## 张祖锦2023年数学专业真题分类70天之第24天
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530、 4、 对于 Dirichlet 级数 $\displaystyle \sum\_\{n=1\}^\infty \frac\{a\_n\}\{n^x\}$, 若级数在 $\displaystyle x=x\_0$ 处收敛, 证明: $\displaystyle \sum\_\{n=1\}^\infty \frac\{a\_n\}\{n^x\}$ 在 $\displaystyle [x\_0,+\infty)$ 上一致收敛. 特别地, 级数在 $\displaystyle x > x\_0+1$ 时绝对收敛. (中国科学院大学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)、 $\displaystyle \sum\_\{n=1\}^\infty \frac\{a\_n\}\{n^\{x\_0\}\}$ 收敛. 当 $\displaystyle x > x\_0$ 时, $\displaystyle \frac\{1\}\{n^\{x-x\_0\}\}$ 关于 $\displaystyle n$ 递减, 且关于 $\displaystyle x$ 一致有界 (为 $\displaystyle 1$). 由 Abel 判别法知 $\displaystyle \sum\_\{n=1\}^\infty \frac\{a\_n\}\{n^x\}$ 在 $\displaystyle [x\_0,+\infty)$ 上一致收敛/ (2)、 当 $\displaystyle x > x\_0+1$ 时, 由第 1 步知 $\displaystyle \sum\_\{n=1\}^\infty \frac\{a\_n\}\{n^\frac\{(x-1)+x\_0\}\{2\}\}$ 收敛, 从而 $\displaystyle \lim\_\{n\to\infty\}\frac\{a\_n\}\{n^\frac\{(x-1)+x\_0\}\{2\}\}=0$,
\begin\{aligned\} \exists\ M > 0,\mathrm\{ s.t.\} \forall\ n\geq 1, &\left|\frac\{a\_n\}\{n^\frac\{(x-1)+x\_0\}\{2\}\}\right|\leq M\\\\ \Rightarrow& \left|\frac\{a\_n\}\{n^x\}\right| =\left|\frac\{a\_n\}\{n^\frac\{(x-1)+x\_0\}\{2\}\}\right| \cdot\frac\{1\}\{n^\frac\{1+x-x\_0\}\{2\}\}\leq \frac\{M\}\{n^\frac\{1+x-x\_0\}\{2\}\}. \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 $\displaystyle \sum\_\{n=1\}^\infty \frac\{a\_n\}\{n^x\}$ 绝对收敛.跟锦数学微信公众号. [在线资料](
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531、 6、 设 $\displaystyle \left\\{u\_n(x)\right\\}$ 为 $\displaystyle [a,b]$ 上的正的递减且收敛于 $\displaystyle 0$ 的函数列, 且每个 $\displaystyle u\_n(x)$ 都是 $\displaystyle [a,b]$ 上的单调函数. 证明: $\displaystyle \sum\_\{n=1\}^\infty (-1)^n u\_n(x)$ 在 $\displaystyle [a,b]$ 上一致收敛. (中国矿业大学(徐州)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) / 由题设, $\displaystyle u\_n(x)\geq u\_\{n+1\}(x)\geq \cdots > 0$, $\displaystyle u\_n(x)$ 关于 $\displaystyle x$ 单调. 由 Leibniz 判别法知 $\displaystyle \sum\_\{n=1\}^\infty (-1)^n u\_n(x)$ 收敛, 且
\begin\{aligned\} &\left|\sum\_\{k=n\}^\infty(-1)^k u\_k(x)\right| =\left|\sum\_\{k=n\}^\infty (-1)^\{k-n\}u\_k(x)\right|\\\\ =&\left|[u\_n(x)-u\_\{n+1\}(x)]+[u\_\{n+2\}(x)-u\_\{n+3\}(x)]+\cdots\right|\\\\ =&[u\_n(x)-u\_\{n+1\}(x)]+[u\_\{n+2\}(x)-u\_\{n+3\}(x)]+\cdots\geq 0\\\\ =&u\_n(x)-\left\[u\_\{n+1\}(x)-u\_\{n+2\}(x)\right\]-\cdots \leq u\_n(x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} 0\leq \sup\_\{x\in [a,b]\}\left|\sum\_\{k=n\}^\infty(-1)^k u\_k(x)\right| \leq& \sup\_\{x\in [a,b]\}u\_n(x)\leq \max\left\\{u\_n(a),u\_n(b)\right\\}\\\\ =&\frac\{u\_n(a)+u\_n(b)+|u\_n(a)-u\_n(b)|\}\{2\}\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\}
这就证明了 $\displaystyle \sum\_\{n=1\}^\infty (-1)^n u\_n(x)$ 在 $\displaystyle [a,b]$ 上一致收敛.跟锦数学微信公众号. [在线资料](
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532、 10、 (15 分) 设函数 $\displaystyle f(x)$ 为定义在 $\displaystyle [0,1]$ 上的连续函数, 令
\begin\{aligned\} f\_n(x)=\int\_0^x f(t^n)\mathrm\{ d\} t, x\in [0,1], n\geq 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: 函数列 $\displaystyle \left\\{f\_n(x)\right\\}$ 在 $\displaystyle [0,1]$ 上一致收敛. (中国人民大学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) / 写出
\begin\{aligned\} &\max\_\{x\in [0,1]\}|f\_n(x)-f(0)x| \leq\max\_\{x\in [0,1]\}\int\_0^x |f(t^n)-f(0)|\mathrm\{ d\} t\\\\ =&\int\_0^1 |f(t^n)-f(0)|\mathrm\{ d\} t =\int\_0^\{1-\delta\}|f(t^n)-f(0)|\mathrm\{ d\} t +\int\_\{1-\delta\}^1 |f(t^n)-f(0)|\mathrm\{ d\} t\\\\ \leq&\max\_\{s\in [0,(1-\delta)^n]\}|f(s)-f(0)| +2\max\_\{[0,1]\}|f|\cdot \delta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle n\to\infty$ 并令用 $\displaystyle f$ 在 $\displaystyle 0$ 处的连续性知
\begin\{aligned\} \varlimsup\_\{n\to\infty\}\max\_\{x\in [0,1]\}|f\_n(x)-f(0)x|\leq 2\max\_\{[0,1]\}|f|\cdot \delta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再令 $\displaystyle \delta\to 0^+$ 得
\begin\{aligned\} &0\leq \varliminf\_\{n\to\infty\}\max\_\{x\in [0,1]\}|f\_n(x)-f(0)x|\leq \varlimsup\_\{n\to\infty\}\max\_\{x\in [0,1]\}|f\_n(x)-f(0)x| \leq 0\\\\ \Rightarrow&\lim\_\{n\to\infty\}\max\_\{x\in [0,1]\}|f\_n(x)-f(0)x|=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了 $\displaystyle f\_n\rightrightarrows f(0)x$, 于 $\displaystyle [0,1]$.跟锦数学微信公众号. [在线资料](
