张祖锦2023年数学专业真题分类70天之第03天

zhangzujin · 发表于 2023-3-5 08:39:54

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## 张祖锦2023年数学专业真题分类70天之第03天 --- 47、 (3)、 $\displaystyle \lim\_\{n\to\infty\}\left(\sqrt\{n+1\}A\_1+\sqrt\{n+2\}A\_2+\cdots+\sqrt\{n+k\}A\_k\right)$, 其中 \begin\{aligned\} A\_1+\cdots+A\_k=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (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\}=&\lim\_\{n\to\infty\} \sum\_\{i=1\}^k \sqrt\{n+i\}A\_i =\lim\_\{n\to\infty\}\left(\sum\_\{i=1\}^\{k-1\}\sqrt\{n+i\}A\_i -\sqrt\{n+k\}\sum\_\{i=1\}^\{k-1\}A\_i\right)\\\\ =&\lim\_\{n\to\infty\}\sum\_\{i=1\}^\{k-1\}A\_i \left(\sqrt\{n+i\}-\sqrt\{n+k\}\right)\\\\ =&\lim\_\{n\to\infty\}\sum\_\{i=1\}^\{k-1\}A\_i\frac\{i-k\}\{\sqrt\{n+i\}+\sqrt\{n+k\}\}=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)/ --- 48、 (5)、 $\displaystyle \lim\_\{n\to\infty\}n^n\left(\mathrm\{e\}^\frac\{1\}\{n\}-1\right)^n$. (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 \left\[\ln n+\ln \left(\mathrm\{e\}^\frac\{1\}\{n\}-1\right)\right\]\right\\}\\\\ \xlongequal[\tiny\mbox\{原理\}]\{\tiny\mbox\{归结\}\}& \mathrm\{exp\}\left\[\lim\_\{x\to 0\}\frac\{-\ln x+\ln (\mathrm\{e\}^x-1)\}\{x\}\right\] =\mathrm\{exp\}\left\[\lim\_\{x\to 0\}\frac\{\ln\frac\{\mathrm\{e\}^x-1\}\{x\}\}\{x\}\right\]\\\\ \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\}& \mathrm\{exp\}\left\[\lim\_\{x\to 0\}\frac\{\frac\{\mathrm\{e\}^x-1\}\{x\}-1\}\{x\}\right\] =\mathrm\{exp\}\left\[\lim\_\{x\to 0\}\frac\{\mathrm\{e\}^x-1-x\}\{x^2\}\right\] \xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\} \mathrm\{e\}^\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\} 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)/ --- 49、 1、计算题 (每题 10 分, 共 80 分). (1)、求极限 $\displaystyle \lim\_\{n\to\infty\}\frac\{(1+\cos 1)^3+(2+\cos 2)^3+\cdots+(n+\cos n)^3\}\{n^4\}$. (吉林大学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\} \frac\{1\}\{n^4\}\sum\_\{k=1\}^n (k-1)^3 \leq\frac\{1\}\{n^4\}\sum\_\{k=1\}^n (k+\cos k)^3 \leq \frac\{1\}\{n^4\}\sum\_\{k=1\}^n (k+1)^3, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} \begin\{aligned\} &\lim\_\{n\to\infty\}\frac\{1\}\{n^4\}\sum\_\{k=1\}^n (k\pm 1)^3 \xlongequal\{\tiny\mbox\{Stolz\}\} \lim\_\{n\to\infty\}\frac\{(n\pm 1)^3\}\{(n+1)^4-n^4\}\\\\ =&\lim\_\{n\to\infty\}\frac\{(n\pm 1)^3\}\{4n^3+6n^2+4n+1\}=\frac\{1\}\{4\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 及迫敛性知原式 $\displaystyle =\frac\{1\}\{4\}$.