## 张祖锦2023年数学专业真题分类70天之第02天
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24、 1、 (15 分) 已知 $\displaystyle L\in (0,1)$, $\displaystyle f: \mathbb\{R\}\to\mathbb\{R\}$ 满足
\begin\{aligned\} |f(x)-f(y)|\leq L|x-y|,\quad \forall\ x,y\in\mathbb\{R\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: 存在唯一的 $\displaystyle x$, 使得 $\displaystyle f(x)=x$. (北京师范大学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](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 这就是压缩映射原理. (1)、 存在性. 任取 $\displaystyle x\_0\in \mathbb\{R\}$, 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 L|x\_n-x\_\{n-1\}|\\\\ &\leq \cdots \leq L^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\}L^k|x\_1-x\_0|\\\\ \leq& \sum\_\{k=n\}^\infty L^k|x\_1-x\_0| =\frac\{L^n\}\{1-L\}|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 L|x-y|\\\\ \Rightarrow& (1-L)|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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25、 10、 已知数列 $\displaystyle \left\\{x\_n\right\\}$ 无界但不是无穷大量, 证明: $\displaystyle \left\\{x\_n\right\\}$ 存在子列收敛, 也存在子列为无穷大量. (北京邮电大学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) / (1)、 由 $\displaystyle \left\\{x\_n\right\\}$ 无界知对
\begin\{aligned\} &M=|x\_1|+1, \exists\ n\_2\geq 1,\mathrm\{ s.t.\} |x\_\{n\_2\}|\geq 2,\\\\ &M=\max\_\{1\leq i\leq n\_2\}|x\_i|+2, \exists\ n\_3 > n\_2,\mathrm\{ s.t.\} |x\_\{n\_3\}|\geq M,\\\\ &M=\max\_\{1\leq i\leq n\_3\}|x\_i|+3, \exists\ n\_4 > n\_3,\mathrm\{ s.t.\} |x\_\{n\_4\}|\geq M, \cdots. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
如此得到 $\displaystyle \left\\{x\_\{n\_k\}\right\\}$ 为无穷大量. (2)、 由 $\displaystyle \left\\{x\_n\right\\}$ 不是无穷大量知
\begin\{aligned\} \exists\ M > 0,\mathrm\{ s.t.\} \forall\ N,\exists\ n\geq N, \mathrm\{ s.t.\} |x\_n|\leq M. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
取
\begin\{aligned\} &N=1, \exists\ n\_1\geq 1,\mathrm\{ s.t.\} |x\_\{n\_1\}|\leq M,\\\\ &N=n\_1+1,\exists\ n\_2\geq N,\mathrm\{ s.t.\} |x\_\{n\_2\}|\leq M,\cdots. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
如此得到 $\displaystyle \left\\{x\_\{n\_k\}\right\\}$ 有界. 据致密性定理知它有一个子列 $\displaystyle \left\\{x\_\{n\_\{k\_i\}\}\right\\}$ 收敛.跟锦数学微信公众号. [在线资料](
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26、 15、 已知 $\displaystyle f\_n(x)=x^n+x, n=1,2,\cdots$. (1)、 证明: 方程 $\displaystyle f\_n(x)=1$ 在 $\displaystyle \left\[\frac\{1\}\{2\},1\right\]$ 上有且仅有一个解 $\displaystyle x\_n$; (2)、 证明: $\displaystyle \left\\{x\_n\right\\}$ 的极限存在, 并求 $\displaystyle \lim\_\{n\to\infty\}x\_n$. (北京邮电大学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\} f\_n\left(\frac\{1\}\{2\}\right)=\left(\frac\{1\}\{2\}\right)^n+\frac\{1\}\{2\}\leq 1 < 2=f(1) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及连续函数介值定理知 $\displaystyle f\_n$ 在 $\displaystyle \left\[\frac\{1\}\{2\},1\right\]$ 内有一个根. 再由