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533、 4、 (15 分) 设 $\displaystyle f(x)$ 为 $\displaystyle \mathbb\{R\}$ 上的连续函数, 且当 $\displaystyle x\neq 0$ 时, 有 $\displaystyle |f(x)| < |x|$. 定义
\begin\{aligned\} f\_1(x)=f(x, f\_2(x)=f\left(f\_1(x)\right), \cdots, f\_\{n+1\}(x)=f\left(f\_n(x)\right), \cdots. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
求证: $\displaystyle \left\\{f\_n(x)\right\\}$ 在任何有限区间 $\displaystyle [-a,a]$ 上一致收敛于 $\displaystyle 0$. (中山大学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)、 我们先证明 Dini 定理. 设 $\displaystyle \left\\{f\_n(x)\right\\}$ 为 $\displaystyle [a,b]$ 上的连续函数列, $\displaystyle f(x)$ 为 $\displaystyle [a,b]$ 上的连续函数, 若
\begin\{aligned\} \forall\ x\in [a,b], f\_n(x)\leq f\_\{n+1\}(x)\left(n=1,2,\cdots\right),\mbox\{且\}\lim\_\{n\to\infty\}f\_n(x)=f(x), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle \left\\{f\_n(x)\right\\}$ 在 $\displaystyle [a,b]$ 上一致收敛于 $\displaystyle f(x)$. 事实上, 设
\begin\{aligned\} 0\leq g\_n(x)=f(x)-f\_n(x)\in C[a,b], \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\} \forall\ x\in [a,b], g\_n(x)\searrow 0.\qquad(\star) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
仅需证明 $\displaystyle g\_n\rightrightarrows 0$. 由 $\displaystyle (\star)$ 知对任意固定的 $\displaystyle \varepsilon > 0$,
\begin\{aligned\} x\in [a,b]&\Rightarrow \lim\_\{n\to\infty\}g\_n(x)=0\\\\ &\Rightarrow \exists\ n\_x, \mathrm\{ s.t.\} 0\leq g\_\{n\_x\}(x) < \varepsilon\\\\ &\Rightarrow \exists\ n\_x, \exists\ \delta\_x > 0,\mathrm\{ s.t.\} \forall\ |x'-x| < \delta\_x, 0\leq g\_\{n\_x\}(x') < \varepsilon\\\\ &\quad \ \left(g\_\{n\_x\}\in C[a,b],\mbox\{保号性\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle \left\[a,b\right\]\subset \bigcup\_\{x\in [a,b]\}U(x;\delta\_x)$ 及有限覆盖定理知
\begin\{aligned\} \left\[a,b\right\]\subset \bigcup\_\{i=1\}^m U(x\_i;\delta\_\{x\_i\}). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
取 $\displaystyle N=\max\left\\{n\_\{x\_1\},\cdots, n\_\{x\_m\}\right\\}$, 则当 $\displaystyle n\geq N, x\in [a,b]$ 时,
\begin\{aligned\} &\exists\ 1\leq i\leq m,\mathrm\{ s.t.\} x\in U(x\_i;\delta\_\{x\_i\})\\\\ \Rightarrow& 0\leq g\_n(x)\leq g\_\{n\_\{x\_i\}\}(x) < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 回到题目. 设 $\displaystyle g\_n(x)=|f\_n(x)-0|\geq 0$, 则
\begin\{aligned\} g\_\{n+1\}(x)=|f\_\{n+1\}(x)|=\left|f\left(f\_n(x)\right)\right| < |f\_n(x)|=g\_n(x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由第 1 步知 $\displaystyle g\_n\rightrightarrows 0$ 于 $\displaystyle [-a,a]$. 此即 $\displaystyle f\_n\rightrightarrows 0$ 于 $\displaystyle [-a,a]$.跟锦数学微信公众号. [在线资料](
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534、 (3)、 证明: 函数项级数 $\displaystyle \sum\_\{n=1\}^\infty \frac\{nx\}\{n^2+(n+2)x\}$ 在 $\displaystyle [0,+\infty)$ 上不一致收敛. (重庆师范大学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) / 设 $\displaystyle f\_n(x)=\frac\{nx\}\{n^2+(n+2)x\}$, 则
\begin\{aligned\} \lim\_\{n\to\infty\}f\_n(x)=\lim\_\{n\to\infty\}\frac\{x\}\{n+\left(1+\frac\{2\}\{n\}\right)x\}=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由 $\displaystyle f\_n'(x)=\frac\{n^3\}\{[n^2+(n+2)x]^2\} > 0$ 知
\begin\{aligned\} &\sup\_\{x\in\mathbb\{R\}\}|f\_n(x)-0|=\sup\_\{x\in\mathbb\{R\}\}f\_n(x)=\lim\_\{x\to+\infty\}f\_n(x)\\\\ =&\lim\_\{x\to+\infty\}\frac\{n\}\{\frac\{n^2\}\{x\}+n+2\} =\frac\{n\}\{n+2\}\to 1. \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 $\displaystyle \left\\{f\_n(x)\right\\}$ 在 $\displaystyle [0,+\infty)$ 上不一致收敛于 $\displaystyle 0$, 进而 $\displaystyle \sum\_\{n=1\}^\infty f\_n(x)$ 在 $\displaystyle [0,+\infty)$ 上不一致收敛.跟锦数学微信公众号. [在线资料](
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535、 3、 求 $\displaystyle f(x)=\frac\{1\}\{x^2-4x+3\}$ 关于 $\displaystyle x-1$ 的幂级数展开式. [张祖锦注: 题目有问题, 函数在 $\displaystyle x=1$ 处都没有定义, 谈何幂级数? 又不是复变的 Laurent 级数] (北京邮电大学2023年数学分析考研试题) [幂级数 ]
[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) / [张祖锦注: 题目有问题, 函数在 $\displaystyle x=1$ 处都没有定义, 谈何幂级数? 又不是复变的 Laurent 级数]跟锦数学微信公众号. [在线资料](
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536、 (3)、 求证如下等式:
\begin\{aligned\} \int\_0^1 \sum\_\{n=1\}^\infty \frac\{x^\{n-1\}\}\{n\}\mathrm\{ d\} x=\sum\_\{n=1\}^\infty \frac\{1\}\{n\}\int\_0^1 x^\{n-1\}\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(大连理工大学2023年数学分析考研试题) [幂级数 ]
[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 $\displaystyle \sum\_\{n=1\}^\infty \frac\{x^\{n-1\}\}\{n\}$ 的收敛半径为 $\displaystyle 1$, 而在 $\displaystyle |x|\leq 1-\varepsilon\ (0 < \varepsilon < 1)$ 上绝对一致收敛. 又由 $\displaystyle \sum\_\{n=1\}^\infty \frac\{x^n\}\{n^2\}$ 的收敛域为 $\displaystyle [-1,1]$ 及幂级数的连续性知
\begin\{aligned\} \mbox\{左端\}=&\lim\_\{\varepsilon\to 0^+\}\int\_0^\{1-\varepsilon\} \sum\_\{n=1\}^\infty \frac\{x^\{n-1\}\}\{n\}\mathrm\{ d\} x =\lim\_\{\varepsilon\to 0^+\}\sum\_\{n=1\}^\infty \int\_0^\{1-\varepsilon\} x^\{n-1\}\mathrm\{ d\} x\\\\ =&\lim\_\{\varepsilon\to 0^+\}\sum\_\{n=1\}^\infty \frac\{(1-\varepsilon)^2\}\{n^2\} =\sum\_\{n=1\}^\infty \frac\{1\}\{n^2\}=\mbox\{右端\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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537、 (2)、 设幂级数 $\displaystyle \sum\_\{n=0\}^\infty a\_nx^n$ 的收敛半径为 $\displaystyle 1$, 左极限 $\displaystyle \lim\_\{x\to 1^-\}\sum\_\{n=0\}^\infty a\_nx^n$ 存在且 $\displaystyle \lim\_\{n\to\infty\}na\_n=0$. 证明: 数项级数 $\displaystyle \sum\_\{n=0\}^\infty a\_n$ 收敛, 且
\begin\{aligned\} \sum\_\{n=0\}^\infty a\_n=\lim\_\{x\to 1^-\}\sum\_\{n=0\}^\infty a\_nx^n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
举例说明仅由’$\sum\_\{n=0\}^\infty a\_nx^n$ 的收敛半径为 $\displaystyle 1$ 及左极限 $\displaystyle \lim\_\{x\to 1^-\}\sum\_\{n=0\}^\infty a\_nx^n$ 存在‘不能断言’数项级数 $\displaystyle \sum\_\{n=0\}^\infty a\_n$ 收敛‘. (电子科技大学2023年数学分析考研试题) [幂级数 ]
[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 $\displaystyle \lim\_\{x\to 1^-\}\sum\_\{n=0\}^\infty a\_nx^n=A$. 再设 $\displaystyle x\_m\nearrow 1$ 待定, 则由 Heine 原理, $\displaystyle \lim\_\{m\to\infty\} \sum\_\{n=0\}^\infty a\_nx\_m^n=A$. 又由 $\displaystyle \{\lim\_\{n\to\infty\}na\_n=0\}$ 知 $\displaystyle \lim\_\{m\to\infty\}\frac\{1\}\{m\}\sum\_\{n=0\}^\{m-1\}n|a\_n|=0.$. 而对任意固定的 $\displaystyle \varepsilon > 0$, $\displaystyle \exists\ N,\ \forall\ m\geq N$,
\begin\{aligned\} \left|\sum\_\{n=0\}^\infty a\_nx\_m^n-A\right| < \frac\{\varepsilon\}\{3\},\quad |ma\_m| < \frac\{\varepsilon\}\{3\},\quad \frac\{1\}\{m\}\sum\_\{n=0\}^\{m-1\}n|a\_n| < \frac\{\varepsilon\}\{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\} &\left|\sum\_\{n=0\}^\{m-1\}a\_n-A\right| \leq \left|\sum\_\{n=0\}^\{m-1\}a\_n -\sum\_\{n=0\}^\infty a\_nx\_m^n\right| +\left|\sum\_\{n=0\}^\infty a\_nx\_m^n-A\right|\\\\ \leq& \left|\sum\_\{n=0\}^\{m-1\} a\_n(1-x\_m^n)\right| +\left|\sum\_\{n=m\}^\infty a\_nx\_m^n\right|+\frac\{\varepsilon\}\{3\}\\\\ \leq& \left|\sum\_\{n=0\}^\{m-1\} a\_n(1-x\_m)(1+x\_m+\cdots+x\_m^\{n-1\})\right| +\frac\{1\}\{m\}\sum\_\{n=m\}^\infty n|a\_n|\cdot x\_m^n+\frac\{\varepsilon\}\{3\}\\\\ \leq& m(1-x\_m)\cdot \frac\{1\}\{m\}\sum\_\{n=0\}^\{m-1\} n|a\_n| +\frac\{1\}\{m\}\sup\_\{n\geq m\} n|a\_n| \cdot \frac\{x\_m^m\}\{1-x\_m\}+\frac\{\varepsilon\}\{3\}\\\\ \leq& \frac\{1\}\{m\}\sum\_\{n=0\}^\{m-1\} n|a\_n| +\sup\_\{n\geq m\} n|a\_n|+\frac\{\varepsilon\}\{3\}\quad \left(\mbox\{取 \}x\_m=1-\frac\{1\}\{m\}\right)\\\\ < &\frac\{\varepsilon\}\{3\}+\frac\{\varepsilon\}\{3\}+\frac\{\varepsilon\}\{3\} =\varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了 $\displaystyle \sum\_\{n=0\}^\infty a\_n=A=\lim\_\{x\to 1^-\}\sum\_\{n=0\}^\infty a\_nx^n$. 仅由’$\sum\_\{n=0\}^\infty a\_nx^n$ 的收敛半径为 $\displaystyle 1$ 及左极限 $\displaystyle \lim\_\{x\to 1^-\}\sum\_\{n=0\}^\infty a\_nx^n$ 存在‘不能断言’数项级数 $\displaystyle \sum\_\{n=0\}^\infty a\_n$ 收敛‘. 比如 $\displaystyle a\_n=(-1)^n$,
\begin\{aligned\} \lim\_\{x\to 1^-\}\sum\_\{n=0\}^\infty a\_nx^n =\lim\_\{x\to 1^-\}\frac\{1\}\{1+x\}=\frac\{1\}\{2\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
但 $\displaystyle \sum\_\{n=1\}^\infty a\_n=\sum\_\{n=1\}^\infty (-1)^n$ 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](
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---
538、 2、 计算级数 $\displaystyle \sum\_\{n=1\}^\infty \frac\{x^\{2n+1\}\}\{2n\}$. (东北大学2023年数学分析考研试题) [幂级数 ]