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)/ --- 50、 (2)、求极限 $\displaystyle \lim\_\{n\to\infty\}n^2\left\[\sin\cos\frac\{1\}\{n\}-\sin 1\right\]$. (吉林大学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\{\sin \cos x-\sin 1\}\{x^2\}\\\\ \xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\}& \lim\_\{x\to 0\}\frac\{\cos \xi\_x(\cos x-1)\}\{x^2\}\left(\cos x < \xi\_x < 1\right)\\\\ \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\}&\cos 1\cdot\left(-\frac\{1\}\{2\}\right)=-\frac\{\cos 1\}\{2\}. \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)/ --- 51、 (5)、求极限 $\displaystyle \lim\_\{n\to\infty\} \frac\{1\}\{n\}\left\[\cos\frac\{\pi\}\{n\}+\cos\frac\{2\pi\}\{n\}+\cdots+\cos\frac\{(n-1)\pi\}\{n\}\right\]$. (吉林大学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\}=&\frac\{1\}\{\pi\}\lim\_\{n\to\infty\}\frac\{\pi\}\{n\}\sum\_\{k=1\}^n\cos\frac\{k\pi\}\{n\} =\frac\{1\}\{\pi\}\int\_0^\pi \cos x\mathrm\{ d\} x=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)/ --- 52、 5、 (12 分) 数列 $\displaystyle \left\\{x\_n\right\\}$ 是方程 \begin\{aligned\} x\cot x=\frac\{\pi\}\{2\}\cot x-10 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 在 $\displaystyle \left(\frac\{\pi\}\{2\},+\infty\right)$ 上的解序列, 试证: \begin\{aligned\} \lim\_\{n\to\infty\}\left\[x\_n-\left(n-\frac\{1\}\{2\}\right)\pi\right\]=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (吉林大学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\} x\cot x=\frac\{\pi\}\{2\}\cot x-10 \Leftrightarrow 10=\left(\frac\{\pi\}\{2\}-x\right)\cot x \Leftrightarrow \frac\{\pi\}\{2\}-x=10\tan x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 设 $\displaystyle f(x)=10\tan x+x-\frac\{\pi\}\{2\}$, 则 \begin\{aligned\} f'(x)=10\sec^2x+1 > 0, f\left(n\pi-\frac\{\pi\}\{2\}+0\right)=-\infty, f\left(n\pi+\frac\{\pi\}\{2\}-0\right)=+\infty. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 由连续函数介值定理及函数的严格单调性知 $\displaystyle f(x)=0$ 在 $\displaystyle \left(n\pi-\frac\{\pi\}\{2\},n\pi+\frac\{\pi\}\{2\}\right)$ 上存在唯一的根 $\displaystyle x\_n$, $\displaystyle n\geq 1$. 再由迫敛性知 $\displaystyle \lim\_\{n\to\infty\}x\_n=+\infty$. 又由 \begin\{aligned\} &\tan \left\[x\_n-\left(n-\frac\{1\}\{2\}\right)\pi\right\]=\tan \left(x\_n+\frac\{\pi\}\{2\}\right) =-\frac\{1\}\{\tan x\_n\}=\frac\{10\}\{x\_n-\frac\{\pi\}\{2\}\}\to 0,\\\\ &0 < x\_n-\left(n-\frac\{1\}\{2\}\right)\pi < \pi \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\} &\tan \left\\{\lim\_\{n\to\infty\}\left\[x\_n-\left(n-\frac\{1\}\{2\}\right)\pi\right\]\right\\} =\lim\_\{n\to\infty\}\tan \left\[x\_n-\left(n-\frac\{1\}\{2\}\right)\pi\right\]=0\\\\ \Rightarrow& \lim\_\{n\to\infty\}\left\[x\_n-\left(n-\frac\{1\}\{2\}\right)\pi\right\]=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)/ --- 53、 (0-20)、证明数列 \begin\{aligned\} \sqrt\{12\}, \sqrt\{12+\sqrt\{12\}\}, \sqrt\{12+\sqrt\{12+\sqrt\{12\}\}\},\cdots \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 收敛, 并求其极限. (吉林师范大学2023年(学科数学)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 \begin\{aligned\} a\_1=\sqrt\{12\}, a\_\{n+1\}=\sqrt\{12+a\_n\}\left(n\geq 1\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 则 \begin\{aligned\} |a\_n-4|=&\left|\sqrt\{12+a\_\{n-1\}\}-\sqrt\{12+4\}\right| =\frac\{|a\_\{n-1\}-4|\}\{\sqrt\{12+a\_\{n-1\}\}+\sqrt\{12+4\}\}\\\\ \leq& \frac\{|a\_\{n-1\}-4|\}\{4\} \leq \cdots \leq \frac\{|a\_1-4|\}\{4^\{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\} 故 $\displaystyle \lim\_\{n\to\infty\}a\_n=4$.