## 张祖锦2023年数学专业真题分类70天之第22天
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484、 6、 (20 分) 判断级数 $\displaystyle \sum\_\{n=1\}^\infty \frac\{(-1)^n \cos n\}\{n\} \left(1+\frac\{1\}\{n\}\right)^n$ 的敛散性. 若收敛, 指出是条件收敛还是绝对收敛. (西南交通大学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) / 由等式
\begin\{aligned\} \frac\{1\}\{2\}+\sum\_\{k=1\}^n \cos kx=\frac\{\sin\left(n+\frac\{1\}\{2\}\right)x\}\{2\sin\frac\{x\}\{2\}\}, x\in [-\pi,\pi], x\neq 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \left|\sum\_\{k=1\}^n (-1)^k \cos k\right| =\left|\sum\_\{k=1\}^n \cos k(\pi+1)\right| \leq \frac\{1\}\{\sin \frac\{\pi+1\}\{2\}\}=\frac\{1\}\{\cos \frac\{1\}\{2\}\}$. 再由 $\displaystyle \frac\{1\}\{n\}\searrow 0$ 及 Dirichlet 判别法知 $\displaystyle \sum\_\{n=1\}^\infty \frac\{(-1)^n \cos n\}\{n\}$ 收敛. 又由 $\displaystyle \left(1+\frac\{1\}\{n\}\right)^n\nearrow \mathrm\{e\}$ 知 Abel 判别法知原级数收敛. 最后,
\begin\{aligned\} \sum\_\{n=1\}^\infty \left|\frac\{(-1)^n \cos n\}\{n\} \left(1+\frac\{1\}\{n\}\right)^n\right| \geq \sum\_\{n=1\}^\infty \frac\{\cos^2n\}\{n\}\left(1+\frac\{1\}\{1\}\right)^1 =\sum\_\{n=1\}^\infty \frac\{1+\cos 2n\}\{n\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
作为一个发散级数与收敛级数的和, 是发散的. 综上即知原级数条件收敛.跟锦数学微信公众号. [在线资料](
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485、 6、 (15 分) 求级数 $\displaystyle \sum\_\{n=0\}^\infty \frac\{a\_n\}\{n+1\}$, 其中 $\displaystyle a\_n=\int\_0^\frac\{\pi\}\{4\}\sin^\{n+1\}x\cos x\mathrm\{ d\} x$. (云南大学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 $\displaystyle a\_n\stackrel\{\sin x=t\}\{=\}\int\_0^1 t^n\mathrm\{ d\} t=\frac\{1\}\{n+1\}$ 知
\begin\{aligned\} \mbox\{原式\}=\sum\_\{n=0\}^\infty \frac\{1\}\{(n+1)^2\}=\sum\_\{k=1\}^\infty \frac\{1\}\{k^2\}=\frac\{\pi^2\}\{6\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
最后一步的理由如下. 设 $\displaystyle f(x)\equiv x(2\pi-x)\sim\frac\{a\_0\}\{2\}+\sum\_\{n=1\}^\infty (a\_n\cos nx+b\_n\sin nx)$, 其中
\begin\{aligned\} a\_0&=\frac\{1\}\{\pi\}\int\_0^\{2\pi \}f(x)\mathrm\{ d\} x=\frac\{4\pi^2\}\{3\},\\\\ n\geq 1\Rightarrow a\_n&=\frac\{1\}\{\pi\}\int\_0^\{2\pi\}f(x)\cos nx\mathrm\{ d\} x=-\frac\{4\}\{n^2\},\\\\ b\_n&=\frac\{1\}\{\pi\}\int\_\{-\pi\}^\pi f(x)\sin nx\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\}
由 $\displaystyle f(0)=f(2\pi)=0$ 知
\begin\{aligned\} f(x)=\frac\{2\pi^2\}\{3\}-\sum\_\{n=1\}^\infty \frac\{4\}\{n^2\}\cos nx,\quad x\in [0,2\pi]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle x=0$ 即得 $\displaystyle \sum\_\{n=1\}^\infty \frac\{1\}\{n^2\}=\frac\{\pi^2\}\{6\}$.跟锦数学微信公众号. [在线资料](
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486、 5、 (10 分) 讨论数项级数 $\displaystyle \sum\_\{n=1\}^\infty (-1)^\{n-1\}\left(\sqrt[n]\{n\}-1\right)$ 的条件收敛和绝对收敛性. (郑州大学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)、 由
\begin\{aligned\} \sqrt[n]\{n\}-1&=\mathrm\{e\}^\frac\{\ln n\}\{n\}-\mathrm\{e\}^0\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\}\mathrm\{e\}^\{\xi\_n\}\frac\{\ln n\}\{n\} \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\} \sqrt[n]\{n\}-1\sim\frac\{\ln n\}\{n\}\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\}
故 $\displaystyle \sum\_\{n=1\}^\infty \left(\sqrt[n]\{n\}-1\right)$ 发散. (2)、 又由