[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\} \mbox\{原式\}=&x\sum\_\{n=1\}^\infty \int\_0^x t^\{2n-1\}\mathrm\{ d\} t =x\int\_0^x \sum\_\{n=1\}^\infty t^\{2n-1\}\mathrm\{ d\} t\\\\ =&x\int\_0^x \frac\{t\}\{1-t^2\}\mathrm\{ d\} t=-x\ln (1-x^2), x\in (-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)/
---
539、 7、 (15 分) 设函数 $\displaystyle f(x)=\frac\{1\}\{1-2x-x^2\}$, 判断级数 $\displaystyle \sum\_\{n=0\}^\infty \frac\{n!\}\{f^\{(n)\}(0)\}$ 是否收敛, 并证明你的结论. (合肥工业大学2023年数学分析考研试题) [幂级数 ]
[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 $\displaystyle \alpha=-1+\sqrt\{2\}, \beta=-1-\sqrt\{2\}$, 则
\begin\{aligned\} f(x)&=\frac\{1\}\{-(x-\alpha)(x-\beta)\} =\frac\{1\}\{\beta-\alpha\} \left(\frac\{1\}\{x-\alpha\}-\frac\{1\}\{x-\beta\}\right)\\\\ &=\frac\{1\}\{\beta-\alpha\}\left(-\frac\{1\}\{\alpha\}\frac\{1\}\{1-\frac\{x\}\{\alpha\}\} +\frac\{1\}\{\beta\} \frac\{1\}\{1-\frac\{x\}\{\beta\}\}\right)\\\\ &=\frac\{1\}\{\beta-\alpha\}\left\[-\frac\{1\}\{\alpha\}\sum\_\{n=0\}^\infty \left(\frac\{x\}\{\alpha\}\right)^n +\frac\{1\}\{\beta\} \sum\_\{n=0\}^\infty \left(\frac\{x\}\{\beta\}\right)^n\right\]\\\\ &=\sum\_\{n=0\}^\infty \frac\{\frac\{1\}\{\beta^\{n+1\}\}-\frac\{1\}\{\alpha^\{n+1\}\}\}\{\beta-\alpha\}x^n, |x| < \min\left\\{|\alpha|,|\beta|\right\\}=\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\}
(2)、 由第 1 步知
\begin\{aligned\} &\left|\frac\{f^\{(n)\}(0)\}\{n!\}\right| =\left|\frac\{\frac\{1\}\{\beta^\{n+1\}\}-\frac\{1\}\{\alpha^\{n+1\}\}\}\{\beta-\alpha\}\right|\\\\ =&\left|\frac\{1\}\{\beta-\alpha\}\left\[(-1)^\{n+1\} \frac\{1\}\{\left(\sqrt\{2\}+1\right)^\{n+1\}\} -\frac\{1\}\{\left(\sqrt\{2\}-1\right)^\{n+1\}\}\right\]\right|\\\\ \geq& \frac\{1\}\{\beta-\alpha\}\left\[\frac\{1\}\{\left(\sqrt\{2\}-1\right)^\{n+1\}\} -\frac\{1\}\{\left(\sqrt\{2\}+1\right)^\{n+1\}\}\right\]\\\\ =&\frac\{1\}\{\beta-\alpha\}\left\[\left(\sqrt\{2\}+1\right)^\{n+1\}-\left(\sqrt\{2\}-1\right)^\{n+1\}\right\]\\\\ =&\frac\{1\}\{\beta-\alpha\} \left\[\sum\_\{k=0\}^\{n+1\}C\_\{n+1\}^k \sqrt\{2\}^k -\sum\_\{k=0\}^\{n+1\} C\_\{n+1\}^k (-1)^k \sqrt\{2\}^k\right\]\\\\ =&\frac\{2\}\{\beta-\alpha\} \left(C\_\{n+1\}^1 \sqrt\{2\} +C\_\{n+1\}^3 \sqrt\{2\}^3 +\cdots +C\_\{n+1\}^\{2\left\[\frac\{n+1\}\{2\}\right\]\} 2^\{\left\[\frac\{n+1\}\{2\}\right\]\}\right)\\\\ \geq&\frac\{2\}\{\beta-\alpha\} C\_\{n+1\}^\{2\left\[\frac\{n+1\}\{2\}\right\]\} 2^\{\left\[\frac\{n+1\}\{2\}\right\]\} \geq \frac\{2\}\{\beta-\alpha\}2^\{\frac\{n+1\}\{2\}-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} \left|\frac\{n!\}\{f^\{(n)\}(0)\}\right|\leq\frac\{\beta-\alpha\}\{\sqrt\{2\}^\{n-1\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由比较判别法即知原级数 (绝对) 收敛.跟锦数学微信公众号. [在线资料](
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---
540、 (4)、 求幂级数 $\displaystyle \sum\_\{n=0\}^\infty \frac\{x^n\}\{2^n(n+1)\}$ 的收敛域及和函数. (河海大学2023年数学分析考研试题) [幂级数 ]
[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 $\displaystyle a\_n=\frac\{1\}\{2^n(n+1)\}$, 则 $\displaystyle \lim\_\{n\to\infty\}\sqrt[n]\{a\_n\}=\frac\{1\}\{2\}$. 故原幂级数的收敛半径为 $\displaystyle R=2$. 又由 $\displaystyle \sum\_\{n=0\}^\infty \frac\{(-1)^n\}\{n+1\}$ 收敛, $\displaystyle \sum\_\{n=0\}^\infty \frac\{1\}\{n+1\}$ 发散知原幂级数的收敛点集为 $\displaystyle [-2,2)$. 当 $\displaystyle x=0$ 时, 原幂级数 $\displaystyle =1$. 当 $\displaystyle x\in [-2,2)\backslash\left\\{0\right\\}$ 时,
\begin\{aligned\} &\quad \sum\_\{n=0\}^\infty \frac\{x^n\}\{2^n(n+1)\} =\left. \sum\_\{n=0\}^\infty \frac\{t^n\}\{n+1\} \right|\_\{x=\frac\{1\}\{2\}\} =\left. \frac\{1\}\{t\}\sum\_\{n=0\}^\infty \frac\{t^\{n+1\}\}\{n+1\} \right|\_\{x=\frac\{1\}\{2\}\}\\\\ & =\left. \frac\{1\}\{t\}\sum\_\{n=0\}^\infty \int\_0^t s^n\mathrm\{ d\} s \right|\_\{x=\frac\{1\}\{2\}\} =\left. \frac\{1\}\{t\}\int\_0^t\frac\{1\}\{1-s\}\mathrm\{ d\} s \right|\_\{x=\frac\{1\}\{2\}\} =-\frac\{2\}\{x\}\ln \left(1-\frac\{x\}\{2\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故原幂级数
\begin\{aligned\} =\left\\{\begin\{array\}\{llllllllllll\} 1,&x=0,\\\\ -\frac\{2\}\{x\}\ln \left(1-\frac\{x\}\{2\}\right),&x\in [-2,0)\cup (0,2). \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\}
