## 张祖锦2023年数学专业真题分类70天之第21天
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461、 1、 计算题. (1)、 求 $\displaystyle \lim\_\{n\to\infty\}\frac\{1^2+3^2+\cdots+(2n-1)^2\}\{n^3\}$. (东北师范大学2023年数学分析考研试题) [数项级数 ]
[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\} &\sum\_\{k=1\}^n (2k-1)^2=\sum\_\{k=1\}^n (4k^2-4k+1)\\\\ =&4\frac\{n(n+1)(2n-1)\}\{6\}-4\frac\{n(n+1)\}\{2\}+n \sim \frac\{4n^3\}\{3\} \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 $\displaystyle =\frac\{4\}\{3\}$.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)/
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462、 10、 设 $\displaystyle a > 0$, 讨论 $\displaystyle \sum\_\{n=1\}^\infty \left(\sqrt[n]\{a\}-\sqrt\{1+\frac\{1\}\{n\}\}\right)$ 的敛散性. (东南大学2023年数学分析考研试题) [数项级数 ]
[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^x-(1+x)^\frac\{1\}\{2\}=\mathrm\{e\}^\{x\ln a\}-(1+x)^\frac\{1\}\{2\} =\left(\ln a-\frac\{1\}\{2\}\right)x+O(x^2) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \sqrt[n]\{a\}-\sqrt\{1+\frac\{1\}\{n\}\}=\frac\{\ln a-\frac\{1\}\{2\}\}\{n\}+O\left(\frac\{1\}\{n^2\}\right)$. 故当且仅当 $\displaystyle \ln a=\frac\{1\}\{2\}\Leftrightarrow a=\sqrt\{\mathrm\{e\}\}$ 时, 原级数发散.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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463、 5、 证明 级数 $\displaystyle \sum\_\{n=1\}^\infty \frac\{(-1)^\{\sqrt\{n\}\}\}\{n\}$ 收敛. (哈尔滨工程大学2023年数学分析考研试题) [数项级数 ]
[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\[\sqrt\{n\}\right\]=k\Leftrightarrow k\leq \sqrt\{n\} < k+1\Leftrightarrow k^2\leq n < (k+1)^2 \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\} \mbox\{原式\}&=\sum\_\{k=1\}^\infty (-1)^k \left\[\frac\{1\}\{k^2\}+\frac\{1\}\{k^2+1\}+\cdots+\frac\{1\}\{(k+1)^2-1\}\right\] \equiv \sum\_\{k=1\}^\infty (-1)^k u\_k. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
$\displaystyle u\_k$ 共有 $\displaystyle 2k+1$ 项, 我们估计如下:
\begin\{aligned\} u\_k&=\sum\_\{i=0\}^\{k-1\} \frac\{1\}\{k^2+i\}+\sum\_\{i=k\}^\{2k\} \frac\{1\}\{k^2+i\} < \frac\{1\}\{k^2\}\cdot k+\frac\{1\}\{k^2+k\}\cdot (k+1)=\frac\{2\}\{k\},\\\\ u\_k& > \sum\_\{i=0\}^\{k-1\}\frac\{1\}\{k^2+k\} +\sum\_\{i=k\}^\{2k\}\frac\{1\}\{k^2+2k\} =\frac\{1\}\{k+1\}+\frac\{k+1\}\{k(k+2)\} > \frac\{2\}\{k+1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle u\_k > \frac\{2\}\{k+1\} > u\_\{k+1\}$. 据 Leibniz 判别法即知 $\displaystyle \sum\_\{k=1\}^\infty (-1)^k u\_k$ 收敛. 又原级数的任一部分和总是夹在新级数 $\displaystyle \sum\_\{k=1\}^\infty (-1)^k u\_k$ 的两部分和之间, 所以原级数也收敛.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)/
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464、 (4)、 对于数列 $\displaystyle \left\\{a\_n\right\\}, \left\\{b\_n\right\\}$, 总有 $\displaystyle a\_n\leq b\_n$. 若 $\displaystyle \sum\_\{n=1\}^\infty b\_n$ 收敛, 则 $\displaystyle \sum\_\{n=1\}^\infty a\_n$ 收敛. (哈尔滨工业大学2023年数学分析考研试题) [数项级数 ]