\begin\{aligned\} f\_n'(x)=nx^\{n-1\}+1 > 0, \frac\{1\}\{2\} < x < 1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle f\_n$ 在 $\displaystyle \left\[\frac\{1\}\{2\},1\right\]$ 内有且只有一个根 $\displaystyle x\_n$. (2)、 由 $\displaystyle \frac\{1\}\{2\} < x < 1\Rightarrow f\_n(x) > f\_\{n+1\}(x)$ 知
\begin\{aligned\} f\_\{n+1\}(x\_\{n+1\})=1=f\_n(x\_n) > f\_\{n+1\}(x\_n)\stackrel\{f\_n' > 0\}\{\Rightarrow\}x\_\{n+1\} > 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 $\displaystyle x\_n\mbox\{严\}\nearrow , x\_n\leq 1$. 由单调有界定理知 $\displaystyle \lim\_\{n\to\infty\}x\_n=l$ 存在, 且
\begin\{aligned\} l^n+l=1\Rightarrow l=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
[若不然, $\displaystyle \frac\{1\}\{2\} < l < 1\Rightarrow \lim\_\{n\to\infty\}(l^n+l)=0+l < 1$, 矛盾.]跟锦数学微信公众号. [在线资料](
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27、 (5)、 设 $\displaystyle \left\\{b\_n\right\\}$ 是正的严格递增数列, 且 $\displaystyle \lim\_\{n\to\infty\}b\_n=+\infty$, 求证:
\begin\{aligned\} \varlimsup\_\{n\to\infty\}\frac\{a\_n\}\{b\_n\}\leq \varlimsup\_\{n\to\infty\}\frac\{a\_n-a\_\{n-1\}\}\{b\_n-b\_\{n-1\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这里假设右侧的上极限存在. (大连理工大学2023年数学分析考研试题) [数列极限 ]
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
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\begin\{aligned\} &\forall\ k\geq n, \frac\{a\_\{k+1\}-a\_k\}\{b\_\{k+1\}-b\_k\}\leq \sup\_\{j\geq n\}\frac\{a\_\{j+1\}-a\_j\}\{b\_\{j+1\}-b\_j\}\\\\ \Rightarrow&\forall\ k\geq n, a\_\{k+1\}-a\_k\leq (b\_\{k+1\}-b\_k)\sup\_\{j\geq n\}\frac\{a\_\{j+1\}-a\_j\}\{b\_\{j+1\}-b\_j\}\\\\ \Rightarrow&\forall\ m > n, a\_m-a\_n\leq (b\_m-b\_n)\sup\_\{j\geq n\}\frac\{a\_\{j+1\}-a\_j\}\{b\_\{j+1\}-b\_j\}\\\\ &\left(\mbox\{对 $\displaystyle k$ 从 $\displaystyle n,\cdots,m-1$ 求和\}\right)\\\\ \Rightarrow&\forall\ m > n, \frac\{a\_m-a\_n\}\{b\_m-b\_n\}\leq \sup\_\{j\geq n\}\frac\{a\_\{j+1\}-a\_j\}\{b\_\{j+1\}-b\_j\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle m\to\infty$ 得
\begin\{aligned\} &\varlimsup\_\{m\to\infty\}\frac\{a\_m\}\{b\_m\}=\varlimsup\_\{m\to\infty\}\frac\{a\_m-a\_n\}\{b\_m\} =\varlimsup\_\{m\to\infty\}\frac\{a\_m-a\_n\}\{b\_m-b\_n\}\cdot \frac\{b\_m-b\_n\}\{b\_m\}\\\\ =&\varlimsup\_\{m\to\infty\}\frac\{a\_m-a\_n\}\{b\_m-b\_n\}\leq \sup\_\{j\geq n\}\frac\{a\_\{j+1\}-a\_j\}\{b\_\{j+1\}-b\_j\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再令 $\displaystyle n\to\infty$ 得
\begin\{aligned\} \varlimsup\_\{m\to\infty\}\frac\{a\_m\}\{b\_m\}\leq \lim\_\{n\to\infty\}\sup\_\{j\geq n\}\frac\{a\_\{j+1\}-a\_j\}\{b\_\{j+1\}-b\_j\} =\varlimsup\_\{n\to\infty\}\frac\{a\_n-a\_\{n-1\}\}\{b\_n-b\_\{n-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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28、 1、 填空题 (共 30 分, 每小题 5 分). (1)、 $\displaystyle \lim\_\{n\to\infty\}\left(\frac\{1\}\{n+1\}+\frac\{1\}\{n+2\}+\cdots+\frac\{1\}\{2n\}\right)=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (电子科技大学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) /