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)/
---
541、 (3)、 设 $\displaystyle f(x)=\frac\{1\}\{1-x-x^2\}, a\_n=\frac\{f^\{(n)\}(0)\}\{n!\}, n\geq 0$. 证明级数 $\displaystyle \sum\_\{n=0\}^\infty \frac\{a\_\{n+1\}\}\{a\_na\_\{n+2\}\}$ 收敛, 并求和. (河海大学2023年数学分析考研试题) [幂级数 ]
[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 $\displaystyle x^2+x-1=0$ 的根为 $\displaystyle \alpha=\frac\{-1-\sqrt\{5\}\}\{2\}, \beta=\frac\{-1+\sqrt\{5\}\}\{2\}$, 则
\begin\{aligned\} f(x)=&-\frac\{1\}\{(x-\alpha)(x-\beta)\} =\frac\{1\}\{\beta-\alpha\}\left(\frac\{1\}\{x-\alpha\}-\frac\{1\}\{x-\beta\}\right)\\\\ =&\frac\{1\}\{\beta-\alpha\}\left(-\frac\{1\}\{\alpha\}\frac\{1\}\{1-\frac\{x\}\{\alpha\}\} +\frac\{1\}\{\beta\}\frac\{1\}\{1-\frac\{x\}\{\beta\}\}\right)\\\\ =&\frac\{1\}\{\beta-\alpha\}\left\[-\frac\{1\}\{\alpha\}\sum\_\{n=0\}^\infty \left(\frac\{x\}\{\alpha\}\right)^n +\frac\{1\}\{\beta\}\sum\_\{n=0\}^\infty \left(\frac\{x\}\{\beta\}\right)^n\right\], |x| < \beta. \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 $\displaystyle |x| < \beta$ 时,
\begin\{aligned\} 1=&(1-x-x^2)f(x)=(1-x-x^2)\sum\_\{n=0\}^\infty a\_nx^n\\\\ =&\sum\_\{n=0\}^\infty a\_nx^n-\sum\_\{n=0\}^\infty a\_nx^\{n+1\}-\sum\_\{n=0\}^\infty a\_nx^\{n+2\}\\\\ =&a\_0+(a\_1-a\_0)x+\sum\_\{n=0\}^\infty (a\_\{n+2\}-a\_\{n+1\}-a\_n)x^\{n+2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
因此,
\begin\{aligned\} a\_0=1, a\_1=a\_0, a\_\{n+2\}=a\_\{n+1\}+a\_n, n\geq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由数学归纳法知 $\displaystyle a\_n\geq n$, 而 $\displaystyle \lim\_\{n\to\infty\}a\_n=+\infty$. 进而
\begin\{aligned\} &\sum\_\{n=0\}^N \frac\{a\_\{n+1\}\}\{a\_na\_\{n+2\}\} =\sum\_\{n=0\}^N \frac\{a\_\{n+2\}-a\_n\}\{a\_na\_\{n+2\}\} =\sum\_\{n=0\}^N \left(\frac\{1\}\{a\_n\}-\frac\{1\}\{a\_\{n+2\}\}\right)\\\\ =&\sum\_\{n=0\}^N \frac\{1\}\{a\_n\}-\sum\_\{n=2\}^\{N+2\}\frac\{1\}\{a\_n\} =\frac\{1\}\{a\_0\}+\frac\{1\}\{a\_1\}-\frac\{1\}\{a\_\{N+1\}\}-\frac\{1\}\{a\_\{N+2\}\} \to \frac\{1\}\{a\_0\}+\frac\{1\}\{a\_1\}=2. \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 $\displaystyle \sum\_\{n=1\}^\infty \frac\{a\_\{n+1\}\}\{a\_na\_\{n+2\}\}=2$.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](
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---
542、 5、 求 $\displaystyle \sum\_\{n=1\}^\infty \frac\{1\}\{n(n+2)\}$. (黑龙江大学2023年数学分析考研试题) [幂级数 ]
[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\} \mbox\{原式\}=&\frac\{1\}\{2\}\sum\_\{n=1\}^\infty \left(\frac\{1\}\{n\}-\frac\{1\}\{n+2\}\right) =\frac\{1\}\{2\}\lim\_\{n\to\infty\}\sum\_\{k=1\}^n \left(\frac\{1\}\{k\}-\frac\{1\}\{k+2\}\right)\\\\ =&\frac\{1\}\{2\}\lim\_\{n\to\infty\}\left(1-\frac\{1\}\{3\}+\frac\{1\}\{2\}-\frac\{1\}\{4\}+\cdots +\frac\{1\}\{n\}-\frac\{1\}\{n+2\}\right)\\\\ =&\frac\{1\}\{2\}\lim\_\{n\to\infty\}\left(1+\frac\{1\}\{2\}-\frac\{1\}\{n+2\}\right) =\frac\{3\}\{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)/
---
543、 3、 求幂级数 $\displaystyle \sum\_\{n=1\}^\infty \left(1+\frac\{1\}\{2\}+\cdots+\frac\{1\}\{n\}\right)^\{-1\}x^n$ 的收敛域. (华东理工大学2023年数学分析考研试题) [幂级数 ]
[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 $\displaystyle a\_n=\left(\sum\_\{k=1\}^n \frac\{1\}\{k\}\right)^\{-1\}$, 则
\begin\{aligned\} \lim\_\{n\to\infty\}\frac\{a\_\{n+1\}\}\{a\_n\}=\lim\_\{n\to\infty\}\frac\{\displaystyle \sum\_\{k=1\}^n \frac\{1\}\{k\}\}\{\displaystyle \sum\_\{k=1\}^\{n+1\}\frac\{1\}\{k\}\} \xlongequal\{\tiny\mbox\{Stolz\}\} \lim\_\{n\to\infty\}\frac\{\displaystyle \frac\{1\}\{n\}\}\{\displaystyle \frac\{1\}\{n+1\}\}=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故收敛半径为 $\displaystyle 1$. 又由
\begin\{aligned\} \sum\_\{k=1\}^n\frac\{1\}\{k\}\geq \sum\_\{k=1\}^n\int\_k^\{k+1\}\frac\{\mathrm\{ d\} x\}\{x\} =\int\_1^\{n+1\}\frac\{\mathrm\{ d\} x\}\{x\}=\ln (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\}
知幂级数在收敛区间的端点处发散, 而收敛域为 $\displaystyle (-1,1)$.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)/
---
544、 3、 计算级数 $\displaystyle \sum\_\{n=1\}^\infty \frac\{n(n+3)\}\{2^n\}$. (华东师范大学2023年数学分析考研试题) [幂级数 ]