\begin\{aligned\} y=x^\frac\{1\}\{x\}\Rightarrow \ln y=\frac\{\ln x\}\{x\}\Rightarrow \frac\{y'\}\{y\}=\frac\{1-\ln x\}\{x^2\} < 0\left(x > \mathrm\{e\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \left\\{\sqrt[n]\{n\}-1\right\\}\_\{n=3\}^\infty$ 递减趋于 $\displaystyle 0$. 据 Leibniz 判别法即知 $\displaystyle \sum\_\{n=1\}^\infty (-1)^\{n-1\}\left(\sqrt[n]\{n\}-1\right)$ 收敛. (3)、 综上即知原级数条件收敛.跟锦数学微信公众号. [在线资料](
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487、 (2)、 级数 $\displaystyle \sum\_\{n=1\}^\infty \frac\{a^n\}\{(1+a)(1+a^2)\cdots(1+a^n)\}$ ($a > 0$) 是收敛的. (中国海洋大学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) / $\displaystyle \surd$. (2-1)、 当 $\displaystyle 0 < a < 1$ 时, $\displaystyle 0 < \frac\{a^n\}\{(1+a)(1+a^2)\cdots(1+a^n)\} < a^n$, 而原级数收敛. (2-2)、 当 $\displaystyle a=1$ 时, $\displaystyle \frac\{a^n\}\{(1+a)(1+a^2)\cdots(1+a^n)\}=\frac\{1\}\{2^n\}$, 而原级数收敛. (2-3)、 当 $\displaystyle a > 1$ 时,
\begin\{aligned\} 0 < \frac\{a^n\}\{(1+a)(1+a^2)\cdots(1+a^n)\} < \frac\{a^n\}\{(1+a)^n\}=\left(\frac\{a\}\{1+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\}
而原级数收敛.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](
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488、 13、 (15 分) 已知 $\displaystyle a\_1=2, a\_\{n+1\}=\frac\{1\}\{2\}\left(a\_n+\frac\{1\}\{a\_n\}\right), n=1,2,\cdots$. 证明: (1)、 $\displaystyle \lim\_\{n\to\infty\}a\_n$ 存在; (2)、 级数 $\displaystyle \sum\_\{n=1\}^\infty \left(\frac\{a\_n\}\{a\_\{n+1\}\}-1\right)$ 收敛. (中国矿业大学(北京)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
\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)、
\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](
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489、 7、 证明: 若正项级数 $\displaystyle \sum\_\{n=1\}^\infty u\_n$ 收敛, 且数列 $\displaystyle \left\\{u\_n\right\\}$ 单调, 则 $\displaystyle \lim\_\{n\to\infty\}nu\_n=0$. (中国矿业大学(徐州)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 $\displaystyle \sum\_\{n=1\}^\infty u\_n$ 收敛知
\begin\{aligned\} \forall\ \varepsilon > 0, \exists\ N,\mathrm\{ s.t.\} \forall\ n\geq N, \sum\_\{k=\left\[\frac\{n\}\{2\}\right\]\}^n u\_k < \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\} \frac\{\varepsilon\}\{2\} > \sum\_\{k=\left\[\frac\{n\}\{2\}\right\]\}^n u\_k > u\_n\sum\_\{k=\left\[\frac\{n\}\{2\}\right\]\}^n 1=u\_n\left(n-\left\[\frac\{n\}\{2\}\right\]\right) > u\_n\cdot\frac\{n\}\{2\}\Rightarrow 0\leq nu\_n < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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490、 2、 (15 分) 判别收敛性. (1)、 (7 分) $\displaystyle \sum\_\{n=1\}^\infty 2022^n\sin\frac\{\pi\}\{2023^n\}$. (中山大学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 $\displaystyle 2022^n\sin\frac\{\pi\}\{2023^n\}\sim \pi\left(\frac\{2022\}\{2023\}\right)^n$ 及比较判别法知原级数收敛.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](
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491、 (2)、 (8 分) $\displaystyle \sum\_\{n=1\}^\infty (-1)^n \frac\{1\}\{2^n\}\left(1+\frac\{1\}\{n\}\right)^\{n^2\}$. (中山大学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 $\displaystyle \lim\_\{n\to\infty\}\left(1+\frac\{1\}\{n\}\right)^n=\mathrm\{e\} > \ell=\frac\{\mathrm\{e\}+2\}\{2\}$ 知当 $\displaystyle n$ 充分大时,
\begin\{aligned\} \frac\{1\}\{2^n\}\left(1+\frac\{1\}\{n\}\right)^\{n^2\} > \frac\{\ell^n\}\{2^n\}=\left(\frac\{\ell\}\{2\}\right)^n\to+\infty. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故原级数的通项的极限不为 $\displaystyle 0$, 而发散.跟锦数学微信公众号. [在线资料](