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)/ --- 54、 1、计算题. (1)、求极限 $\displaystyle \lim\_\{n\to\infty\}\frac\{\sqrt[3]\{n^2\}\cos\left(n+\frac\{1\}\{n\}\right) \sin\frac\{1\}\{n+1\}\}\{\sqrt[3]\{n^2\}+2023n-2022\}$. (暨南大学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) / 由 \begin\{aligned\} \left|\sqrt[3]\{n^2\}\cos\left(n+\frac\{1\}\{n\}\right) \sin\frac\{1\}\{n+1\}\right|\leq n^\frac\{2\}\{3\} \frac\{1\}\{n\}, \sqrt[3]\{n^2\}+2023n-2022\sim 2023n \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](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 55、 2、证明题. (1)、已知 $\displaystyle \left\\{a\_n\right\\}$ 满足 $\displaystyle \lim\_\{n\to\infty\}a\_n=a$, 证明: $\displaystyle \lim\_\{n\to\infty\}\frac\{a\_1+2a\_2+\cdots+na\_n\}\{1+2+\cdots+n\}=a$. (暨南大学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 $\displaystyle x\_n=\frac\{a\_1+2a\_2+\cdots+na\_n\}\{1+2+\cdots+n\}$. 当 $\displaystyle a=0$ 时, 由 $\displaystyle \lim\_\{n\to\infty\}a\_n=0$ 知 $\displaystyle \forall\ \varepsilon > 0,\ \exists\ N\_1 > 0,\mathrm\{ s.t.\} n > N\_1\Rightarrow |a\_n| < \frac\{\varepsilon\}\{2\}.$ 对上述 $\displaystyle N\_1$, 由 $\displaystyle \lim\_\{n\to\infty\}\frac\{N\_1(N\_1+1)M\}\{n(n+1)\}=0$ 知 \begin\{aligned\} \exists\ N > N\_1,\mathrm\{ s.t.\} n\geq N\Rightarrow \frac\{N\_1(N\_1+1)M\}\{n(n+1)\} < \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 $\displaystyle n\geq N$ 时, \begin\{aligned\} |x\_n|=&\left|\frac\{a\_1+2a\_2+\cdots+na\_n\}\{1+2+\cdots+n\}\right|\\\\ & \leq \left|\frac\{a\_1+\cdots+N\_1a\_\{N\_1\}\}\{1+2+\cdots+n\}\right|+ \left|\frac\{(N\_1+1)a\_\{N\_1+1\}+\cdots+\cdots+na\_n\}\{1+2+\cdots+n\}\right|\\\\ & \leq \frac\{N\_1(N\_1+1)M\}\{n(n+1)\}+\frac\{(N\_1+1)+\cdots+n\}\{1+2+\cdots+n\}\max\_\{N\_1+1\leq i\leq n\}|a\_i| \leq \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\} 这表明 $\displaystyle \lim\_\{n\to\infty\}x\_n=0$. 当 $\displaystyle a\neq 0$ 时, 设 $\displaystyle b\_n=a\_n-a$, 则 $\displaystyle \lim\_\{n\to\infty\}b\_n=0$. 由已证知 \begin\{aligned\} 0=\lim\_\{n\to\infty\}\frac\{b\_1+2b\_2+\cdots+nb\_n\}\{1+2+\cdots+n\} =\lim\_\{n\to\infty\}\frac\{a\_1+2a\_2+\cdots+na\_n\}\{1+2+\cdots+n\}-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](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 56、 (2)、设 $\displaystyle f\_n(x)=\cos x+\cos^2x+\cdots+\cos^nx$. 证明: 对任意的正整数 $\displaystyle n$, $\displaystyle f\_n(x)=1$ 在 $\displaystyle \left\[0,\frac\{\pi\}\{3\}\right)$ 内有且仅有一个根 $\displaystyle x\_n$, 进一步证明 $\displaystyle \lim\_\{n\to\infty\}x\_n$ 存在, 且为 $\displaystyle \frac\{\pi\}\{3\}$. (暨南大学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) / (2-1)、由 $\displaystyle f(0)=n > 1=\sum\_\{k=1\}^n\frac\{1\}\{2^k\}=f\left(\frac\{\pi\}\{3\}\right)$ 及连续函数介值定理知 $\displaystyle \exists\ x\_n\in \left\[0,\frac\{\pi\}\{3\}\right),\mathrm\{ s.t.