2, \frac\{1\}\{n\}\sum\_\{k=1\}^\{N\_1\}|b\_k-b| < \frac\{\varepsilon\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是当 $\displaystyle n\geq N=\max\left\\{N\_1,N\_2\right\\}$ 时,
\begin\{aligned\} &\left|\frac\{b\_1+\cdots+b\_n\}\{n\}-b\right|\leq \frac\{|b\_1-b|+\cdots+|b\_n-b|\}\{n\}\\\\ \leq& \frac\{1\}\{n\}\sum\_\{k=1\}^\{N\_1\}|b\_k-b| +\frac\{1\}\{n\}\sum\_\{k=N\_1+1\}^n |b\_k-b| < \frac\{\varepsilon\}\{2\}+\frac\{\varepsilon\}\{2\}=\varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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](
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---
90、 2、 (15 分) 求极限. (1)、 (8 分) $\displaystyle \lim\_\{n\to\infty\}\left(\frac\{n\}\{n^2+1^2\} +\frac\{n\}\{n^2+2^2\}+\cdots+\frac\{n\}\{n^2+n^2\}\right)$. (中国科学技术大学2023年数学分析考研试题) [数列极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}=&\lim\_\{n\to\infty\}\sum\_\{k=1\}^n \frac\{n\}\{n^2+k^2\} =\lim\_\{n\to\infty\}\frac\{1\}\{n\}\sum\_\{k=1\}^n \frac\{1\}\{1+\left(\frac\{k\}\{n\}\right)^2\}\\\\ =&\int\_0^1 \frac\{1\}\{1+x^2\}\mathrm\{ d\} x=\frac\{\pi\}\{4\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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---
91、 1、 (1)、 求极限 $\displaystyle \lim\_\{n\to\infty\}\left(\frac\{2+\sqrt[n]\{64\}\}\{3\}\right)^\{2n+1\}$. (中国科学院大学2023年数学分析考研试题) [数列极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}=&\mathrm\{exp\}\left\[\lim\_\{n\to\infty\}(2n+1)\ln \frac\{2+\sqrt[n]\{64\}\}\{3\}\right\] \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \mathrm\{exp\}\left\[\lim\_\{n\to\infty\}(2n+1)\frac\{\sqrt[n]\{64\}-1\}\{3\}\right\]\\\\ =&\mathrm\{exp\}\left\[\lim\_\{n\to\infty\}(2n+1)\frac\{\mathrm\{e\}^\{\frac\{1\}\{n\}\ln 64\}-1\}\{3\}\right\] \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \mathrm\{exp\}\left\[\lim\_\{n\to\infty\}(2n+1) \frac\{\ln 64\}\{3n\}\right\]\\\\ =&\mathrm\{exp\}\left\[\frac\{2\}\{3\}\cdot 6\ln 2\right\]=16. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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---
92、 (2)、 证明: 数列
\begin\{aligned\} a\_n=1+\frac\{1\}\{2\}+\cdots+\frac\{1\}\{n\}-\ln n \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
收敛. (中国科学院大学2023年数学分析考研试题) [数列极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由
\begin\{aligned\} a\_\{n+1\}-a\_n=&\frac\{1\}\{n+1\}-\ln \left(1+\frac\{1\}\{n\}\right) < 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle a\_n\searrow$. 又由
\begin\{aligned\} \ln \left(1+\frac\{1\}\{k\}\right) < \frac\{1\}\{k\}, 1\leq k\leq n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
相加即知
\begin\{aligned\} 0 < \ln (n+1)-\ln n < a\_n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
据单调有界定理知 $\displaystyle \lim\_\{n\to\infty\}a\_n$ 存在.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)/
发表于 2023-3-5 08:40:30