[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\} \mbox\{原式\}=&\sum\_\{n=1\}^\infty \frac\{n(n+1)\}\{2^n\}+\sum\_\{n=1\}^\infty \frac\{n\}\{2^\{n-1\}\} =\left. \frac\{1\}\{2\}\left(\sum\_\{n=1\}^\infty t^\{n+1\}\right)'' \right|\_\{t=\frac\{1\}\{2\}\} +\left. \left(\sum\_\{n=1\}^\infty t^n\right)' \right|\_\{t=\frac\{1\}\{2\}\}\\\\ =&\left. \frac\{1\}\{2\}\cdot\frac\{2\}\{(1-t)^3\} \right|\_\{t=\frac\{1\}\{2\}\} +\left. \frac\{1\}\{(1-t)^2\} \right|\_\{t=\frac\{1\}\{2\}\} =8+4=12. \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)/
---
545、 (2)、 $\displaystyle \sum\_\{k=0\}^n C\_\alpha^k C\_\beta^\{n-k\}=C\_\{\alpha+\beta\}^n$, 其中
\begin\{aligned\} C\_\alpha^k=\frac\{\alpha(\alpha-1)\cdots (\alpha-k+1)\}\{k!\}, \quad C\_\alpha^0=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(华中科技大学2023年数学分析考研试题) [幂级数 ]
[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\} (1+x)^\gamma=\sum\_\{n=0\}^\infty C\_\gamma^n x^n, |x| < 1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} (1+x)^\{\alpha+\beta\}=&\sum\_\{n=0\}^\infty C\_\{\alpha+\beta\}^n x^n,\\\\ =&(1+x)^\alpha (1+x)^\beta=\sum\_\{k=0\}^\infty C\_\alpha^k x^k\sum\_\{l=0\}^\infty C\_\beta^l x^l. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
比较两端 $\displaystyle x^n$ 的系数即知 $\displaystyle C\_\alpha^k=\frac\{\alpha(\alpha-1)\cdots (\alpha-k+1)\}\{k!\}$.跟锦数学微信公众号. [在线资料](
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---
546、 4、 (12 分) 求幂级数 $\displaystyle \sum\_\{n=1\}^\infty n^2x^\{n-1\}$ 的收敛域, 并求和函数. (吉林大学2023年数学分析考研试题) [幂级数 ]
[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) / 易知幂级数的收敛域为 $\displaystyle (-1,1)$. 当 $\displaystyle x\in(-1,1)$ 时,
\begin\{aligned\} \mbox\{原式\}&=x\sum\_\{n=1\}^\infty n(n-1)x^\{n-2\}+\sum\_\{n=1\}^\infty nx^\{n-1\}\\\\ &=x\sum\_\{n=1\}^\infty (x^n)''+\sum\_\{n=1\}^\infty (x^n)'\\\\ &=\frac\{2x\}\{(1-x)^3\}+\frac\{1\}\{(1-x)^2\}=\frac\{1+x\}\{(1-x)^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)/
---
547、 (0-24)、 求幂级数 $\displaystyle \sum\_\{n=1\}^\infty \frac\{x^\{4n\}\}\{4n+1\}$ 的收敛域及和函数. (吉林师范大学2023年(学科数学)数学分析考研试题) [幂级数 ]
[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) / 已知原幂级数的收敛域为 $\displaystyle [-1,1)$. 当 $\displaystyle x\in [-1,0)\cup (0,1)$ 时,
\begin\{aligned\} \mbox\{原式\}=&\frac\{1\}\{x\}\sum\_\{n=1\}^\infty \frac\{x^\{4n+1\}\}\{4n+1\} =\frac\{1\}\{x\}\sum\_\{n=1\}^\infty \int\_0^x t^\{4n\}\mathrm\{ d\} t =\frac\{1\}\{x\}\int\_0^x \frac\{t^4\}\{1-t^4\}\mathrm\{ d\} t\\\\ =&\frac\{1\}\{x\}\int\_0^x \left\[-1-\frac\{1\}\{4(t-1)\}+\frac\{1\}\{4(t+1)\} +\frac\{1\}\{2(1+t^2)\}\right\]\mathrm\{ d\} t\\\\ =&-x+\frac\{\arctan x\}\{2x\}+\frac\{1\}\{4x\}\ln \frac\{1+x\}\{1-x\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故原式 $\displaystyle =\left\\{\begin\{array\}\{llllllllllll\}-x+\frac\{\arctan x\}\{2x\}+\frac\{1\}\{4x\}\ln \frac\{1+x\}\{1-x\}, &x\in [-1,0)\cup (0,1),\\\\ 0,&x=0.\end\{array\}\right.$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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---
548、 (7)、 已知 $\displaystyle f(x)=\sum\_\{n=0\}^\infty \frac\{x^\{4n\}\}\{(4n)!\}$. 求 $\displaystyle f^\{(4)\}(3)-f(3)$. (暨南大学2023年数学分析考研试题) [幂级数 ]
[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) / $\displaystyle f(x)$ 的收敛半径为 $\displaystyle +\infty$, 而可逐项求导:
\begin\{aligned\} f^\{(4)\}(x)=\sum\_\{n=1\}^\infty \frac\{x^\{4(n-1)\}\}\{\left(4(n-1)\right)!\}=f(x)\Rightarrow \mbox\{原式\}=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)/
---
549、 9、 (20 分) 幂级数 $\displaystyle f(x)=\sum\_\{n=1\}^\infty a\_nx^n$ 在 $\displaystyle [0,1]$ 上收敛, $\displaystyle a\_1\neq 0$. 若 $\displaystyle f(x)$ 满足 $\displaystyle 0\leq f(x)\leq x, x\in [0,1]$, 且 $\displaystyle f(x)-x$ 不恒等于零. (1)、 (5 分) 求证: $\displaystyle 0 < a\_1\leq 1$; (2)、 (10 分) 设 $\displaystyle 0 < x\_0 < 1, x\_n=f(x\_\{n-1\}), n\geq 1$. 求证: $\displaystyle \exists\ \delta > 0$, 当 $\displaystyle x\_0 < \delta$ 时, $\displaystyle \lim\_\{n\to\infty\}x\_n=0$; (3)、 (5 分) 设 $\displaystyle a\_2\neq 0$, 令 $\displaystyle x\_n$ 如上且 $\displaystyle x\_0 < \delta$. 求证: $\displaystyle \sum\_\{n=1\}^\infty x\_n$ 收敛当且仅当 $\displaystyle a\_1 < 1$. (南京大学2023年数学分析考研试题) [幂级数 ]
[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)、 $\displaystyle 0\leq f(x)\leq x$ 中令 $\displaystyle x=0$ 知 $\displaystyle f(0)=0$, 而