\} f(x\_n)=1$. 又由 \begin\{aligned\} f\_n'(x)=-\sin x-2\cos x\sin x-\cdots -n\cos^\{n-1\} x\sin x < 0,x\in \left\[0,\frac\{\pi\}\{3\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle x\_n$ 是 $\displaystyle f\_n(x)=1$ 的唯一根. (2-2)、由 $\displaystyle f\_n\leq f\_\{n+1\}$ 及 $\displaystyle f\_n\mbox\{严\}\searrow $ 知 \begin\{aligned\} f\_\{n+1\}(x\_\{n+1\})=1=f\_n(x\_n)\leq f\_\{n+1\}(x\_n)\Rightarrow x\_\{n+1\}\geq 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 $\displaystyle \left\\{x\_n\right\\}$ 递增有上界. 据单调有界定理知 $\displaystyle \lim\_\{n\to\infty\}x\_n=l$ 存在. 由 $\displaystyle x\_n\nearrow$ 知 $\displaystyle 0=x\_1 < x\_2\leq x\_n < \frac\{\pi\}\{3\}$, \begin\{aligned\} &0 < x\_2 < l < \frac\{1\}\{2\} \Rightarrow \frac\{1\}\{2\}\leq \cos x\_n\leq \cos x\_2 < 1\\\\ \Rightarrow& \frac\{1\}\{2^n\}\leq \cos^n x\_n\leq \cos^n x\_2\Rightarrow \lim\_\{n\to\infty\}\cos^nx\_n=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 在 \begin\{aligned\} 1=\cos x\_n+\cos^2x\_n+\cdots+\cos^n x\_n=\frac\{\cos x\_n(1-\cos^nx\_n)\}\{1-\cos x\_n\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 中令 $\displaystyle n\to\infty$ 得 $\displaystyle 1=\frac\{\cos l\}\{1-\cos l\}\Rightarrow \cos l=\frac\{1\}\{2\}\Rightarrow l=\frac\{\pi\}\{3\}$.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)/ --- 57、 1、计算题. (1)、求 $\displaystyle \lim\_\{n\to\infty\}\left(\frac\{1\}\{\sqrt\{n^2+1^2\}\}+\frac\{1\}\{\sqrt\{n^2+2^2\}\} +\cdots+\frac\{1\}\{\sqrt\{n^2+n^2\}\}\right)$. (南昌大学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\{1\}\{\sqrt\{n^2+k^2\}\} =\lim\_\{n\to\infty\}\frac\{1\}\{n\}\sum\_\{k=1\}^n \frac\{1\}\{\sqrt\{1+\left(\frac\{k\}\{n\}\right)^2\}\} =\int\_0^1 \frac\{\mathrm\{ d\} x\}\{\sqrt\{1+x^2\}\}\\\\ & \stackrel\{x=\tan \theta\}\{=\}\int\_0^\frac\{\pi\}\{4\}\sec \theta\mathrm\{ d\} \theta=\left.\ln (\sec\theta+\tan \theta)\right|\_0^\frac\{\pi\}\{4\} =\ln \left(\sqrt\{2\}+1\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)/ --- 58、 8、设 $\displaystyle \sum\_\{n=1\}^\infty a\_n$ 收敛, 则下列级数是否一定收敛? 若收敛, 请证明; 若不收敛, 请举例说明. (1)、 $\displaystyle \sum\_\{n=1\}^\infty a\_n^2$; (2)、 $\displaystyle \sum\_\{n=1\}^\infty \sqrt\{a\_n\}\ (a\_n\geq 0)$; (3)、 $\displaystyle \sum\_\{n=1\}^\infty \sqrt\{a\_na\_\{n+1\}\}\ (a\_n\geq 0)$. (南昌大学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)、 $\displaystyle \sum\_\{n=1\}^\infty a\_n^2$ 未必收敛, 比如 $\displaystyle a\_n=\frac\{(-1)^n\}\{\sqrt\{n\}\}$. (2)、 $\displaystyle \sum\_\{n=1\}^\infty \sqrt\{a\_n\}\ (a\_n\geq 0)$ 未必收敛, 比如 $\displaystyle a\_n=\frac\{1\}\{n^2\}$. (3)、 $\displaystyle \sum\_\{n=1\}^\infty \sqrt\{a\_na\_\{n+1\}\}\ (a\_n\geq 0)$ 一定收敛. 