[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 $\displaystyle a\_n=-1, b\_n=\frac\{1\}\{n^2\}$.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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465、 6、 证明级数 $\displaystyle \sum\_\{n=1\}^\infty \left(\frac\{1\}\{\sqrt\{n\}\}-\sqrt\{\ln \frac\{n+1\}\{n\}\}\right)$ 收敛, 并且其和小于 $\displaystyle 1$. (哈尔滨工业大学2023年数学分析考研试题) [数项级数 ]
[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\}\{\sqrt\{k\}\}-\sqrt\{\ln \frac\{k+1\}\{k\}\} =&\frac\{1\}\{\sqrt\{k\}\}-\sqrt\{\ln (k+1)-\ln k\}\\\\ \xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\}& \frac\{1\}\{\sqrt\{k\}\}-\sqrt\{\frac\{1\}\{\xi\_k\}\}\left(k < \xi\_k < k+1\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知原级数为正项级数, 且
\begin\{aligned\} &\sum\_\{k=1\}^n \left(\frac\{1\}\{\sqrt\{k\}\}-\sqrt\{\ln \frac\{k+1\}\{k\}\}\right) =\sum\_\{k=1\}^n \left(\frac\{1\}\{\sqrt\{k\}\}-\sqrt\{\frac\{1\}\{\xi\_k\}\}\right)\\\\ < &\sum\_\{k=1\}^n \left(\frac\{1\}\{\sqrt\{k\}\}-\sqrt\{\frac\{1\}\{k+1\}\}\right) =1-\frac\{1\}\{\sqrt\{n+1\}\} < 1.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故原级数的部分和有界, 从而原级数收敛. 进一步,
\begin\{aligned\} \sum\_\{n=1\}^\infty \left(\frac\{1\}\{\sqrt\{n\}\}-\sqrt\{\ln \frac\{n+1\}\{n\}\}\right) =&1-\sqrt\{\ln 2\}+\sum\_\{n=2\}^\infty\left(\frac\{1\}\{\sqrt\{n\}\}-\sqrt\{\ln \frac\{n+1\}\{n\}\}\right)\\\\ \stackrel\{\mbox\{类似 $\displaystyle (I)$\}\}\{\leq\}&1-\sqrt\{\ln 2\}+\frac\{1\}\{\sqrt\{2\}\} < 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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)/
---
466、 (4)、 级数 $\displaystyle \sum\_\{n=1\}^\infty a\_n$ 收敛, 则 $\displaystyle a\_n=o\left(\frac\{1\}\{n\}\right)$. (河海大学2023年数学分析考研试题) [数项级数 ]
[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 $\displaystyle a\_n=\left\\{\begin\{array\}\{llllllllllll\}\frac\{1\}\{n^2\},&n\neq k^2\left(k=1,2,\cdots\right),\\\\ \frac\{1\}\{k\},&n=k^2\end\{array\}\right.$. 我们有
\begin\{aligned\} \sum\_\{n=1\}^\infty a\_n=\sum\_\{n\not\in \left\\{1,4,9,\cdots\right\\}\}\frac\{1\}\{n^2\} +\sum\_\{n\in\left\\{1,4,9,\cdots\right\\}\}\frac\{1\}\{n\} \leq \sum\_\{n=1\}^\infty \frac\{1\}\{n^2\}+\sum\_\{n=1\}^\infty \frac\{1\}\{n^2\} < \infty, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
但
\begin\{aligned\} k^2a\_\{k^2\}=k^2\cdot \frac\{1\}\{k\}=k\not\to 0\left(k\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\}
故 $\displaystyle \lim\_\{n\to\infty\}na\_n=0$ 不成立.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)/
---
467、 13、 设 $\displaystyle \sum\_\{n=1\}^\infty x\_n$ 收敛, $\displaystyle \lim\_\{n\to\infty\}\frac\{y\_n\}\{x\_n\}=1$, 问: $\displaystyle \sum\_\{n=1\}^\infty y\_n$ 是否收敛? (黑龙江大学2023年数学分析考研试题) [数项级数 ]
[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
\begin\{aligned\} x\_n=\frac\{(-1)^n\}\{\sqrt\{n\}\}, y\_n=\frac\{(-1)^n\}\{\sqrt\{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)/
---
468、 10、 已知 $\displaystyle \left\\{a\_n\right\\}$ 满足 $\displaystyle a\_1\geq a\_2\geq \cdots \geq a\_n\geq 0$. 证明: $\displaystyle \sum\_\{k=1\}^\infty a\_k$ 收敛当且仅当 $\displaystyle \sum\_\{i=0\}^\infty 2^i a\_\{2^i\}$ 收敛. (华南理工大学2023年数学分析考研试题) [数项级数 ]