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492、 5、 (10 分) 函数 $\displaystyle f(x)$ 在 $\displaystyle x=0$ 的某一邻域内具有二阶连续导数, 且 $\displaystyle \lim\_\{x\to 0\}\frac\{f(x)\}\{x\}=0$. 证明级数 $\displaystyle \sum\_\{n=1\}^\infty f\left(\frac\{1\}\{n\}\right)$ 绝对收敛. (重庆大学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) /
\begin\{aligned\} f(0)&=\lim\_\{x\to 0\}f(x)=\lim\_\{x\to 0\}\frac\{f(x)\}\{x\}\cdot x=0, f'(0)=\lim\_\{x\to 0\}\frac\{f(x)-f(0)\}\{x-0\}=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle f$ 在 $\displaystyle 0$ 的领域 $\displaystyle U(0;\delta)$ 内具有二阶连续导函数, 则
\begin\{aligned\} \sum\_\{n=1\}^\infty \left|f\left(\frac\{1\}\{n\}\right)\right| \xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\}& \sum\_\{n=1\}^\{\left\[\frac\{2\}\{\delta\}\right\]\} \left|f\left(\frac\{1\}\{n\}\right)\right| +\sum\_\{n=\{\left\[\frac\{2\}\{\delta\}\right\]\}+1\}^\infty\left|\frac\{f''(\xi\_n)\}\{2!\}\frac\{1\}\{n^2\}\right|\\\\ \leq&\sum\_\{n=1\}^\{\left\[\frac\{2\}\{\delta\}\right\]\} \left|f\left(\frac\{1\}\{n\}\right)\right| +\frac\{1\}\{2\}\max\_\{\left\[-\frac\{\delta\}\{2\},\frac\{\delta\}\{2\}\right\]\}|f''|\sum\_\{n=\{\left\[\frac\{2\}\{\delta\}\right\]\}+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\}
跟锦数学微信公众号. [在线资料](
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493、 13、 讨论函数列 $\displaystyle f\_n(x)=\frac\{nx\}\{1+n^2x^2\}$ 在 $\displaystyle (-\infty,+\infty)$ 上的一致收敛性. (北京邮电大学2023年数学分析考研试题) [函数项级数 ]
[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) / 当 $\displaystyle x=0$ 时, $\displaystyle f\_n(x)=0$. 当 $\displaystyle x\neq 0$ 时, $\displaystyle \lim\_\{n\to\infty\}f\_n(x)\xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \lim\_\{n\to\infty\}\frac\{nx\}\{n^2x^2\} =0$. 故 $\displaystyle \lim\_\{n\to\infty\}f\_n(x)=0$. 由
\begin\{aligned\} \sup\_\{x\in\mathbb\{R\}\}|f\_n(x)-0|\geq f\_n\left(\frac\{1\}\{n\}\right)=\frac\{1\}\{2\}\not\to 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle f\_n$ 在 $\displaystyle \mathbb\{R\}$ 上不一致收敛.跟锦数学微信公众号. [在线资料](
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494、 1、 简答题. (1)、 设 $\displaystyle f\_n(x)=xn^\{-x\}\ (n=1,2,3,\cdots)$. 问 $\displaystyle \left\\{f\_n(x)\right\\}$ 在 $\displaystyle [0,+\infty)$ 上是否一致收敛, 为什么? (大连理工大学2023年数学分析考研试题) [函数项级数 ]
[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 $\displaystyle f\_n'(x)=n^x(1-x\ln n)$ 知
\begin\{aligned\} \max\_\{x\geq 0\}|f\_n(x)-0|=f\_n\left(\frac\{1\}\{\ln n\}\right)=\frac\{1\}\{\mathrm\{e\}\ln 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\}
故 $\displaystyle f\_n\rightrightarrows 0$ 于 $\displaystyle [0,+\infty)$.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)/
---
495、 6、 讨论级数 $\displaystyle \sum\_\{n=0\}^\infty \frac\{nx\}\{1+n^5x^2\}$ 在 $\displaystyle (-\infty,+\infty)$ 上的一致收敛性. (东北大学2023年数学分析考研试题) [函数项级数 ]
[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 $\displaystyle f\_n(x)=\frac\{nx\}\{1+n^5x\}$, 则
\begin\{aligned\} f\_n(x)=\frac\{n(1-n^5x^2)\}\{(1+n^5x^2)^2\} \Rightarrow \max\_\mathbb\{R\} |f|=\left|f\left(\pm \frac\{1\}\{n^\frac\{5\}\{2\}\}\right)\right| =\frac\{1\}\{2n^\frac\{3\}\{2\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle \sum\_\{n=1\}^\infty \frac\{1\}\{2n^\frac\{3\}\{2\}\} < \infty$ 及优级数判别法知 $\displaystyle \sum\_\{n=0\}^\infty f\_n(x)$ 在 $\displaystyle \mathbb\{R\}$ 上一致收敛.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)/