\begin\{aligned\} \mbox\{原式\}&=\lim\_\{n\to\infty\}\sum\_\{k=1\}^n\frac\{1\}\{n+k\} =\lim\_\{n\to\infty\}\frac\{1\}\{n\}\sum\_\{k=1\}^n\frac\{1\}\{1+\frac\{k\}\{n\}\}\\\\ &=\int\_0^1 \frac\{1\}\{1+x\}\mathrm\{ d\} x=\ln 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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29、 4、 综合题 (共 30 分, 每小题 15 分). (1)、 设函数 $\displaystyle f\in C^2[0,1], f'(0)=1, f''(0)\neq 0$ 且 $\displaystyle 0 < f(x) < x, x\in (0,1)$. 令
\begin\{aligned\} a\_1\in (0,1), \quad a\_\{n+1\}=f(a\_n)\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\}
(1-1)、 证明: 数列 $\displaystyle \left\\{a\_n\right\\}$ 收敛, 并求 $\displaystyle \lim\_\{n\to\infty\}a\_n$; (1-2)、 试问数列 $\displaystyle \left\\{na\_n\right\\}$ 是否一定收敛? 若不一定收敛, 请举出反例; 若收敛, 求其极限 $\displaystyle \lim\_\{n\to\infty\}na\_n$. (电子科技大学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) / (1-1)、 由
\begin\{aligned\} 0 < a\_1 < 1, 0 < a\_n < 1\Rightarrow 0 < a\_\{n+1\}=f(a\_n) < a\_n < 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 $\displaystyle 0 < a\_n < 1$, 且 $\displaystyle a\_n\mbox\{严\}\searrow $. 由单调有界定理知 $\displaystyle \lim\_\{n\to\infty\}a\_n=\ell$ 存在, 且
\begin\{aligned\} \ell=f(\ell), 0\leq \ell < 1\stackrel\{f(x) < x, \forall\ 0 < x < 1\}\{\Rightarrow\}\ell=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1-2)、
\begin\{aligned\} \lim\_\{n\to\infty\}na\_n=&\lim\_\{n\to\infty\}\frac\{n\}\{\frac\{1\}\{a\_n\}\} \xlongequal\{\tiny\mbox\{Stolz\}\} \lim\_\{n\to\infty\}\frac\{1\}\{\frac\{1\}\{a\_\{n+1\}\}-\frac\{1\}\{a\_n\}\}\\\\ \xlongequal[\tiny\mbox\{原理\}]\{\tiny\mbox\{归结\}\}&\lim\_\{x\to 0\}\frac\{1\}\{\frac\{1\}\{f(x)\}-\frac\{1\}\{x\}\} =\lim\_\{x\to 0\}\frac\{xf(x)\}\{x-f(x)\}\\\\ \xlongequal\{\tiny\mbox\{L'Hospital\}\}&\lim\_\{x\to 0\}\frac\{f(x)+xf'(x)\}\{1-f'(x)\}\\\\ \xlongequal\{\tiny\mbox\{L'Hospital\}\}& \lim\_\{x\to 0\}\frac\{2f'(x)+xf''(x)\}\{-f''(x)\}=-\frac\{2\}\{f''(0)\}. \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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30、 5、 若数列 $\displaystyle \left\\{x\_n\right\\}, \left\\{y\_n\right\\}$ 满足: $\displaystyle \left\\{y\_n\right\\}$ 有界, $\displaystyle x\_n=\sum\_\{k=1\}^n \frac\{y\_k\}\{k(k+1)\}$, 求证: $\displaystyle \left\\{x\_n\right\\}$ 收敛. (东北大学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) / 设 $\displaystyle |y\_k|\leq M, \forall\ k\geq 1$, 则
\begin\{aligned\} |x\_\{n+p\}-x\_n|&=\left|\sum\_\{k=n+1\}^\{n+p\} \frac\{y\_k\}\{k(k+1)\}\right| \leq M\sum\_\{k=n+1\}^\{n+p\}\left(\frac\{1\}\{k\}-\frac\{1\}\{k+1\}\right)\\\\ &=M\left(\frac\{1\}\{n+1\}-\frac\{1\}\{n+p+1\}\right) < \frac\{M\}\{n+1\}\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 收敛准则即知 $\displaystyle \left\\{x\_n\right\\}$ 收敛.跟锦数学微信公众号. [在线资料](
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---
31、 6、 已知正项级数 $\displaystyle \sum\_\{n=1\}^\infty \alpha\_n$ 收敛, 数列 $\displaystyle \left\\{x\_n\right\\}$ 满足