[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\} \sum\_\{n=1\}^\infty a\_n&=a\_1+(a\_2+a\_3)+(a\_4+\cdots+a\_7) +(a\_8+\cdots+a\_\{15\})+\cdots\\\\ &\leq a\_1+2a\_2+2^2a\_4+a^3a\_8+\cdots =\sum\_\{n=0\}^\infty 2^na\_\{2^n\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \sum\_\{n=0\}^\infty 2^na\_\{2^n\}$ 收敛 $\displaystyle \Rightarrow \sum\_\{n=1\}^\infty a\_n$ 收敛. 又由
\begin\{aligned\} \sum\_\{n=1\}^\infty a\_n&=a\_1+a\_2+(a\_3+a\_4)+(a\_5+\cdots+a\_8)+(a\_9+\cdots+a\_\{16\})+\cdots\\\\ &\geq a\_1+a\_2+2a\_\{2^2\}+2^2a\_\{2^3\} +2^3a\_\{2^4\}+\cdots\\\\ &=\frac\{a\_1\}\{2\}+\frac\{1\}\{2\}\left(a\_1+2a\_2+2^2a\_\{2^2\}+2^3a\_\{2^3\} +2^4a\_\{2^4\}+\cdots\right)\\\\ &=\frac\{a\_1\}\{2\}+\frac\{1\}\{2\}\sum\_\{n=0\}^\infty 2^na\_\{2^n\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \sum\_\{n=1\}^\infty a\_n$ 收敛 $\displaystyle \Rightarrow \sum\_\{n=0\}^\infty 2^na\_\{2^n\}$ 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)/
---
469、 4、 (10 分) 判断级数 $\displaystyle \sum\_\{n=1\}^\infty \frac\{(-1)^n\}\{n^\{p+\frac\{1\}\{n\}\}\}$ 的敛散性, 其中 $\displaystyle p > 0$. 若收敛, 是条件收敛还是绝对收敛? (华南师范大学2023年数学分析考研试题) [数项级数 ]
[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)、 当 $\displaystyle 0 < p\leq 1$ 时, 设 $\displaystyle g(x)=\frac\{1\}\{x^\{p+\frac\{1\}\{x\}\}\}$, 则
\begin\{aligned\} &\ln g(x)=-\left(p+\frac\{1\}\{x\}\right) \ln x\\\\ \Rightarrow&\frac\{g'(x)\}\{g(x)\}=\frac\{1\}\{x^2\}\ln x-\left(p+\frac\{1\}\{x\}\right)\frac\{1\}\{x\} -\frac\{1\}\{x\}\left\[p+\frac\{1\}\{x\}-\frac\{\ln x\}\{x\}\right\] < 0, x\gg 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故当 $\displaystyle x$ 充分大时, $\displaystyle g\searrow 0$. 据 Leibniz 判别法知原级数收敛. 又由
\begin\{aligned\} \frac\{1\}\{n^\{p+\frac\{1\}\{n\}\}\}\sim \frac\{1\}\{n^p\}, \sum\_\{n=1\}^\infty \frac\{1\}\{n^p\}=+\infty \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知原级数条件收敛. (2)、 当 $\displaystyle p > 1$ 时, 由
\begin\{aligned\} \frac\{1\}\{n^\{p+\frac\{1\}\{n\}\}\} < \frac\{1\}\{n^p\}, \sum\_\{n=1\}^\infty \frac\{1\}\{n^p\} < \infty \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知原级数绝对收敛.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
470、 3、 (12 分) 证明数项级数 $\displaystyle \sum\_\{n=1\}^\infty \cos\left(\pi \sqrt\{n^2+n\}\right) \arcsin\frac\{n\}\{n+1\}$ 是条件收敛的. (吉林大学2023年数学分析考研试题) [数项级数 ]
[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\} &\cos\left(\pi \sqrt\{n^2+n\}\right)=(-1)^n \cos\pi \left(\sqrt\{n^2+n\}-n\right)\\\\ =&(-1)^n \cos\frac\{\pi n\}\{\sqrt\{n^2+n\}+n\} =(-1)^n \cos\frac\{\pi\}\{\sqrt\{1+\frac\{1\}\{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
\begin\{aligned\} \mbox\{原式\}=\sum\_\{n=1\}^\infty (-1)^n \cos\frac\{\pi\}\{\sqrt\{1+\frac\{1\}\{n\}\}+1\}\arcsin \frac\{1\}\{1+\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\}
由 Leibniz 判别法知 $\displaystyle \sum\_\{n=1\}^\infty (-1)^n \cos\frac\{\pi\}\{\sqrt\{1+\frac\{1\}\{n\}\}+1\}$ 收敛. 又由 Abel 判别法知原式收敛. 最后, 由