---
496、 7、 证明: 当 $\displaystyle x > 0$ 时,
\begin\{aligned\} \ln \sqrt\{x\}=\sum\_\{n=1\}^\infty \frac\{1\}\{2n-1\}\left(\frac\{x-1\}\{1+x\}\right)^\{2n-1\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
并讨论 $\displaystyle \sum\_\{n=1\}^\infty \frac\{1\}\{2n-1\}\left(\frac\{x-1\}\{1+x\}\right)^\{2n-1\}$ 关于 $\displaystyle x\in (0,+\infty)$ 是否一致收敛. (东北师范大学2023年数学分析考研试题) [函数项级数 ]
[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\}\{1-s\}=\sum\_\{n=0\}^\infty s^n, |s| < 1\\\\ \Rightarrow&-\ln(1-s)=\sum\_\{n=0\}^\infty \frac\{s^\{n+1\}\}\{n+1\}=\sum\_\{n=1\}^\infty \frac\{s^n\}\{n\}, |s| < 1\\\\ \Rightarrow&\ln(1+s)=\sum\_\{n=1\}^\infty \frac\{(-1)^\{n-1\}s^n\}\{n\} \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\} &\ln \sqrt\{\frac\{1+s\}\{1-s\}\}=\frac\{1\}\{2\}[\ln(1+s)-\ln(1-s)]\\\\ =&\frac\{1\}\{2\}\left\[\sum\_\{n=1\}^\infty \frac\{(-1)^\{n-1\}s^n\}\{n\}+\sum\_\{n=1\}^\infty \frac\{s^n\}\{n\}\right\] =\sum\_\{k=1\}^\infty \frac\{s^\{2k-1\}\}\{2k-1\}, |s| < 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle \frac\{x-1\}\{1+x\}=s$, 则 $\displaystyle x > 0\Leftrightarrow |s| < 1$, 而上式即蕴含
\begin\{aligned\} \ln \sqrt\{x\}=\sum\_\{n=1\}^\infty \frac\{1\}\{2n-1\}\left(\frac\{x-1\}\{1+x\}\right)^\{2n-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\}
往用反证法证明 $\displaystyle \sum\_\{n=1\}^\infty \frac\{1\}\{2n-1\}\left(\frac\{x-1\}\{1+x\}\right)^\{2n-1\}$ 关于 $\displaystyle x\in (0,+\infty)$ 不一致收敛. 若不然, $\displaystyle \forall\ \varepsilon > 0,\exists\ N,\mathrm\{ s.t.\} \forall\ m > n\geq N$,
\begin\{aligned\} &\forall\ x > 0, \left|\sum\_\{k=n\}^m \frac\{1\}\{2k-1\}\left(\frac\{x-1\}\{1+x\}\right)^\{2k-1\}\right| < \varepsilon\\\\ \stackrel\{x\to+\infty\}\{\Rightarrow\}& \sum\_\{k=n\}^m \frac\{1\}\{2k-1\}\leq\varepsilon \stackrel\{m\to\infty\}\{\Rightarrow\} \sum\_\{k=n\}^\infty \frac\{1\}\{2k-1\}\leq \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这与 $\displaystyle \sum\_\{n=1\}^\infty \frac\{1\}\{2k-1\}$ 发散矛盾. 故有结论.跟锦数学微信公众号. [在线资料](
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---
497、 11、 设 $\displaystyle \alpha > 0$, 讨论 $\displaystyle \sum\_\{n=1\}^\infty x^\alpha \mathrm\{e\}^\{-nx\}$ 在 $\displaystyle (0,+\infty)$ 上的一致收敛性. (东南大学2023年数学分析考研试题) [函数项级数 ]
[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 $\displaystyle f\_n(x)=x^\alpha \mathrm\{e\}^\{-nx\}$, 则由 $\displaystyle f\_n'(x)=\mathrm\{e\}^\{-nx\}x^\{\alpha-1\}(\alpha-nx)$ 知
\begin\{aligned\} \sup\_\{x\geq 0\}f\_n(x)=f\_n\left(\frac\{\alpha\}\{n\}\right)=\mathrm\{e\}^\{-\alpha\} \left(\frac\{\alpha\}\{n\}\right)^\alpha. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 当 $\displaystyle \alpha > 1$ 时, 由 $\displaystyle \sum\_\{n=1\}^\infty \mathrm\{e\}^\{-\alpha\}\left(\frac\{\alpha\}\{n\}\right)^\alpha$ 收敛及优级数判别法知 $\displaystyle \sum\_\{n=1\}^\infty x^\alpha \mathrm\{e\}^\{-nx\}$ 在 $\displaystyle (0,+\infty)$ 上一致收敛. (2)、 当 $\displaystyle 0 < \alpha\leq 1$ 时, 由