\begin\{aligned\} |x\_\{n+1\}-x\_n|\leq \alpha\_n, \forall\ n\geq 1. \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. (东北师范大学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](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由正项级数 $\displaystyle \sum\_\{n=1\}^\infty \alpha\_n$ 收敛及 Cauchy 收敛准则知
\begin\{aligned\} &\forall\ \varepsilon > 0,\exists\ N,\mathrm\{ s.t.\}\forall\ n\geq N, \forall\ p\geq 1,\\\\ &\varepsilon > \sum\_\{k=n\}^\{n+p-1\}\alpha\_k \geq \sum\_\{k=n\}^\{n+p-1\}|x\_\{k+1\}-x\_k| \geq |x\_\{n+p\}-x\_n|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由数列极限的 Cauchy 收敛准则即知 $\displaystyle \left\\{x\_n\right\\}$ 收敛.跟锦数学微信公众号. [在线资料](
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32、 3、 求 $\displaystyle \lim\_\{n\to\infty\}\frac\{\sqrt[n]\{n(n+1)\cdots(2n-1)\}\}\{n\}$. (东南大学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\}=&\lim\_\{n\to\infty\} \sqrt[n]\{\left(1+\frac\{0\}\{n\}\right) \left(1+\frac\{1\}\{n\}\right)\cdots \left(1+\frac\{n-1\}\{n\}\right)\}\\\\ =&\mathrm\{exp\}\left\[ \lim\_\{n\to\infty\}\frac\{1\}\{n\} \sum\_\{k=0\}^\{n-1\}\ln\left(1+\frac\{k\}\{n\}\right)\right\]\\\\ =&\mathrm\{exp\}\left\[\int\_0^1 \ln(1+x)\mathrm\{ d\} x\right\] =\mathrm\{exp\}\left(2\ln 2-1\right)=\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\}
其中倒数第二步中, 积分可以通过分部积分方法求解得到.跟锦数学微信公众号. [在线资料](
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---
33、 13、 设 $\displaystyle \left\\{x\_n\right\\}$ 有界, $\displaystyle \left\\{x\_n+2x\_\{n+2\}\right\\}$ 收敛, 证明: $\displaystyle \left\\{x\_n\right\\}$ 收敛. (东南大学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) / 设 $\displaystyle y\_n=x\_n+2x\_\{n+2\}$, 则 $\displaystyle \lim\_\{n\to\infty\}y\_n=b$ 存在. 于是
\begin\{aligned\} x\_n=y\_n-2x\_\{n+2\}\Rightarrow& \left\\{\begin\{array\}\{llllllllllll\}\varlimsup\_\{n\to\infty\}x\_n=b+\varlimsup\_\{n\to\infty\}(-2x\_\{n+2\}) =b-2\varliminf\_\{n\to\infty\}x\_n\\\\ \varliminf\_\{n\to\infty\}x\_n=b+\varliminf\_\{n\to\infty\}(-2x\_\{n+2\}) =b-2\varlimsup\_\{n\to\infty\}x\_n\end\{array\}\right.\\\\ \Rightarrow&\varlimsup\_\{n\to\infty\}x\_n=\varliminf\_\{n\to\infty\}x\_n=\frac\{b\}\{3\}\Rightarrow \lim\_\{n\to\infty\}x\_n=\frac\{b\}\{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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---
34、 1、 (20 分) 设 $\displaystyle a\_1 > 0$, 且 $\displaystyle a\_\{n+1\}=\frac\{3(1+a\_n)\}\{3+a\_n\}, n=1,2,\cdots$. 证明: 数列 $\displaystyle \left\\{a\_n\right\\}$ 收敛, 并求出极限. (福州大学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) / (1)、 先证明压缩数列一定收敛. 设数列 $\displaystyle \left\\{a\_n\right\\}$ 满足压缩性条件, 即存在常数 $\displaystyle k\in (0,1)$, 使得
\begin\{aligned\} |a\_\{n+1\}-a\_n|\leq k|a\_n-a\_\{n-1\}|\left(n=2,3,\cdots\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle \left\\{a\_n\right\\}$ 收敛. 事实上, 由
\begin\{aligned\} |a\_\{n+1\}-a\_n|\leq k|a\_n-a\_\{n-1\}|\leq \cdots \leq k^\{n-1\}|a\_2-a\_1| \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\} |a\_\{n+p\}-a\_n|&\leq \sum\_\{i=n\}^\{n+p-1\}|a\_\{i+1\}-a\_i| \leq \sum\_\{i=n\}^\infty k^\{i-1\}|a\_2-a\_1|\\\\ & =\frac\{k^\{n-1\}\}\{1-k\}|a\_2-a\_1|\to 0\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\}