\begin\{aligned\} &\cos\frac\{\pi\}\{\sqrt\{1+\frac\{1\}\{n\}\}+1\}\arcsin \frac\{1\}\{1+\frac\{1\}\{n\}\} \sim\sin\left(\frac\{\pi\}\{2\}-\frac\{\pi\}\{\sqrt\{1+\frac\{1\}\{n\}\}+1\}\right)\cdot\frac\{\pi\}\{2\}\\\\ \sim&\left(\frac\{\pi\}\{2\}-\frac\{\pi\}\{\sqrt\{1+\frac\{1\}\{n\}\}+1\}\right)\cdot\frac\{\pi\}\{2\} =\left(\frac\{\pi\}\{2\}\right)^2 \left(1-\frac\{2\}\{\sqrt\{1+\frac\{1\}\{n\}\}+1\}\right)\\\\ =&\left(\frac\{\pi\}\{2\}\right)^2 \frac\{\sqrt\{1+\frac\{1\}\{n\}\}-1\}\{\sqrt\{1+\frac\{1\}\{n\}\}+1\} \sim\frac\{\pi^2\}\{8\}\left(\sqrt\{1+\frac\{1\}\{n\}\}-1\right) =\frac\{\pi^2\frac\{1\}\{n\}\}\{8\left(\sqrt\{1+\frac\{1\}\{n\}\}+1\right)\}\sim\frac\{\pi^2\}\{16n\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知原式条件收敛.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)/
---
471、 (0-16)、 判断级数 $\displaystyle \sum\_\{n=1\}^\infty \frac\{n!\}\{20^n\}$ 的敛散性. (吉林师范大学2023年(学科数学)数学分析考研试题) [数项级数 ]
[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 $\displaystyle a\_n=\frac\{n!\}\{20^n\}$, 则
\begin\{aligned\} \lim\_\{n\to\infty\}\frac\{a\_\{n+1\}\}\{a\_n\}=\lim\_\{n\to\infty\}\frac\{(n+1)!\}\{20^\{n+1\}\}\cdot \frac\{20^n\}\{n!\} =\lim\_\{n\to\infty\}\frac\{n+1\}\{20\}=+\infty. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由比较判别法知原级数发散.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)/
---
472、 7、 设
\begin\{aligned\} a\_1=1, a\_2=2, a\_\{n+2\}=3a\_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\}
记 $\displaystyle x\_n=\frac\{1\}\{a\_n\}$, 证明: $\displaystyle \left\\{x\_n\right\\}$ 收敛, 并求 $\displaystyle \lim\_\{n\to\infty\}x\_n$, 判断级数 $\displaystyle \sum\_\{n=1\}^\infty x\_n$ 的敛散性. (南京航空航天大学2023年数学分析考研试题) [数项级数 ]
[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) / 递推式的特征方程为 $\displaystyle \lambda^2-3\lambda+1=0\Rightarrow\lambda =\lambda\_\pm =\frac\{3\pm \sqrt\{5\}\}\{2\}$. 故可设
\begin\{aligned\} a\_n=c\_1\lambda\_+^n+c\_2\lambda\_-^n \stackrel\{a\_1=1, a\_2=2\}\{\Rightarrow\}& c\_1=\frac\{5-\sqrt\{5\}\}\{10\}, c\_2=\frac\{5+\sqrt\{5\}\}\{10\}\\\\ \Rightarrow& a\_n=\frac\{5-\sqrt\{5\}\}\{10\}\lambda\_+^n+\frac\{5+\sqrt\{5\}\}\{10\}\lambda\_-^n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle \lambda\_+ > 1 > \lambda\_- > 0$ 知 $\displaystyle \lim\_\{n\to\infty\}a\_n=+\infty$, $\displaystyle \lim\_\{n\to\infty\}x\_n=0$, 且
\begin\{aligned\} a\_n\sim \frac\{5-\sqrt\{5\}\}\{10\}\lambda\_+^n\Rightarrow x\_n\sim \frac\{10\}\{5-\sqrt\{5\}\}\frac\{1\}\{\lambda\_+^n\} \Rightarrow \sum\_\{n=1\}^\infty x\_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)/
---
473、 6、 (15 分) 设 $\displaystyle \left\\{a\_n\right\\}$ 是单调递减正整数列, $\displaystyle \varlimsup\_\{n\to\infty\}na\_n > 0$. 证明级数 $\displaystyle \sum\_\{n=1\}^\infty a\_n$ 发散. (南开大学2023年数学分析考研试题) [数项级数 ]