\begin\{aligned\} &\sup\_\{x > 0\}\sum\_\{k=n\}^\infty x^\alpha \mathrm\{e\}^\{-kx\} =\sup\_\{x > 0\} \frac\{x^\alpha \mathrm\{e\}^\{-nx\}\}\{1-\mathrm\{e\}^\{-x\}\}\\\\ \geq& \lim\_\{x\to 0^+\}\frac\{x^\alpha \mathrm\{e\}^\{-nx\}\}\{1-\mathrm\{e\}^\{-x\}\} \xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} \lim\_\{x\to 0^+\}\frac\{x^\alpha\}\{x\}=\left\\{\begin\{array\}\{llllllllllll\}1,&\alpha=1\\\\ +\infty,&0 < \alpha < 1\end\{array\}\right. \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 $\displaystyle \sum\_\{n=1\}^\infty x^\alpha \mathrm\{e\}^\{-nx\}$ 在 $\displaystyle (0,+\infty)$ 上不一致收敛.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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---
498、 4、 (20 分) 证明: $\displaystyle \zeta(x)=\sum\_\{n=1\}^\infty \frac\{1\}\{n^x\}$ 在 $\displaystyle (1,+\infty)$ 上连续, 并有连续的各阶导函数. (福州大学2023年数学分析考研试题) [函数项级数 ]
[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) / 对 $\displaystyle \forall\ a > 1$,
\begin\{aligned\} &\frac\{1\}\{n^x\}=n^\{-x\}, \left(\frac\{1\}\{n^x\}\right)' =-n^\{-x\}\ln n, \left(\frac\{1\}\{n^x\}\right)''=n^\{-x\}\ln^2n, \cdots, \\\\ &\left(\frac\{1\}\{n^x\}\right)^\{(k)\} =(-1)^k n^\{-x\}\ln^kn. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由
\begin\{aligned\} x\geq a\Rightarrow \left|\left(\frac\{1\}\{n^x\}\right)^\{(k)\}\right|\leq\frac\{\ln^kn\}\{n^x\}\leq \frac\{\ln^kn\}\{n^a\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及优级数判别法知 $\displaystyle \sum\_\{n=1\}^\infty \left(\frac\{1\}\{n^x\}\right)^\{(k)\}$ 在 $\displaystyle [a,+\infty)$ 上一致收敛. 故
\begin\{aligned\} \zeta^\{(k)\}(x)=\sum\_\{n=1\}^\infty \left(\frac\{1\}\{n^x\}\right)^\{(k)\} =\sum\_\{n=1\}^\infty \frac\{(-1)^k \ln^kn\}\{n^x\},\quad x\geq a, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
且连续. 由 $\displaystyle a > 1$ 的任意性即知
\begin\{aligned\} \zeta^\{(k)\}(x)=\sum\_\{n=1\}^\infty \frac\{(-1)^k \ln^kn\}\{n^x\},\quad x > 1, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
且连续.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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---
499、 8、 (15 分) 讨论 $\displaystyle \sum\_\{n=1\}^\infty \left(1+\frac\{1\}\{2\}+\cdots+\frac\{1\}\{n\}\right)\frac\{\sin nx\}\{n\}$ 的敛散性. (福州大学2023年数学分析考研试题) [函数项级数 ]
[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) / $\displaystyle \left(1+\frac\{1\}\{2\}+\cdots+\frac\{1\}\{n\}\right)\frac\{\sin nx\}\{n\}$ 关于 $\displaystyle x$ 是奇函数, 且是 $\displaystyle 2\pi$ 周期函数. 故仅需讨论 $\displaystyle x\in [0,\pi)$ 的情形. 当 $\displaystyle x=0$ 时, 通项为 $\displaystyle 0$, 而收敛. 当 $\displaystyle 0 < x < \pi$ 时, 由
\begin\{aligned\} &\left|2\sin\frac\{x\}\{2\}\sum\_\{k=1\}^n \sin kx\right| =\left|\sum\_\{k=1\}^n \left\[\cos \left(k-\frac\{1\}\{2\}\right)x-\cos \left(k+\frac\{1\}\{2\}\right)x\right\]\right|\\\\ =&\left|\cos \frac\{x\}\{2\}-\cos\left(n+\frac\{1\}\{2\}\right)x\right|\leq 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\} \left|\sum\_\{k=1\}^n \sin kx\right|\leq \frac\{1\}\{\sin\frac\{x\}\{2\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle a\_n=\frac\{1+\frac\{1\}\{2\}+\cdots+\frac\{1\}\{n\}\}\{n\}$, 则
\begin\{aligned\} &na\_n=1+\frac\{1\}\{2\}+\cdots+\frac\{1\}\{n\} \Rightarrow (n+1)a\_\{n+1\}-na\_n=\frac\{1\}\{n+1\}\\\\ \Rightarrow&n(a\_\{n+1\}-a\_n)=\frac\{1\}\{n+1\}-a\_\{n+1\}=\frac\{1\}\{n+1\}-\frac\{1+\frac\{1\}\{2\}+\cdots+\frac\{1\}\{n+1\}\}\{n+1\} < 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle a\_n\searrow$, 且 $\displaystyle \lim\_\{n\to\infty\}a\_n\xlongequal\{\tiny\mbox\{Stolz\}\} \lim\_\{n\to\infty\}\frac\{\frac\{1\}\{n\}\}\{1\}=0$. 由 Leibniz 判别法知原函数项级数收敛. 综上即知对 $\displaystyle \forall\ x\in\mathbb\{R\}$, 原函数项级数收敛.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](