据 Cauchy 收敛准则即知 $\displaystyle \left\\{a\_n\right\\}$ 收敛. (2)、 由数学归纳法易知 $\displaystyle a\_n > 0$. 设
\begin\{aligned\} f(x)=\frac\{3(1+x)\}\{3+x\}\Rightarrow f'(x)=\frac\{6\}\{(3+x)^2\}\leq \frac\{2\}\{3\}, x > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} |a\_\{n+1\}-a\_n|=|f(a\_n)-f(a\_\{n-1\})|=|f'(\xi\_n)(a\_n-a\_\{n-1\})| \leq\frac\{2\}\{3\}|a\_n-a\_\{n-1\}|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
据第 1 步即知 $\displaystyle \lim\_\{n\to\infty\}a\_n=l$ 存在, 且
\begin\{aligned\} l=\frac\{3(1+l)\}\{3+l\}\Rightarrow l=\sqrt\{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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---
35、 2、 证明题 (每题 10 分, 共 60 分). (1)、 设 $\displaystyle a\_1,\cdots,a\_k$ 是 $\displaystyle k$ 个正数, 证明:
\begin\{aligned\} \lim\_\{n\to\infty\}\sqrt[n]\{a\_1^n+\cdots+a\_k^n\}=\max\left\\{a\_1,\cdots,a\_k\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(广西大学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 $\displaystyle A=\max\left\\{a\_1,\cdots,a\_k\right\\}$, 则由 $\displaystyle A\leq \sqrt[n]\{a\_1^n+\cdots+a\_k^n\}\leq A n^\frac\{1\}\{n\}$ 及迫敛性知结论成立.跟锦数学微信公众号. [在线资料](
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---
36、 2、 (10 分) 设数列 $\displaystyle \left\\{a\_n\right\\}$ 收敛到正数 $\displaystyle a$, 且 $\displaystyle \left\\{a\_n\right\\}$ 恒正. 证明: $\displaystyle \lim\_\{n\to\infty\}\sqrt[n]\{a\_1\cdots a\_n\}=a$. (哈尔滨工程大学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 Stolz 公式知
\begin\{aligned\} \lim\_\{n\to\infty\}\frac\{n\}\{\displaystyle\sum\_\{k=1\}^n \frac\{1\}\{a\_k\}\}=&\lim\_\{n\to\infty\}\frac\{1\}\{\displaystyle\frac\{1\}\{a\_n\}\}=\lim\_\{n\to\infty\}a\_n=a,\\\\ \lim\_\{n\to\infty\}\frac\{\displaystyle\sum\_\{k=1\}^n a\_k\}\{n\}=&\lim\_\{n\to\infty\}\frac\{a\_n\}\{1\}=\lim\_\{n\to\infty\}a\_n=a. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再由均值公式
\begin\{aligned\} \frac\{n\}\{\displaystyle\sum\_\{k=1\}^n \frac\{1\}\{a\_k\}\}\leq \sqrt[n]\{a\_1\cdots a\_n\}\leq \frac\{\displaystyle\sum\_\{k=1\}^n a\_k\}\{n\} \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 $\displaystyle \lim\_\{n\to\infty\}\sqrt[n]\{a\_1\cdots a\_n\}=a$.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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---
37、 (2)、 数列 $\displaystyle \left\\{a\_n\right\\}$ 收敛的充要条件是: 对任意的正整数 $\displaystyle p$, 有 $\displaystyle \lim\_\{n\to\infty\}|a\_\{n+p\}-a\_n|=0$. (哈尔滨工业大学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) / $\displaystyle \times$. 比如 $\displaystyle a\_n=\sqrt\{n\}$ 满足
\begin\{aligned\} \lim\_\{n\to\infty\}|a\_\{n+p\}-a\_n|=\lim\_\{n\to\infty\}\left(\sqrt\{n+p\}-\sqrt\{n\}\right) =\lim\_\{n\to\infty\}\frac\{1\}\{\sqrt\{n+p\}+\sqrt\{n\}\}=0, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
但 $\displaystyle \left\\{a\_n\right\\}$ 发散.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)/