[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\} &0 < a=\varlimsup\_\{n\to\infty\}a\_n=\inf\_\{n\geq 1\}\sup\_\{k\geq n\}ka\_k\\\\ \Rightarrow&\forall\ n\geq 1, \exists\ k\geq n,\mathrm\{ s.t.\} ka\_k > \frac\{a\}\{2\}\\\\ \Rightarrow&\exists\ n\_k\mbox\{严\}\nearrow ,\mathrm\{ s.t.\} n\_ka\_\{n\_k\} > \frac\{a\}\{2\} \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\} &\sum\_\{k=m+1\}^\{m+p\}\sum\_\{i=n\_\{k-1\}+1\}^\{n\_k\}a\_i \geq \sum\_\{k=m+1\}^\{m+p\} a\_\{n\_k\}(n\_k-n\_\{k-1\})\\\\ \geq& \sum\_\{k=m+1\}^\{m+p\} \frac\{a\}\{n\_k\}(n\_k-n\_\{k-1\}) \geq\frac\{a\}\{2n\_\{m+p\}\} \sum\_\{k=m+1\}^\{m+p\}(n\_k-n\_\{k-1\})\\\\ =&\frac\{a\}\{2n\_\{m+p\}\}(n\_\{m+p\}-n\_m) =\frac\{a\}\{2\}\left(1-\frac\{n\_m\}\{n\_\{m+p\}\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle p\to\infty$ 得 $\displaystyle \sum\_\{i=n\_m+1\}^\infty a\_i\geq \frac\{a\}\{2\}$. 故 $\displaystyle \sum\_\{n=1\}^\infty a\_n$ 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)/
---
474、 (2)、 (20 分) 设 $\displaystyle a\_1=2, a\_\{n+1\}=\frac\{1\}\{2\}\left(a\_n+\frac\{1\}\{a\_n\}\right), n=1,2,\cdots$. 证明: (2-1)、 $\displaystyle \lim\_\{n\to\infty\}a\_n$ 存在; (2-2)、 级数 $\displaystyle \sum\_\{n=1\}^\infty \left(\frac\{a\_n\}\{a\_\{n+1\}\}-1\right)$ 收敛. (山东大学2023年数学分析考研试题) [数项级数 ]
[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
\begin\{aligned\} a\_1=2,\quad n\geq 2\Rightarrow a\_n=\frac\{1\}\{2\}\left(a\_\{n-1\}+\frac\{1\}\{a\_\{n-1\}\}\right)\geq 1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle a\_n\geq 1$. 又由
\begin\{aligned\} a\_\{n+1\}-a\_n=\frac\{1\}\{2\}\left(\frac\{1\}\{a\_n\}-a\_n\right) =\frac\{1-a\_n^2\}\{2a\_n\}\leq 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle a\_n\searrow$. 据单调有界定理知 $\displaystyle \lim\_\{n\to\infty\}a\_n=l$ 存在, 且
\begin\{aligned\} l=\frac\{1\}\{2\}\left(l+\frac\{1\}\{l\}\right)\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\}
(2-2)、
\begin\{aligned\} 0\leq&\sum\_\{n=1\}^\infty \left(\frac\{a\_n\}\{a\_\{n+1\}\}-1\right) =\sum\_\{n=1\}^\infty \frac\{a\_n-a\_\{n+1\}\}\{a\_\{n+1\}\}\\\\ \leq& \sum\_\{n=1\}^\infty (a\_n-a\_\{n+1\}) =a\_1-\lim\_\{n\to\infty\}a\_n=a\_1-1=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](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
475、 5、 (15 分) 已知 $\displaystyle p\in\mathbb\{R\}$, 讨论级数 $\displaystyle \sum\_\{n=1\}^\infty \frac\{\ln n\}\{n^p\}$ 的敛散性. (陕西师范大学2023年数学分析考研试题) [数项级数 ]
[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)、 当 $\displaystyle p > 1$ 时, 由
\begin\{aligned\} &\lim\_\{n\to\infty\}\frac\{\frac\{\ln n\}\{n^p\}\}\{\frac\{1\}\{n^\frac\{1+p\}\{2\}\}\} =\lim\_\{n\to\infty\}\frac\{\ln n\}\{n^\frac\{p-1\}\{2\}\} \xlongequal[\tiny\mbox\{原理\}]\{\tiny\mbox\{归结\}\} \lim\_\{x\to+\infty\}\frac\{2\}\{p-1\}\frac\{\ln x^\frac\{p-1\}\{2\}\}\{x^\frac\{p-1\}\{2\}\}\\\\ =&\frac\{2\}\{p-1\}\lim\_\{t\to+\infty\}\frac\{\ln t\}\{t\}\xlongequal\{\tiny\mbox\{L'Hospital\}\} \cdots=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及比较判别法知原级数收敛. (2)、 当 $\displaystyle p\leq 1$ 时, 由
\begin\{aligned\} \lim\_\{n\to\infty\}\frac\{\frac\{\ln n\}\{n^p\}\}\{\frac\{1\}\{n\}\} =\lim\_\{n\to\infty\}n^\{1-p\}\ln n=+\infty \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及比较判别法知原级数发散.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)/