\begin\{aligned\} a\_1=f'(0)=\lim\_\{x\to0^+\} \frac\{f(x)\}\{x\}\in [0,1]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由 $\displaystyle a\_1\neq 0$ 知 $\displaystyle 0 < a\_1\leq 1$. (2)、 由 $\displaystyle 0\leq x\_n=f(x\_\{n-1\})\leq x\_\{n-1\}$ 知 $\displaystyle x\_n\searrow, \geq0$. 据单调有界定理知 $\displaystyle \lim\_\{n\to\infty\}x\_n=\ell$ 存在, 且 $\displaystyle \ell=f(\ell)$. (2-1)、 若 $\displaystyle 0 < a\_1=f'(0) < 1$, 则
\begin\{aligned\} &\exists\ \delta > 0,\mathrm\{ s.t.\} \forall\ 0 < x < \delta, |f'(x)| < \frac\{1+a\_1\}\{2\}\\\\ \Rightarrow& |f(x)|=|f(x)-f(0)|\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} |f'(\xi\_x)|\cdot |x| < |x|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故当 $\displaystyle 0 < x\_0 < \delta$ 时, $\displaystyle \ell=0$. (2-2)、 若 $\displaystyle a\_1=1$, 则
\begin\{aligned\} \exists\ 0 < \delta < 1,\mathrm\{ s.t.\} \forall\ 0 < x < \delta, f(x) < x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
从而 $\displaystyle \ell=f(\ell)\Rightarrow \ell=0$. 用反证法证明:
\begin\{aligned\} &\forall\ 0 < \delta < 1, \exists\ 0 < x\_\delta < \delta,\mathrm\{ s.t.\} f(x\_\delta)=x\_\delta \stackrel\{f(x\_\delta)\leq x\_\delta\}\{\Rightarrow\} f(x\_\delta)=x\_\delta\\\\ \Rightarrow&\exists\ x\_k\mbox\{严\}\searrow 0,\mathrm\{ s.t.\} f(x\_k)=x\_k \stackrel\{g(x)=f(x)-x\}\{\Rightarrow\} g(x\_k)=0\\\\ \stackrel\{\mbox\{Rolle\}\}\{\Longrightarrow\}&\exists\ y\_k\mbox\{严\}\searrow 0, \mathrm\{ s.t.\} g'(y\_k)=0 \stackrel\{\mbox\{Rolle\}\}\{\Longrightarrow\} \exists\ z\_k\mbox\{严\}\searrow 0, \mathrm\{ s.t.\} g''(z\_k)=0\\\\ \Rightarrow& \cdots \Rightarrow g^\{(k)\}(0)=0, \forall\ k\geq 0 \Rightarrow f(x)\equiv x,\mbox\{与题设矛盾, 故有结论\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、 (3-1)、 $\displaystyle \Leftarrow$: 由
\begin\{aligned\} x\_n=&|x\_n-0|=|f(x\_\{n-1\})-f(0)| \xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} |f'(\xi\_n)|\cdot x\_\{n-1\}\\\\ \leq&\frac\{1+a\_1\}\{2\}x\_\{n-1\} \leq \cdots \leq \left(\frac\{1+a\_1\}\{2\}\right)^n x\_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 $\displaystyle \sum\_\{n=1\}^\infty x\_n$ 收敛. (3-2)、 $\displaystyle \Leftarrow$: 用反证法. 若 $\displaystyle a\_1=1$, 则
\begin\{aligned\} &x+a\_2x+\cdots=f(x)\leq x \Rightarrow 1+a\_2+a\_3x+\cdots \leq 1\\\\ \stackrel\{x\to 0\}\{\Rightarrow\}&a\_2\leq 0\stackrel\{a\_2\neq 0\}\{\Rightarrow\}a\_2 < 0\\\\ \Rightarrow& f(x)=x+a\_2x^2+\cdots \geq x+\frac\{3a\_2\}\{2\}x^2, x\in (0,\delta). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
若 $\displaystyle x\_n\geq \frac\{c\}\{n\}$, 则由 $\displaystyle f\mbox\{严\}\nearrow , x\in (0,\delta)$ 知
\begin\{aligned\} x\_\{n+1\}=f(x\_n)\geq \frac\{c\}\{n\}+\frac\{3a\_2\}\{2n^2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令它 $\displaystyle \geq \frac\{c\}\{n+1\}$, 则等价于
\begin\{aligned\} \frac\{c\}\{n(n+1)\}\geq \frac\{-3a\_2\}\{2n^2\}\Leftrightarrow \frac\{c\}\{-3a\_2\}\geq \frac\{n+1\}\{2n\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故令 $\displaystyle c=-3a\_2$ 即可由数学归纳法推得 $\displaystyle x\_n\geq \frac\{c\}\{n\}$. 而 $\displaystyle \sum\_\{n=1\}^\infty x\_n$ 发散. 与必要性假设矛盾. 故有结论.跟锦数学微信公众号. [在线资料](
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---
550、 6、 若数列 $\displaystyle \left\\{a\_n\right\\}$ 满足
\begin\{aligned\} a\_1=1, \quad (n+1)a\_\{n+1\}=\left(n+\frac\{1\}\{2\}\right)a\_n, n\geq 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
求证: 当 $\displaystyle |x| < 1$ 时, 幂级数 $\displaystyle \sum\_\{n=1\}^\infty a\_nx^n$ 收敛, 并求其和函数. (南京师范大学2023年数学分析考研试题) [幂级数 ]
[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\} \lim\_\{n\to\infty\}\frac\{a\_\{n+1\}\}\{a\_n\}=\lim\_\{n\to\infty\}\frac\{n+\frac\{1\}\{2\}\}\{n+1\}=1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知幂级数 $\displaystyle \sum\_\{n=1\}^\infty a\_nx^n$ 的收敛半径为 $\displaystyle 1$, 而当 $\displaystyle |x| < 1$ 时收敛. 设 $\displaystyle f(x)=\sum\_\{n=1\}^\infty a\_nx^n$, 则
\begin\{aligned\} &f'(x)=\sum\_\{n=1\}^\infty na\_nx^\{n-1\} =a\_1+\sum\_\{n=2\}^\infty \left(n-\frac\{1\}\{2\}\right)a\_\{n-1\}x^\{n-1\}\\\\ =&a\_1+x\sum\_\{n=2\}^\infty (n-1)a\_\{n-1\}x^\{n-2\} +\frac\{1\}\{2\}\sum\_\{n=2\}^\infty a\_\{n-1\}x^\{n-1\} =1+xf'(x)+\frac\{1\}\{2\}f(x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