---
38、 2、 设 $\displaystyle f(x)$ 在 $\displaystyle x=a$ 处可微, 且 $\displaystyle f(a)\neq 0$. 求极限
\begin\{aligned\} \lim\_\{n\to\infty\}\left\[\frac\{f\left(a+\frac\{1\}\{n\}\right)\}\{f(a)\}\right\]^n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(哈尔滨工业大学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) / 不妨设 $\displaystyle f(a) > 0$, 否则用 $\displaystyle -f$ 代替 $\displaystyle f$ 后考虑. 于是
\begin\{aligned\} \mbox\{原式\}&=\mathrm\{exp\}\left\\{\lim\_\{n\to\infty\}n\left\[\ln f\left(a+\frac\{1\}\{n\}\right)-\ln f(a)\right\]\right\\}\\\\ &=\mathrm\{exp\}\left\[\lim\_\{x\to 0\}\frac\{\ln f(a+x)-\ln f(a)\}\{x\}\right\]\\\\ & =\mathrm\{exp\}\left\[\left(\ln f(x)\right)'|\_\{x=a\}\right\] =\mathrm\{exp\}\left\[\frac\{f'(a)\}\{f(a)\}\right\]. \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)/
---
39、 1、 判断题. 证明说明理由, 错误举出反例. 每题 5 分, 共 20 分. (1)、 数列 $\displaystyle \left\\{a\_n\right\\}$ 满足 $\displaystyle \forall\ p\in\mathbb\{Z\}\_+$, 成立 $\displaystyle \lim\_\{n\to\infty\}(a\_\{n+p\}-a\_n)=0$, 则 $\displaystyle \left\\{a\_n\right\\}$ 为 Cauchy 列. (河海大学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) / $\displaystyle \times$. 比如 $\displaystyle a\_n=\sqrt\{n\}$ 满足
\begin\{aligned\} \forall\ p\geq 1, \lim\_\{n\to\infty\}(a\_\{n+p\}-a\_n)=\lim\_\{n\to\infty\}\frac\{1\}\{\sqrt\{n+p\}+\sqrt\{n\}\}=0, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
但 $\displaystyle \left\\{a\_n\right\\}$ 发散, 而不是 Cauchy 列.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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---
40、 (2)、 求 $\displaystyle \lim\_\{n\to\infty\}\left(\frac\{\sqrt[n]\{a\}+\sqrt[n]\{b\}\}\{2\}\right)^n$, 其中 $\displaystyle a > 0, b > 0$. (河南大学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\}=&\mathrm\{exp\}\left\[\lim\_\{n\to\infty\}n \ln \frac\{\sqrt[n]\{a\}+\sqrt[n]\{b\}\}\{2\}\right\] \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \mathrm\{exp\}\left\[\lim\_\{n\to\infty\}n \left(\frac\{\sqrt[n]\{a\}+\sqrt[n]\{b\}\}\{2\}-1\right)\right\]\\\\ \xlongequal[\tiny\mbox\{原理\}]\{\tiny\mbox\{归结\}\}&\mathrm\{exp\}\left\[\lim\_\{x\to 0\} \frac\{a^x+b^x-2\}\{2x\}\right\] \xlongequal\{\tiny\mbox\{L'Hospital\}\} \mathrm\{exp\}\left\[\lim\_\{x\to 0\}\frac\{a^x\ln a+b^x\ln b\}\{2\}\right\] =\sqrt\{ab\}. \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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---
41、 (2)、 求 $\displaystyle \lim\_\{n\to\infty\}n^2\left(\arcsin \frac\{a\}\{n\}-\arcsin \frac\{a\}\{n+1\}\right)$, 其中 $\displaystyle a\neq 0$. (华东理工大学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) /
\begin\{aligned\} \mbox\{原式\}\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} \lim\_\{n\to\infty\}n^2\frac\{1\}\{\sqrt\{1-\xi\_n^2\}\}\left(\frac\{a\}\{n\}-\frac\{a\}\{n+1\}\right) =a. \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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---