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---
500、 7、 设函数列 $\displaystyle f\_n(x)=\left(\frac\{\sin x\}\{x\}\right)^n$, 说明 $\displaystyle \left\\{f\_n(x)\right\\}$ 在 $\displaystyle (0,1)$ 上的一致收敛性. (哈尔滨工业大学2023年数学分析考研试题) [函数项级数 ]
[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) / 当 $\displaystyle 0 < x < 1$ 时, $\displaystyle 0 < \sin x < x$, 而 $\displaystyle 0 < \frac\{\sin x\}\{x\} < 1$, $\displaystyle \lim\_\{n\to\infty\}f\_n(x)=0$. 又由
\begin\{aligned\} \sup\_\{0 < x < 1\}|f\_n(x)-0|=\sup\_\{0 < x < 1\}, \left(\frac\{\sin x\}\{x\}\right)^n \geq \lim\_\{x\to 0^+\}\left(\frac\{\sin x\}\{x\}\right)^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 $\displaystyle \left\\{f\_n(x)\right\\}$ 在 $\displaystyle (0,1)$ 上不一致收敛.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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---
501、 8、 (15 分) 设 $\displaystyle f(x)=\sum\_\{n=1\}^\infty \left(x+\frac\{1\}\{n\}\right)^n$. 请确定 $\displaystyle f(x)$ 的定义域, 并讨论 $\displaystyle f(x)$ 在定义域内的连续性. (合肥工业大学2023年数学分析考研试题) [函数项级数 ]
[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)、 为保证指数函数的底 $\displaystyle x+\frac\{1\}\{n\} > 0$, 必须 $\displaystyle x\geq 0$. 由 $\displaystyle \lim\_\{n\to\infty\} \sqrt[n]\{\left(x+\frac\{1\}\{n\}\right)^n\}=x$ 及正项级数的 Cauchy 根值判别法知当 $\displaystyle 0\leq x < 1$ 时, $\displaystyle \sum\_\{n=1\}^\infty \left(x+\frac\{1\}\{n\}\right)^n$ 收敛; 当 $\displaystyle x > 1$ 时, $\displaystyle \sum\_\{n=1\}^\infty \left(x+\frac\{1\}\{n\}\right)^n$ 发散. 又当 $\displaystyle x=1$ 时, 级数通项 $\displaystyle \left(1+\frac\{1\}\{n\}\right)^n\to \mathrm\{e\}\left(n\to\infty\right)$, 而 $\displaystyle \sum\_\{n=1\}^\infty \left(x+\frac\{1\}\{n\}\right)^n$ 发散. 综上, $\displaystyle \sum\_\{n=1\}^\infty \left(x+\frac\{1\}\{n\}\right)^n$ 的收敛域为 $\displaystyle [0,1)$. (2)、 设 $\displaystyle S(x)=\sum\_\{n=1\}^\infty \left(x+\frac\{1\}\{n\}\right)^n$, 则对 $\displaystyle \forall\ 0 < a < 1$, 取 $\displaystyle b=\frac\{a+1\}\{2\} < 1$, 则当 $\displaystyle n\geq \frac\{1\}\{b-a\}$ 时,
\begin\{aligned\} 0\leq x\leq a\Rightarrow \left(x+\frac\{1\}\{n\}\right)^n \leq \left(a+\frac\{1\}\{n\}\right)^n\leq b^n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
据 Weierstrass 判别法知 $\displaystyle \sum\_\{n=1\}^\infty \left(x+\frac\{1\}\{n\}\right)^n$ 在 $\displaystyle [0,a]$ 上一致收敛. 而 $\displaystyle S(x)$ 在 $\displaystyle [0,a]$ 上连续. 由 $\displaystyle a$ 的任意性知 $\displaystyle S(x)$ 在 $\displaystyle [0,1)$ 上连续.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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---
502、 (3)、 叙述函数项级数 $\displaystyle \sum\_\{n=1\}^\infty a\_n(x)b\_n(x)$ 在 $\displaystyle D$ 上一致收敛的 Abel 判别法. (河海大学2023年数学分析考研试题) [函数项级数 ]
[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) / 这就是 Abel 判别法的表述. 若 $\displaystyle \sum\_\{n=1\}^\infty a\_n(x)$ 关于 $\displaystyle x\in D$ 一致收敛, $\displaystyle b\_n(x)$ 关于 $\displaystyle n$ 单调且 (关于 $\displaystyle x\in D$) 一致有界, 则 $\displaystyle \sum\_\{n=1\}^\infty a\_n(x)b\_n(x)$ 在 $\displaystyle D$ 上一致收敛.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)/
---