事实上, \begin\{aligned\} \sum\_\{n=1\}^\infty \sqrt\{a\_na\_\{n+1\}\}\stackrel\{\tiny\mbox\{Schwarz\}\}\{\leq\} \sum\_\{n=1\}^\infty \frac\{a\_n+a\_\{n+1\}\}\{2\} \leq \frac\{1\}\{2\}\sum\_\{n=1\}^\infty a\_n+\frac\{1\}\{2\}\sum\_\{n=1\}^\infty a\_n=\sum\_\{n=1\}^\infty a\_n < \infty. \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)/ --- 59、 1、 (10 分) 求极限 \begin\{aligned\} \lim\_\{n\to\infty\}\left(\frac\{5^\frac\{1\}\{n\}\}\{n+1\} +\frac\{5^\frac\{2\}\{n\}\}\{n+\frac\{1\}\{2\}\}+\cdots+\frac\{5^\frac\{n\}\{n\}\}\{n+\frac\{1\}\{n\}\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (南京大学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\} \frac\{n\}\{n+1\}\cdot\frac\{1\}\{n\}\sum\_\{k=1\}^n 5^\frac\{k\}\{n\} < \sum\_\{k=1\}^n \frac\{5^\frac\{k\}\{n\}\}\{n+\frac\{1\}\{k\}\} < \frac\{1\}\{n\}\sum\_\{k=1\}^n 5^\frac\{k\}\{n\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 及迫敛性知原式 $\displaystyle =\int\_0^1 5^x\mathrm\{ d\} x=\frac\{4\}\{\ln 5\}$.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)/ --- 60、 (3)、设 $\displaystyle 0 < k < 1$, $\displaystyle a > 0$, 求极限 \begin\{aligned\} \lim\_\{n\to\infty\}(a^\frac\{1\}\{n\}+a^\frac\{1\}\{n-1\}k+\cdots+ak^\{n-1\}+k^n). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (南京师范大学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) / (1)、先证明一个结论. 设 $\displaystyle 0 < \kappa < 1$, $\displaystyle \lim\_\{n\to\infty\}a\_n=a$, 则 \begin\{aligned\} \lim\_\{n\to\infty\}\left(a\_n+\kappa a\_\{n-1\}+\cdots+\kappa^\{n-1\}a\_1+\kappa^n a\_0\right)=\frac\{a\}\{1-\kappa\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 事实上, 由 $\displaystyle \lim\_\{n\to\infty\}a\_n=a$ 知 \begin\{aligned\} &\exists\ M > 0,\mathrm\{ s.t.\} n\geq 1\Rightarrow |a\_n|\leq M;\\\\ &\forall\ \varepsilon > 0,\ \exists\ N,\mathrm\{ s.t.\} n\geq N\Rightarrow |a\_n-a| < (1-\kappa)\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\} 对上述 $\displaystyle N$, 又由 $\displaystyle \lim\_\{n\to\infty\} M\frac\{1-\kappa^\{-N\}\}\{1-\kappa^\{-1\}\}=0$ 知 \begin\{aligned\} \exists\ N\_1\geq N,\mathrm\{ s.t.\} n\geq N\_1 \Rightarrow M\frac\{1-\kappa^\{-N\}\}\{1-\kappa^\{-1\}\} < \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\} 于是对 $\displaystyle n\geq N\_1$, \begin\{aligned\} &\left|a\_n+\kappa a\_\{n-1\}+\cdots+\kappa^na\_0-\frac\{1-\kappa^n\}\{1-\kappa\}a\right|\\\\ \leq& |a\_n-a| +\kappa|a\_\{n-1\}-a| +\cdots +\kappa^n |a\_0-a|\\\\ =&\sum\_\{k=N\}^n \kappa^\{n-k\}|a\_k-a| +\sum\_\{k=0\}^\{N-1\} \kappa^\{n-k\}|a\_k-a|\\\\ \leq& \sum\_\{k=N\}^n \kappa^\{n-k\} \sup\_\{k\geq N\}|a\_k-a| +M\sum\_\{k=0\}^\{N-1\} \kappa^\{n-k\}\\\\ =& \frac\{1-\kappa^\{n-N+1\}\}\{1-\kappa\} \sup\_\{k\geq N\}|a\_k-a| +M\frac\{1-\kappa^\{-N\}\}\{1-\kappa^\{-1\}\}\\\\ \leq& \frac\{1\}\{1-\kappa\}\sup\_\{k\geq