42、 10、 设 $\displaystyle \alpha > 0$, 若
\begin\{aligned\} nx\_n=1+o(n^\{-\alpha\}). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: 数列 $\displaystyle \left\\{x\_1+\cdots+x\_n-\ln n\right\\}$ 收敛. (华东师范大学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](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle y\_n=x\_1+\cdots+x\_n-\ln n$, 则
\begin\{aligned\} |y\_\{n+p\}-y\_n|=&\left|\sum\_\{k=n+1\}^\{n+p\}x\_k-\ln \left(1+\frac\{p\}\{n\}\right)\right|\\\\ =&\left|\sum\_\{k=n+1\}^\{n+p\}\left\[\frac\{1\}\{k\}+o\left(\frac\{1\}\{k^\{1+\alpha\}\}\right)\right\]-\ln \left(1+\frac\{p\}\{n\}\right)\right|\\\\ \leq&\left|\sum\_\{k=n+1\}^\{n+p\}\frac\{1\}\{k\}-\ln \left(1+\frac\{p\}\{n\}\right)\right| +o\left(\sum\_\{k=n+1\}^\{n+p\}\frac\{1\}\{k^\{1+\alpha\}\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 由
\begin\{aligned\} \sum\_\{k=n+1\}^\{n+p\}\frac\{1\}\{k\}& < \sum\_\{k=n+1\}^\{n+p\}\int\_\{k-1\}^k \frac\{\mathrm\{ d\} x\}\{x\} =\int\_n^\{n+p\}\frac\{\mathrm\{ d\} x\}\{x\} =\ln \left(1+\frac\{p\}\{n\}\right),\\\\ \sum\_\{k=n+1\}^\{n+p\}\frac\{1\}\{k\}& > \sum\_\{k=n+1\}^\{n+p\}\int\_k^\{k+1\}\frac\{\mathrm\{ d\} x\}\{x\} =\int\_\{n+1\}^\{n+p+1\}\frac\{\mathrm\{ d\} x\}\{x\} =\ln \left(1+\frac\{p\}\{n+1\}\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\} &\left|\sum\_\{k=n+1\}^\{n+p\}\frac\{1\}\{k\}-\ln \left(1+\frac\{p\}\{n\}\right)\right| < \ln \left(1+\frac\{p\}\{n\}\right)-\ln \left(1+\frac\{p\}\{n+1\}\right)\\\\ =&\ln \frac\{(n+p)(n+1)\}\{n(n+p+1)\} < \ln \left(1+\frac\{1\}\{n\}\right) < \frac\{1\}\{n\}\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\}
(2)、 由
\begin\{aligned\} \sum\_\{k=n+1\}^\{n+p\}\frac\{1\}\{k^\{1+\alpha\}\}& < \sum\_\{k=n+1\}^\{n+p\}\int\_\{k-1\}^k \frac\{\mathrm\{ d\} x\}\{x^\{1+\alpha\}\} =\int\_n^\{n+p\}\frac\{\mathrm\{ d\} x\}\{x^\{1+\alpha\}\}\\\\ &=\frac\{1\}\{\alpha\}\left\[\frac\{1\}\{n^\alpha\}-\frac\{1\}\{(n+p)^\alpha\}\right\] < \frac\{1\}\{\alpha n^\alpha\} \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\} o\left(\sum\_\{k=n+1\}^\{n+p\}\frac\{1\}\{k^\{1+\alpha\}\}\right) =o\left(\frac\{1\}\{n^\alpha\}\right)\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 收敛准则即知 $\displaystyle \left\\{y\_n\right\\}$ 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](
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---
43、 12、 若数列 $\displaystyle \left\\{a\_n\right\\}$ 满足
\begin\{aligned\} (2-a\_n)a\_\{n+1\}=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: (1)、 存在正整数 $\displaystyle k$, 使得 $\displaystyle a\_k\leq 1$; (2)、 $\displaystyle \left\\{a\_n\right\\}$ 的极限存在, 并求其极限值; (3)、 若 $\displaystyle a\_1\neq 1$, 则 $\displaystyle a\_n\ (n=1,2,\cdots)$ 两两不等; (4)、 满足题设且 $\displaystyle a\_1\neq 1$ 的数列 $\displaystyle \left\\{a\_n\right\\}$ 存在. (华东师范大学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](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle b\_n=\frac\{1\}\{1-a\_n\}$, 则 $\displaystyle a\_n=\frac\{b\_n-1\}\{b\_n\}$,