---
476、 (2)、 已知正项级数 $\displaystyle \sum\_\{n=1\}^\infty x\_n$ 收敛, 级数 $\displaystyle \sum\_\{n=1\}^\infty \sqrt\{x\_nx\_\{n+1\}\}$ 是否收敛? 反之, 若级数 $\displaystyle \sum\_\{n=1\}^\infty \sqrt\{x\_nx\_\{n+1\}\}$ 收敛, $\displaystyle \sum\_\{n=1\}^\infty x\_n$ 是否收敛? (上海财经大学2023年数学分析考研试题) [数项级数 ]
[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
\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\}
知 $\displaystyle \sum\_\{n=1\}^\infty \sqrt\{x\_nx\_\{n+1\}\}$ 收敛. (2-2)、 $\displaystyle \sum\_\{n=1\}^\infty x\_n$ 未必收敛. 比如 $\displaystyle a\_n=\left\\{\begin\{array\}\{llllllllllll\}\frac\{1\}\{k\},&n=2k,\\\\ \frac\{1\}\{k^3\},&n=2k-1.\end\{array\}\right.$ 则
\begin\{aligned\} \sum\_\{n=1\}^\infty a\_n > \sum\_\{k=1\}^\infty \frac\{1\}\{k\}=+\infty, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
但
\begin\{aligned\} &\sqrt\{a\_na\_\{n+1\}\}=\left\\{\begin\{array\}\{llllllllllll\}\sqrt\{\frac\{1\}\{k\} \frac\{1\}\{(k+1)^3\}\},&n=2k,\\\\ \sqrt\{\frac\{1\}\{k^3\}\cdot \frac\{1\}\{k\}\},&n=2k-1\end\{array\}\right.\\\\ \Rightarrow&\sum\_\{n=1\}^\infty \sqrt\{a\_na\_\{n+1\}\} < \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)/
---
477、 (3)、 级数 $\displaystyle \sum\_\{n=1\}^\infty a\_n$ 收敛, 则任意加括号, 只要不改变其先后次序, 所得的新级数仍然收敛. (上海交通大学2023年数学分析考研试题) [数项级数 ]
[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) / $\displaystyle \times$. $\displaystyle \sum\_\{n=1\}^\infty a\_n$ 条件收敛时, 结论不成立. 这就是著名的 Riemann 重排定理.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)/
---
478、 6、 (20 分) 设 $\displaystyle a\_n > 0$, 且 $\displaystyle \sum\_\{n=1\}^\infty a\_n$ 收敛, 令 $\displaystyle r\_n=\sum\_\{m=n\}^\infty a\_m$. (1)、 证明 $\displaystyle \sum\_\{n=1\}^\infty \frac\{a\_n\}\{r\_n\}$ 发散; (2)、 证明 $\displaystyle \sum\_\{n=1\}^\infty \frac\{a\_n\}\{\sqrt\{r\_n\}\}$ 收敛. (上海交通大学2023年数学分析考研试题) [数项级数 ]
[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)、 对 $\displaystyle \forall\ n\geq 1$,
\begin\{aligned\} &\sum\_\{k=n\}^\{n+p-1\}\frac\{a\_k\}\{r\_k\} =\sum\_\{k=n\}^\{n+p-1\}\frac\{r\_k-r\_\{k+1\}\}\{r\_k\}\\\\ \geq& \frac\{1\}\{r\_n\}\sum\_\{k=n\}^\{n+p-1\}(r\_k-r\_\{k+1\}) =\frac\{r\_n-r\_\{n+p\}\}\{r\_n\}=1-\frac\{r\_\{n+p\}\}\{r\_n\}\xrightarrow\{p\to\infty\}1. \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 $\displaystyle \sum\_\{n=1\}^\infty \frac\{a\_n\}\{r\_n\}$ 发散. (2)、 $\displaystyle \sum\_\{n=1\}^\infty \frac\{a\_n\}\{\sqrt\{r\_n\}\}$ 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)/
---
479、 6、 (本题 10 分) 讨论级数 $\displaystyle \sum\_\{n=1\}^\infty \frac\{1\}\{n\}\left\[\mathrm\{e\}-\left(1+\frac\{1\}\{n\}\right)^n\right\]^p$ 的敛散性. (四川大学2023年数学分析考研试题) [数项级数 ]