N\}|a\_k-a| +M\frac\{1-\kappa^\{-N\}\}\{1-\kappa^\{-1\}\} \leq\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\} 这表明 \begin\{aligned\} &\lim\_\{n\to\infty\}\left\[a\_n+\kappa a\_\{n-1\}+\cdots+\kappa^na\_0-\frac\{1-\kappa^n\}\{1-\kappa\}a \right\]=0\\\\ \Rightarrow &\lim\_\{n\to\infty\}(a\_n+\kappa a\_\{n-1\}+\cdots+\kappa^na\_0) =\lim\_\{n\to\infty\}\frac\{1-\kappa^n\}\{1-\kappa\}a=\frac\{a\}\{1-\kappa\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (2)、在第 1 步中取 $\displaystyle a\_n=a^\frac\{1\}\{n\}\xrightarrow\{n\to\infty\}1$, 则原式 $\displaystyle =\frac\{1\}\{1-k\}$.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)/ --- 61、 3、解答如下问题: (1)、用定积分计算 $\displaystyle \lim\_\{n\to\infty\}\frac\{1\}\{n\}\left(\arctan \frac\{1\}\{n\}+\arctan\frac\{2\}\{n\}+\cdots+\arctan\frac\{n\}\{n\}\right)$; (南京师范大学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) / \begin\{aligned\} \mbox\{原式\}=&\int\_0^1\arctan x\mathrm\{ d\} x\xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} \arctan 1-\int\_0^1 \frac\{1\}\{1+x^2\}\cdot x\mathrm\{ d\} x =\frac\{\pi\}\{4\}-\frac\{\ln 2\}\{2\}. \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)/ --- 62、 9、命题: 对于数列 $\displaystyle \left\\{a\_n\right\\}$, 若 $\displaystyle \lim\_\{n\to\infty\}|a\_\{n+1\}-a\_n|=0$, 则 $\displaystyle \left\\{a\_n\right\\}$ 收敛. (1)、举例说明上述命题是假命题. (2)、在原命题上修改, 使其为真命题, 并证明. (南京师范大学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) / (1)、命题是假命题. 比如 $\displaystyle a\_n=\sqrt\{n\}$, 则 \begin\{aligned\} \lim\_\{n\to\infty\}\left(a\_\{n+1\}-a\_n\right)=\lim\_\{n\to\infty\}\frac\{1\}\{\sqrt\{n+1\}+\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 $\displaystyle \left\\{a\_n\right\\}$ 无界而发散. (2)、若 $\displaystyle a\_\{n+p\}-a\_n\xrightarrow\{n\to\infty\}0$ 关于 $\displaystyle p$ 一致, 则由 Cauchy 收敛准则知 $\displaystyle \left\\{a\_n\right\\}$ 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](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 63、 1、设 $\displaystyle 0 < \kappa < 1$, $\displaystyle \lim\_\{n\to\infty\}a\_n=a$, 证明: \begin\{aligned\} \lim\_\{n\to\infty\}\left(a\_n+\kappa a\_\{n-1\}+\cdots+\kappa^\{n-1\}a\_1+\kappa^n a\_0\right)=\frac\{a\}\{1-\kappa\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} (厦门大学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\} &\exists\ M > 0,\mathrm\{ s.t.\} n\geq 1\Rightarrow |a\_n|\leq M;\\\\ &\forall\ \varepsilon > 0,\ \exists\ N,\mathrm\{ s.t.\} n\geq N\Rightarrow |a\_n-a| < (1-\kappa)\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\} 对上述 $\displaystyle N$, 又由 $\displaystyle \lim\_\{n\to\infty\} M\frac\{1-\kappa^\{-N\}\}\{1-\kappa^\{-1\}\}=0$ 知 \begin\{aligned\} \exists\ N\_1\geq N,\mathrm\{ s.t.