503、 2、 (18 分) 考查函数项级数 $\displaystyle \sum\_\{n=1\}^\infty \frac\{\sin nx\}\{\sqrt\{n\}\}, x\in [-\pi,\pi]$. (1)、 证明函数项级数在 $\displaystyle x=0, \pm \pi$ 处绝对收敛, 在 $\displaystyle x\in(-\pi,0)\cup (0,\pi)$ 处条件收敛; (2)、 设 $\displaystyle S(x)$ 为函数项级数的和函数, 证明: $\displaystyle S(x)$ 在 $\displaystyle (-\pi,0)\cup (0,\pi)$ 上连续; (3)、 证明: $\displaystyle S(x)$ 在 $\displaystyle x=0$ 处不连续. [张祖锦注: 第 3 问是证明和函数在原点处不连续, 故而仅仅证明函数项级数在 $\displaystyle [-\pi,\pi]$ 上不一致收敛是不够的.] (河南大学2023年数学分析考研试题) [函数项级数 ]
[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)、 设题中函数项级数为 $\displaystyle S(x)$, 则 $\displaystyle S(x)$ 是奇函数, 而仅需讨论原函数项级数在 $\displaystyle [0,\pi]$ 上的绝对收敛性与条件收敛性. (1-1)、 当 $\displaystyle x=0$ 或 $\displaystyle \pi$ 时, 原函数项级数通项为 $\displaystyle 0$, 而绝对收敛. (1-2)、 当 $\displaystyle 0 < x < \pi$ 时, 由
\begin\{aligned\} \left|\sum\_\{k=1\}^n \sin kx\right| &=\frac\{\left|\sum\_\{k=1\}^n \left\[\cos\frac\{2k-1\}\{2\}x-\cos \frac\{2k+1\}\{2\}x\right\]\right|\}\{2\sin\frac\{x\}\{2\}\}\\\\ & =\frac\{\left|\cos\frac\{x\}\{2\}-\cos\frac\{2n+1\}\{2\}x\right|\}\{2\sin\frac\{x\}\{2\}\} \leq\frac\{1\}\{\left|\sin\frac\{x\}\{2\}\right|\},\qquad(I)\\\\ \frac\{1\}\{\sqrt\{n\}\}&\searrow 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及 Dirichlet 判别法知原函数项级数收敛. 又由
\begin\{aligned\} &\sum\_\{n=1\}^\infty \frac\{|\sin nx|\}\{\sqrt\{n\}\} \geq \sum\_\{n=1\}^\infty \frac\{\sin^2nx\}\{\sqrt\{n\}\} =\frac\{1\}\{2\}\left\[\sum\_\{n=1\}^\infty \frac\{1\}\{\sqrt\{n\}\} -\sum\_\{n=1\}^\infty \frac\{\cos 2nx\}\{\sqrt\{n\}\}\right\],\\\\ &\left|\sum\_\{k=1\}^n \cos 2kx\right|=\frac\{\left|\sum\_\{k=1\}^n \left\[\sin (2k+1)x-\sin (2k-1)x\right\]\right|\}\{2\sin x\} \leq \frac\{1\}\{\sin x\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知原函数项级数条件收敛. (2)、 对 $\displaystyle \forall\ \delta > 0$, $\displaystyle (I)\Rightarrow \sup\_\{x\in [\delta,\pi-\delta]\}\left|\sum\_\{k=1\}^n \sin kx\right| \leq \frac\{1\}\{\sin \frac\{\delta\}\{2\}\}$. 由 Dirichlet 判别法知 $\displaystyle \sum\_\{n=1\}^\infty \frac\{\sin nx\}\{\sqrt\{n\}\}$ 在 $\displaystyle [\delta,\pi-\delta]$ 上一致收敛. 这表明 $\displaystyle S(x)$ 在 $\displaystyle [\delta,\pi-\delta]$ 上连续. 由 $\displaystyle \delta$ 的任意性知 $\displaystyle S$ 在 $\displaystyle (0,\pi)$ 上连续. 又由 $\displaystyle S$ 是奇函数知结论成立. (3)、 对 $\displaystyle \forall\ N\in\mathbb\{Z\}\_+$,
\begin\{aligned\} &S\left(\frac\{\pi\}\{N\}\right)=\sum\_\{n=1\}^\infty \frac\{\sin\frac\{n\pi\}\{N\}\}\{\sqrt\{n\}\} \stackrel\{n=qN+r, 0\leq r\leq N-1\}\{=\}\sum\_\{r=0\}^\{N-1\} \sin\frac\{r\pi\}\{N\}\sum\_\{q=0\}^\infty \frac\{(-1)^q\}\{\sqrt\{qN+r\}\}\\\\ =&\sum\_\{r=0\}^\{N-1\} \sin\frac\{r\pi\}\{N\}\sum\_\{k=0\}^\infty \left\[\frac\{(-1)^\{2k\}\}\{\sqrt\{2kN+r\}\}+\frac\{(-1)^\{2k+1\}\}\{\sqrt\{(2k+1)N+1\}\}\right\]\\\\ \geq& \sum\_\{r=0\}^\{N-1\} \sin\frac\{r\pi\}\{N\}\left(\frac\{1\}\{\sqrt\{r\}\}-\frac\{1\}\{\sqrt\{N+r\}\}\right)\\\\ =&\sqrt\{N\}\sum\_\{r=0\}^\{N-1\} \sin\frac\{r\pi\}\{N\}\left(\frac\{1\}\{\sqrt\{\frac\{r\}N\}\}-\frac\{1\}\{\sqrt\{1+\frac\{r\}\{N\}\}\}\right)\cdot\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
\begin\{aligned\} &\lim\_\{N\to\infty\}\sum\_\{r=0\}^\{N-1\} \sin\frac\{r\pi\}\{N\}\left(\frac\{1\}\{\sqrt\{\frac\{r\}N\}\}-\frac\{1\}\{\sqrt\{1+\frac\{r\}\{N\}\}\}\right)\cdot\frac\{1\}\{N\}\\\\ =&\int\_0^1 \sin(\pi x)\left(\frac\{1\}\{\sqrt\{x\}\}-\frac\{1\}\{\sqrt\{1+x\}\}\right)\mathrm\{ d\} x\equiv I > 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知当 $\displaystyle N$ 充分大时,