我们得到常微分方程
\begin\{aligned\} f'(x)+\frac\{1\}\{2(x-1)\}f(x)+\frac\{1\}\{x-1\}=0.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
先求齐次方程
\begin\{aligned\} &\frac\{\mathrm\{ d\} y\}\{\mathrm\{ d\} x\}+\frac\{1\}\{2(x-1)\}y=0\Leftrightarrow \frac\{\mathrm\{ d\} y\}\{y\}+\frac\{1\}\{2(x-1)\}\mathrm\{ d\} x=0\\\\ \Leftrightarrow& \ln |y|+\frac\{1\}\{2\}\ln (1-x)=C\_1\Leftrightarrow y=\frac\{C\}\{\sqrt\{1-x\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由常数变易法知可设 $\displaystyle f(x)=\frac\{C(x)\}\{\sqrt\{1-x\}\}$. 代入 $\displaystyle (I)$ 得
\begin\{aligned\} &\frac\{C'(x)\}\{\sqrt\{1-x\}\}+\frac\{1\}\{x-1\}=0\Leftrightarrow C'(x)=\frac\{1\}\{\sqrt\{1-x\}\} \Leftrightarrow C(x)=-2\sqrt\{1-x\}+C\\\\ \Rightarrow& f(x)=\frac\{C\}\{\sqrt\{1-x\}\}-2 \stackrel\{f'(0)=a\_1=1\}\{\Rightarrow\}\frac\{C\}\{2\}=1\Rightarrow C=2\Rightarrow f(x)=\frac\{2\}\{\sqrt\{1-x\}\}-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)/
---
551、 7、 求幂级数 $\displaystyle \sum\_\{n=0\}^\infty \frac\{1\}\{2n+1\}\left(\frac\{x\}\{1+x\}\right)^\{2n+1\}$ 的收敛域与和函数. (太原理工大学2023年数学分析考研试题) [幂级数 ]
[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 $\displaystyle \sum\_\{n=0\}^\infty \frac\{1\}\{2n+1\}t^\{2n+1\}$ 的收敛域为 $\displaystyle [-1,1)$, 且
\begin\{aligned\} \sum\_\{n=0\}^\infty \frac\{1\}\{2n+1\}t^\{2n+1\} =\int\_0^t \sum\_\{n=0\}^\infty s^\{2n\}\mathrm\{ d\} s =\int\_0^t \frac\{\mathrm\{ d\} s\}\{1-s^2\}=\frac\{1\}\{2\}\ln\frac\{1+t\}\{1-t\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \sum\_\{n=0\}^\infty \frac\{1\}\{2n+1\}\left(\frac\{x\}\{1+x\}\right)^\{2n+1\}$ 的收敛域为 $\displaystyle \left\[-\frac\{1\}\{2\},+\infty\right)$, 和函数为 $\displaystyle \frac\{\ln(1+2x)\}\{2\}$.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](
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---
552、 4、 设 $\displaystyle f(x)$ 是仅有正实根的多项式, 且 $\displaystyle \frac\{f'(x)\}\{f(x)\}=-\sum\_\{n=0\}^\infty c\_nx^n$. 5、 证明 $\displaystyle c\_n > 0\ (n > 0)$; 6、 证明极限 $\displaystyle \lim\_\{n\to\infty\}\frac\{1\}\{\sqrt[n]\{c\_n\}\}$ 存在, 且为 $\displaystyle f(x)$ 的根的最小值. (武汉大学2023年数学分析考研试题) [幂级数 ]
[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
\begin\{aligned\} f(x)=A(x-a\_1)^\{r\_1\}\cdots (x-a\_k)^\{r\_k\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle 0 < a\_1 < \cdots < a\_k$, $\displaystyle r\_i\geq 1$, 则
\begin\{aligned\} f'(x)&=Ar\_1(x-a\_1)^\{r\_1-1\}\cdots (x-a\_k)^\{r\_k\}\\\\ &\quad +Ar\_k(x-a\_1)^\{r\_1\}\cdots (x-a\_k)^\{r\_k-1\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而
\begin\{aligned\} \frac\{f'(x)\}\{f(x)\} &=\sum\_\{i=1\}^k \frac\{r\_i\}\{x-a\_i\} =-\sum\_\{i=1\}^k \frac\{r\_i\}\{a\_i\}\frac\{1\}\{1-\frac\{x\}\{a\_i\}\}\\\\ &=-\sum\_\{i=1\}^k \frac\{r\_i\}\{a\_i\} \sum\_\{n=0\}^\infty \left(\frac\{x\}\{a\_i\}\right)^n =-\sum\_\{n=0\}^\infty \left(\sum\_\{i=1\}^k \frac\{r\_i\}\{a\_i^\{n+1\}\}\right)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
\begin\{aligned\} c\_n&=\sum\_\{i=1\}^k \frac\{r\_i\}\{a\_i^\{n+1\}\} > 0,\\\\ \lim\_\{n\to\infty\}\frac\{1\}\{\sqrt[n]\{c\_n\}\}&=\mathrm\{exp\}\left\[\lim\_\{n\to\infty\}\frac\{1\}\{n\}\ln \frac\{1\}\{c\_n\}\right\] =\mathrm\{exp\}\left\[\lim\_\{n\to\infty\}\frac\{\ln c\_1+\ln \frac\{c\_1\}\{c\_2\}+\cdots+\ln \frac\{c\_\{n-1\}\}\{c\_n\}\}\{n\}\right\]\\\\ &\xlongequal\{\tiny\mbox\{Stolz\}\} \mathrm\{exp\}\left\[\lim\_\{n\to\infty\}\ln \frac\{c\_\{n-1\}\}\{c\_n\}\right\]=\lim\_\{n\to\infty\}\frac\{c\_n\}\{c\_\{n+1\}\} =\lim\_\{n\to\infty\}\frac\{\frac\{r\_1\}\{a\_1^\{n+1\}\}+\cdots+\frac\{r\_k\}\{a\_k^\{n+1\}\}\}\{ \frac\{r\_1\}\{a\_1^\{n+2\}\} +\cdots+\frac\{r\_k\}\{a\_k^\{n+2\}\} \}\\\\ &=\lim\_\{n\to\infty\} \frac\{r\_1a\_1+\cdots+r\_k\left(\frac\{a\_1\}\{a\_k\}\right)^\{n+1\}a\_1\}\{ r\_1+\cdots+r\_k\left(\frac\{a\_1\}\{a\_k\}\right)^\{n+2\} \}\left(\mbox\{分子分母同时除以 $\displaystyle a\_1^\{n+2\}$\}\right)\\\\ &=\frac\{r\_1a\_1+0+\cdots+0\}\{r\_1+0+\cdots+0\} =a\_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)/
发表于 2023-3-5 08:53:09