\begin\{aligned\} &1=(2-a\_n)a\_\{n+1\} =\frac\{1+b\_n\}\{b\_n\}\cdot \frac\{b\_\{n+1\}-1\}\{b\_\{n+1\}\}\\\\ \Rightarrow&b\_\{n+1\}=b\_n+1 \Rightarrow b\_n=b\_1+\sum\_\{k=1\}^\{n-1\}(b\_\{k+1\}-b\_k) =b\_1+n-1\\\\ \Rightarrow&a\_n=1-\frac\{1\}\{b\_1+n-1\} =1-\frac\{1-a\_1\}\{1+(n-1)(1-a\_1)\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 用反证法. 若 $\displaystyle \forall\ k, a\_k > 1$, 则
\begin\{aligned\} 0 > 1-a\_n=\frac\{1-a\_1\}\{1+(n-1)(1-a\_1)\}\Rightarrow a\_1 > 1, n < \frac\{a\_1\}\{a\_1-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这与 $\displaystyle n$ 可任意大矛盾. 故有结论. (2)、 由 $\displaystyle a\_n$ 的表达式即知 $\displaystyle \lim\_\{n\to\infty\}a\_n=1$. (3)、 设
\begin\{aligned\} f(t)=1-\frac\{1-a\_1\}\{1+(t-1)(1-a\_1)\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle f'(t)=\frac\{(1-a\_1)^2\}\{(na\_1-a\_1-n)^2\} > 0$. 故 $\displaystyle f\mbox\{严\}\nearrow $,
\begin\{aligned\} n < m\Rightarrow a\_n=f(n) < f(m)=a\_m. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(4)、 取 $\displaystyle a\_0=0, a\_n=1-\frac\{1\}\{n\}$ 就是一个例子.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)/
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44、 2、 解答如下问题: (1)、 已知 $\displaystyle \lim\_\{n\to\infty\}\frac\{a\_1+\cdots+a\_n\}\{n\}=a$ (有限数), 证明: $\displaystyle \lim\_\{n\to\infty\}\frac\{a\_n\}\{n\}=0$. (华南理工大学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) / 设 $\displaystyle b\_n=\frac\{a\_1+\cdots+a\_n\}\{n\}$, 则 $\displaystyle a\_n=nb\_n-(n-1)b\_\{n-1\}$, 则
\begin\{aligned\} \lim\_\{n\to\infty\}\frac\{a\_n\}\{n\}=\lim\_\{n\to\infty\}\left(b\_n-\frac\{n-1\}\{n\}b\_\{n-1\}\right)=a-1\cdot a=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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
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45、 (2)、 利用柯西收敛准则证明数列 $\displaystyle a\_n=1+\frac\{1\}\{2^k\}+\cdots+\frac\{1\}\{n^k\}\ (0,k\leq 1)$ 发散. (华南理工大学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\} |a\_\{2n-1\}-a\_\{n-1\}|=\sum\_\{m=n\}^\{2n-1\}\frac\{1\}\{m^k\} \geq \sum\_\{m=n\}^\{2n-1\}\int\_m^\{m+1\}\frac\{\mathrm\{ d\} x\}\{x\}=\int\_n^\{2n\}\frac\{\mathrm\{ d\} x\}\{x\}=\ln 2 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及 Cauchy 收敛准则知 $\displaystyle \left\\{a\_n\right\\}$ 发散.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)/
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46、 (2)、 $\displaystyle \lim\_\{n\to\infty\}\left(1+\frac\{1\}\{n\}+\frac\{1\}\{n^2\}\right)^n$. (华南师范大学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](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}&=\mathrm\{exp\}\left\[\lim\_\{n\to\infty\}n\ln\left(1+\frac\{1\}\{n\}+\frac\{1\}\{n^2\}\right)\right\] =\mathrm\{exp\}\left\[\lim\_\{x\to 0\}\frac\{\ln(1+x+x^2)\}\{x\}\right\]\\\\ &\xlongequal[t\to 0]\{\ln (1+t)\sim t\}\mathrm\{exp\}\left\[\lim\_\{x\to 0\}\frac\{x+x^2\}\{x\}\right\] =\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)/
发表于 2023-3-5 08:35:20