\begin\{aligned\} S\left(\frac\{\pi\}\{N\}\right)\geq\frac\{I\}\{2\}\sqrt\{N\}\Rightarrow \lim\_\{N\to\infty\}S\left(\frac\{\pi\}\{N\}\right)=+\infty\neq 0=S(0). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle S$ 在原点处不连续.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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504、 9、 讨论函数列 $\displaystyle f\_n(x)=n(x^n-x^\{2n\})$ 在 $\displaystyle [0,1]$ 上的一致收敛性. (黑龙江大学2023年数学分析考研试题) [函数项级数 ]
[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) / 易知 $\displaystyle \lim\_\{n\to\infty\}f\_n(x)=0$. 由 $\displaystyle f\_n'(x)=n^2x^\{n-1\}(1-2x^n)$ 知
\begin\{aligned\} \max\_\{x\in [0,1]\}|f\_n(x)-0|=f\_n\left(\frac\{1\}\{2^\frac\{1\}\{n\}\}\right)=\frac\{n\}\{4\}\not\to 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f\_n\not\rightrightarrows 0$ 于 $\displaystyle [0,1]$.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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505、 10、 讨论 $\displaystyle S(x)=\sum\_\{n=1\}^\infty n\mathrm\{e\}^\{-nx\}$ 在 $\displaystyle (0,+\infty)$ 上的连续性. (黑龙江大学2023年数学分析考研试题) [函数项级数 ]
[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) / 对 $\displaystyle \forall\ \delta > 0$, 由
\begin\{aligned\} \sup\_\{x\geq \delta\} n\mathrm\{e\}^\{-nx\}=n\mathrm\{e\}^\{-n \delta\}, \sum\_\{n=1\}^\infty n\mathrm\{e\}^\{-n\delta\} < \infty \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及 Weierstrass 判别法知 $\displaystyle \sum\_\{n=1\}^\infty n\mathrm\{e\}^\{-nx\}$ 在 $\displaystyle [\delta,+\infty)$ 上一致收敛, 而和函数 $\displaystyle S(x)$ 连续. 由 $\displaystyle \delta$ 的任意性知 $\displaystyle S(x)$ 在 $\displaystyle (0,+\infty)$ 上连续.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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506、 6、 求证级数 $\displaystyle \sum\_\{n=1\}^\infty (-1)^n \frac\{\mathrm\{e\}^\{x^2\}+\sqrt\{n\}\}\{n^\frac\{3\}\{2\}\}$ 在任何有界区间 $\displaystyle [a,b]$ 上是一致收敛, 但在任何一点 $\displaystyle x\_0$ 处不绝对收敛. (湖南大学2023年数学分析考研试题) [函数项级数 ]
[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) / (0-7)、 由 $\displaystyle \sup\_\{x\in [a,b]\}\left|(-1)^n \frac\{\mathrm\{e\}^\{x^2\}\}\{n^\frac\{3\}\{2\}\}\right| \leq \frac\{\mathrm\{e\}^\{\max\\{|a|^2,|b|^2\\}\}\}\{n^\frac\{3\}\{2\}\}$ 及优级数判别法知 $\displaystyle \sum\_\{n=1\}^\infty (-1)^n \frac\{\mathrm\{e\}^\{x^2\}\}\{n^\frac\{3\}\{2\}\}$ 在 $\displaystyle [a,b]$ 上一致收敛. 又由 $\displaystyle \sum\_\{n=1\}^\infty (-1)^n \frac\{1\}\{\sqrt\{n\}\}$ 收敛知其关于 $\displaystyle x\in [a,b]$ 一致收敛. 故级数 $\displaystyle \sum\_\{n=1\}^\infty (-1)^n \frac\{\mathrm\{e\}^\{x^2\}+\sqrt\{n\}\}\{n^\frac\{3\}\{2\}\}$ 在 $\displaystyle [a,b]$ 上是一致收敛. (0-8)、 由
\begin\{aligned\} \sum\_\{n=1\}^\infty \left|(-1)^n \frac\{\mathrm\{e\}^\{x^2\}+\sqrt\{n\}\}\{n^\frac\{3\}\{2\}\}\right| =\sum\_\{n=1\}^\infty \frac\{\mathrm\{e\}^\{x^2\}+\sqrt\{n\}\}\{n^\frac\{3\}\{2\}\}\geq \sum\_\{n=1\}^\infty \frac\{1\}\{\sqrt\{n\}\}=+\infty \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知级数 $\displaystyle \sum\_\{n=1\}^\infty (-1)^n \frac\{\mathrm\{e\}^\{x^2\}+\sqrt\{n\}\}\{n^\frac\{3\}\{2\}\}$ 在任何一点 $\displaystyle x$ 处不绝对收敛.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)/
发表于 2023-3-5 08:52:16