\} n\geq N\_1 \Rightarrow M\frac\{1-\kappa^\{-N\}\}\{1-\kappa^\{-1\}\} < \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\} 于是对 $\displaystyle n\geq N\_1$, \begin\{aligned\} &\left|a\_n+\kappa a\_\{n-1\}+\cdots+\kappa^na\_0-\frac\{1-\kappa^n\}\{1-\kappa\}a\right|\\\\ \leq& |a\_n-a| +\kappa|a\_\{n-1\}-a| +\cdots +\kappa^n |a\_0-a|\\\\ =&\sum\_\{k=N\}^n \kappa^\{n-k\}|a\_k-a| +\sum\_\{k=0\}^\{N-1\} \kappa^\{n-k\}|a\_k-a|\\\\ \leq& \sum\_\{k=N\}^n \kappa^\{n-k\} \sup\_\{k\geq N\}|a\_k-a| +M\sum\_\{k=0\}^\{N-1\} \kappa^\{n-k\}\\\\ =& \frac\{1-\kappa^\{n-N+1\}\}\{1-\kappa\} \sup\_\{k\geq N\}|a\_k-a| +M\frac\{1-\kappa^\{-N\}\}\{1-\kappa^\{-1\}\}\\\\ \leq& \frac\{1\}\{1-\kappa\}\sup\_\{k\geq N\}|a\_k-a| +M\frac\{1-\kappa^\{-N\}\}\{1-\kappa^\{-1\}\} \leq\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\} 这表明 \begin\{aligned\} &\lim\_\{n\to\infty\}\left\[a\_n+\kappa a\_\{n-1\}+\cdots+\kappa^na\_0-\frac\{1-\kappa^n\}\{1-\kappa\}a \right\]=0\\\\ \Rightarrow &\lim\_\{n\to\infty\}(a\_n+\kappa a\_\{n-1\}+\cdots+\kappa^na\_0) =\lim\_\{n\to\infty\}\frac\{1-\kappa^n\}\{1-\kappa\}a=\frac\{a\}\{1-\kappa\}. \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)/ --- 64、 2、 (60 分) 证明题. (1)、 (20 分) 设 $\displaystyle a\_n\geq 0, \varlimsup\_\{n\to\infty\}\sqrt[n]\{a\_n\}\leq 1$. 证明: 对任意的 $\displaystyle l > 1$, 有 $\displaystyle \lim\_\{n\to\infty\}\frac\{a\_n\}\{l^n\}=0$. (山东大学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\} &\frac\{1+l\}\{2\} > l\geq \varlimsup\_\{n\to\infty\}\sqrt[n]\{a\_n\}=\inf\_\{n\geq 1\}\sup\_\{k\geq n\}\sqrt[k]\{a\_k\}\\\\ \Rightarrow&\exists\ n,\mathrm\{ s.t.\} \forall\ k\geq n, \sqrt[k]\{a\_k\} < \frac\{1+l\}\{2\}\\\\ \Rightarrow&\forall\ k\geq n, 0\leq \frac\{a\_k\}\{l^k\} < \left(\frac\{\frac\{1+l\}\{2\}\}\{l\}\right)^k \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 $\displaystyle \lim\_\{n\to\infty\}\frac\{a\_n\}\{l^n\}=0$.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](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 65、 1、 (15 分) 用定积分计算极限 $\displaystyle \lim\_\{n\to\infty\}n\left(\frac\{1\}\{n^2+1\}+\frac\{1\}\{n^2+2^2\}+\cdots+\frac\{1\}\{2n^2\}\right)$. (山西大学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) / \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\{\mathrm\{ d\} x\}\{1+x^2\}=\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](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 66、 2、 (10 分) 设 $\displaystyle x\_1=10, x\_\{n+1\}=\sqrt\{6+x\_n\}, n=1,2,\cdots$. 试证数列 $\displaystyle \left\\{x\_n\right\\}$ 极限存在, 并求此极限. (陕西师范大学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 \begin\{aligned\} x\_1=10 > 3, x\_n > 3 \Rightarrow x\_\{n+1\}=\sqrt\{6+x\_n\} > 3 \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 $\displaystyle x\_n > 3$. 又由 \begin\{aligned\} x\_\{n+1\}-x\_n&=\sqrt\{6+x\_n\}-\sqrt\{6+x\_\{n-1\}\} =\frac\{x\_n-x\_\{n-1\}\}\{\sqrt\{6+x\_n\}+\sqrt\{6+x\_\{n-1\}\}\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle x\_\{n+1\}-x\_n$ 与 $\displaystyle x\_n-x\_\{n-1\}$ 同号, 而与 \begin\{aligned\} x\_\{n-2\}-x\_\{n-3\},\cdots, x\_2-x\_1=\sqrt\{6+10\}-10=4-10 < 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 同号. 