[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\}\left\[\mathrm\{e\}-\mathrm\{e\}^\{n\ln \left(1+\frac\{1\}\{n\}\right)\}\right\]^p =\frac\{1\}\{n\}\mathrm\{e\}^p \left\\{1-\mathrm\{e\}^\{n\left\[\ln \left(1+\frac\{1\}\{n\}\right)-\frac\{1\}\{n\}\right\]\}\right\\}^p\\\\ \sim& \mathrm\{e\}^p\frac\{1\}\{n\}\left\\{n\left\[\frac\{1\}\{n\}-\ln \left(1+\frac\{1\}\{n\}\right)\right\]\right\\}^p\left(\mathrm\{e\}^t-1\sim t, t\to 0\right)\\\\ \sim& \mathrm\{e\}^p\frac\{1\}\{n\}\frac\{1\}\{(2n)^p\}\left(t-\ln (1+t)\sim \frac\{t^2\}\{2\}, t\to 0\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知当 $\displaystyle p\leq 0$ 时, 原级数发散; 当 $\displaystyle p > 0$ 时, 原级数收敛.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)/
---
480、 11、 (10 分) 设 $\displaystyle \left\\{a\_n\right\\}$ 是正项数列, 且级数 $\displaystyle \sum\_\{n=1\}^\infty a\_n$ 发散. (1)、 若 $\displaystyle \left\\{a\_n\right\\}$ 单调递减, 证明 $\displaystyle \sum\_\{n=1\}^\infty \frac\{a\_n\}\{1+na\_n\}$ 发散; (2)、 举例说明 $\displaystyle \sum\_\{n=1\}^\infty \frac\{a\_n\}\{1+na\_n\}$ 可以收敛. (苏州大学2023年数学分析考研试题) [数项级数 ]
[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)、 设 $\displaystyle S\_n=\sum\_\{k=1\}^n a\_k$, 则由 $\displaystyle \sum\_\{n=1\}^\infty a\_n$ 发散知 $\displaystyle \lim\_\{n\to\infty\}S\_n=+\infty$. 又由 $\displaystyle a\_n\searrow$ 知 $\displaystyle na\_n\leq S\_n\leq na\_1$. 于是
\begin\{aligned\} &\sum\_\{k=n+1\}^\{n+p\}\frac\{a\_k\}\{1+ka\_k\} \geq \sum\_\{k=n+1\}^\{n+p\}\frac\{a\_k\}\{1+S\_k\} \geq \frac\{1\}\{1+S\_\{n+p\}\}\sum\_\{k=n+1\}^\{n+p\}a\_k\\\\ =&\frac\{S\_\{n+p\}-S\_n\}\{1+S\_\{n+p\}\} =\frac\{1\}\{\frac\{1\}\{S\_\{n+p\}\}+1\}-\frac\{S\_n\}\{1+S\_\{n+p\}\} \xrightarrow\{p\to\infty\}1. \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 $\displaystyle \sum\_\{n=1\}^\infty \frac\{a\_n\}\{1+na\_n\}$ 发散. (2)、 取 $\displaystyle a\_n=\left\\{\begin\{array\}\{llllllllllll\}1,&n=k^2\\\\ 0,&n\neq k^2\end\{array\}\right.(k=1,2,\cdots)$, 则 $\displaystyle \sum\_\{n=1\}^\infty a\_n=\sum\_\{k=1\}^\infty 1=+\infty$, 但
\begin\{aligned\} \sum\_\{n=1\}^\infty \frac\{a\_n\}\{1+na\_n\}=\sum\_\{k=1\}^\infty \frac\{1\}\{1+k^2\} < \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)/
---
481、 12、 设正项级数 $\displaystyle \sum\_\{n=1\}^\infty a\_n$ 收敛, 记 $\displaystyle R\_n=\sum\_\{k=n+1\}^\infty a\_k$. 证明: (1)、 级数 $\displaystyle \sum\_\{n=1\}^\infty \frac\{a\_n\}\{\sqrt[3]\{R\_n^2\} +\sqrt[3]\{R\_nR\_\{n-1\}\}+\sqrt[3]\{R\_\{n-1\}^2\}\}$ 收敛; (2)、 级数 $\displaystyle \sum\_\{n=1\}^\infty \frac\{a\_n\}\{R\_n+R\_\{n-1\}\}$ 发散. (太原理工大学2023年数学分析考研试题) [数项级数 ]
[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 $\displaystyle \sum\_\{n=1\}^\infty a\_n$ 收敛知 $\displaystyle \lim\_\{n\to\infty\}R\_n=0$, 而
\begin\{aligned\} \mbox\{原式\}=&\sum\_\{n=1\}^\infty \frac\{R\_\{n-1\}-R\_n\}\{\sqrt[3]\{R\_n^2\} +\sqrt[3]\{R\_nR\_\{n-1\}\}+\sqrt[3]\{R\_\{n-1\}^2\}\} =\sum\_\{n=1\}^\infty \left(\sqrt[3]\{R\_\{n-1\}\}-\sqrt[3]\{R\_\{n\}\}\right)\\\\ =&\lim\_\{n\to\infty\}\sum\_\{k=1\}^n \left(\sqrt[3]\{R\_\{k-1\}\}-\sqrt[3]\{R\_\{k\}\}\right) =\lim\_\{n\to\infty\}\left(\sqrt[3]\{R\_0\}-\sqrt[3]\{R\_n\}\right)\\\\ =&\sqrt[3]\{R\_0\}=\sqrt[3]\{\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\}
(2)、 由