故 $\displaystyle x\_n\searrow$. 据单调有界定理知 $\displaystyle \lim\_\{n\to\infty\}x\_n=l$ 存在, 且 \begin\{aligned\} l=\sqrt\{6+l\}\Rightarrow l=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](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/ --- 67、 1、简答题. 成立请给出证明, 错误请举出反例. (1)、已知 $\displaystyle x\_0 > 1, x\_n=x\_\{n-1\}+\frac\{1\}\{x\_\{n-1\}\}, n\geq 1$. 数列 $\displaystyle \left\\{\frac\{x\_n\}\{\sqrt\{n\}\}\right\\}$ 是否收敛? (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) / 收敛. 由归纳法知 $\displaystyle x\_n > 0$, 从而 $\displaystyle x\_n-x\_\{n-1\}=\frac\{1\}\{x\_\{n-1\}\} > 0\Rightarrow x\_n\nearrow$. 若 $\displaystyle \lim\_\{n\to\infty\}x\_n=\ell$ 存在, 则 $\displaystyle \ell=\ell+\frac\{1\}\{\ell\}$, 矛盾. 故 $\displaystyle \lim\_\{n\to\infty\}x\_n=+\infty$. 再由 \begin\{aligned\} &x\_n^2=\left(x\_\{n-1\}+\frac\{1\}\{x\_\{n-1\}\}\right)^2 =2+x\_\{n-1\}^2+\frac\{1\}\{x\_\{n-1\}^2\}\\\\ \Rightarrow&x\_n^2-x\_\{n-1\}^2=2+\frac\{1\}\{x\_\{n-1\}^2\}\to 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\} 知 \begin\{aligned\} \lim\_\{n\to\infty\}\frac\{x\_n^2\}\{2n\}\xlongequal\{\tiny\mbox\{Stolz\}\}\lim\_\{n\to\infty\}\frac\{x\_n^2-x\_\{n-1\}^2\}\{2\}=1 \Rightarrow \lim\_\{n\to\infty\}\frac\{x\_n\}\{\sqrt\{n\}\}=\sqrt\{2\}. \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)/ --- 68、 3、求极限. (1)、求 $\displaystyle \lim\_\{n\to\infty\}\frac\{(1-\sqrt\{\cos x\})(1-\sqrt[3]\{\cos x\}) \cdots (1-\sqrt[n]\{\cos x\})\}\{(1-\cos x)^\{n-1\}\}$. (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) / \begin\{aligned\} \mbox\{原式\}&=\lim\_\{n\to\infty\}\frac\{1\}\{\boxed\{\begin\{array\}\{c\}(1+\sqrt\{x\})(1+\sqrt[3]\{x\}+\sqrt[3]\{\cos^2x\})\cdots\\\\ (1+\sqrt[n]\{\cos x\}+\cdots+\sqrt[n]\{\cos^\{n-1\}x\})\end\{array\}\}\} =\frac\{1\}\{2\cdot 3\cdots n\}=\frac\{1\}\{n!\}. \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)/ --- 69、 (2)、求 $\displaystyle \lim\_\{n\to\infty\}n\sin(2\pi \mathrm\{e\} n!)$. (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) / 由 \begin\{aligned\} &\quad n\sin (2\pi \mathrm\{e\} n!)\\\\ &\xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\} \sin \left\[2\pi n! \left(1+\frac\{1\}\{1!\}+\frac\{1\}\{2!\}+\cdots+\frac\{1\}\{n!\} +\frac\{1\}\{(n+1)!\} +\frac\{\mathrm\{e\}^\{\xi\_n\}\}\{(n+2)!\}\right)\right\]\\\\ &=n \sin 2\pi \left(\frac\{1\}\{n+1\}+\frac\{\mathrm\{e\}^\{\xi\_n\}\}\{(n+2)(n+1)\}\right)\\\\ &\sim n \cdot 2\pi \left(\frac\{1\}\{n+1\}+\frac\{\mathrm\{e\}^\{\xi\_n\}\}\{(n+2)(n+1)\}\right) \xrightarrow\{n\to\infty\}2\pi \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知原式 $\displaystyle =2\pi$.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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