\begin\{aligned\} &\sum\_\{k=n+1\}^\{n+p\}\frac\{a\_k\}\{R\_k+R\_\{k-1\}\} =\sum\_\{k=n+1\}^\{n+p\}\frac\{R\_\{k-1\}-R\_k\}\{R\_\{k-1\}+R\_k\}\\\\ \geq& \frac\{1\}\{R\_n\}\sum\_\{k=n+1\}^\{n+p\}(R\_\{k-1\}-R\_k) =\frac\{1\}\{R\_n\}(R\_n-R\_\{n+p\}) =1-\frac\{R\_\{n+p\}\}\{R\_n\}\xrightarrow\{p\to\infty\}1 \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 $\displaystyle \sum\_\{n=1\}^\infty \frac\{a\_n\}\{R\_n+R\_\{n-1\}\}$ 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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482、 6、 设 $\displaystyle n$ 是一正整数, 证明: (1)、 方程 $\displaystyle f\_n(x)=x^n+nx-1=0$ 在 $\displaystyle (0,+\infty)$ 内有唯一的正实根 $\displaystyle a\_n$; (2)、 $\displaystyle \sum\_\{n=1\}^\infty (-1)^na\_\{2n+1\}$ 条件收敛. (天津大学2023年数学分析考研试题) [数项级数 ]
[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 $\displaystyle f\_n(0)=-1, f\_n(1)=n > 0$ 知 $\displaystyle f\_n$ 在 $\displaystyle (0,1)$ 中有一个零点 $\displaystyle a\_n$. 又由 $\displaystyle f\_n'(x)=nx^\{n-1\}+n > 0$ 知 $\displaystyle f$ 在正实根唯一, 就是 $\displaystyle a\_n$. (2)、 由
\begin\{aligned\} &f\_\{2n+1\}(x)-f\_\{2n-1\}(x)=[x^\{2n+1\}+(2n+1)x]-[x^\{2n-1\}+(2n-1)x]\\\\ =&-x^\{2n-1\}(1-x^2)+2x \geq -x(1-x^2)+2x=x(1+x^2) > 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\} f\_\{2n+1\}(a\_\{2n+1\})=0=f\_\{2n-1\}(a\_\{2n-1\})\leq f\_\{2n+1\}(a\_\{2n-1\}) \stackrel\{f\_\{2n+1\}\mbox\{严\}\nearrow \}\{\Rightarrow\}a\_\{2n+1\}\leq a\_\{2n-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由
\begin\{aligned\} f\_n(a\_n)=0 < \frac\{1\}\{n^n\}=f\_n\left(\frac\{1\}\{n\}\right)\Rightarrow a\_n < \frac\{1\}\{n\}\Rightarrow \lim\_\{n\to\infty\}a\_n=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及 Leibniz 判别法知 $\displaystyle \sum\_\{n=1\}^\infty (-1)^n a\_\{2n-1\}$ 收敛. 最后由
\begin\{aligned\} 1-na\_n=a\_n^n < \frac\{1\}\{n^n\}\xrightarrow\{n\to\infty\}0\Rightarrow \lim\_\{n\to\infty\}na\_n=1\Rightarrow a\_n\sim\frac\{1\}\{n\} \Rightarrow \sum\_\{n=1\}^\infty a\_\{2n-1\}=+\infty \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \sum\_\{n=1\}^\infty (-1)^na\_\{2n+1\}$ 条件收敛.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)/
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483、 (2)、 讨论级数 $\displaystyle \sum\_\{n=1\}^\infty \frac\{a^nn!\}\{n^n\}$ 的敛散性 ($a > 0$). (武汉理工大学2023年数学分析考研试题) [数项级数 ]
[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 $\displaystyle x\_n=\frac\{a^n n!\}\{n^n\}$, 则
\begin\{aligned\} \lim\_\{n\to\infty\}\frac\{x\_\{n+1\}\}\{x\_n\}=\lim\_\{n\to\infty\}\frac\{a\}\{\left(1+\frac\{1\}\{n\}\right)^n\}=\frac\{a\}\{\mathrm\{e\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故当 $\displaystyle a < \mathrm\{e\}$ 时, 原级数收敛; 当 $\displaystyle a > \mathrm\{e\}$ 时, 原级数发散; 当 $\displaystyle a=\mathrm\{e\}$ 时, 由 Stirling 公式
\begin\{aligned\} n!\sim \sqrt\{2\pi n\}\left(\frac\{n\}\{\mathrm\{e\}\}\right)^n\Rightarrow x\_n\sim \sqrt\{2\pi n\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知原级数发散. 综上即知原级数收敛 $\displaystyle \Leftrightarrow 0 < a < \mathrm\{e\}$.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 08:51:49
