Section01_常数项级数

定义

1. $\{a_{n}\}$ 为常数项数列，$\displaystyle \sum_{n=1}^{\infty}a_{n}$ 成为常数项级数
2. $S_{n} = a_{1} + a_{2} +\cdots + a_{n}$ 称为部分和

Notes

1. $S_{n}$ 与 $\displaystyle \sum_{n=1}^{\infty}a_{n}$ 不同
2. $\displaystyle \lim_{n\to \infty}S_{n}$ 与 $\displaystyle \sum_{n=1}^{\infty}a_{n}$ 相同

1. 收敛与发散

   1. 若 $\displaystyle \lim_{n\to \infty}S_{n} = S$，则称 $\displaystyle \sum_{n=1}^{\infty}a_{n}$ 收敛于 $S$
   2. 若 $\displaystyle \lim_{n\to \infty}S_{n} \ \nexists$，则称 $\displaystyle \sum_{n=1}^{\infty}a_{n}$ 发散

2. 例1 $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n(n+1)}$ $$\begin{array}{ll} \because & \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \\ \therefore & S_{n} = \frac{1}{1\times 2} + \frac{1}{2\times 3} + \cdots + \frac{1}{n(n+1)} \\ & = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \cdots + \frac{1}{n} - \frac{1}{n+1} \\ & = 1 - \frac{1}{n+1}\\ \therefore &\displaystyle \lim_{n\to \infty}S_{n} = 1 \\ \therefore & \displaystyle \sum_{n=1}^{\infty}\frac{1}{n(n+1)} = 1 \\ \end{array}$$

性质

1. $\displaystyle \sum_{n=1}^{\infty}a_{n}= A, \sum_{n=1}^{\infty}b_{n} = B$，则 $\displaystyle \sum_{n=1}^{\infty}(a_{n}\pm b_{n}) = A\pm B$

2. $\displaystyle \sum_{n=1}^{\infty}a_{n} = S$，则 $\displaystyle \sum_{n=1}^{\infty}k a_{n} = kS$

   Notes

   • 若$k \ne 0$，则$\displaystyle \sum_{n=1}^{\infty}ka_{n}$ 和 $\displaystyle \sum_{n=1}^{\infty}a_{n}$ 的敛散性相同

3. 级数加减，改变有限项，敛散性不变

   • 如 $\displaystyle \sum_{n=1}^{\infty}\frac{1}{2^{n}} = \frac{1}{2} + \frac{1}{2^{2}} +\cdots = 1$
     • Case 1: $2 + 1 + \frac{1}{2} + \frac{1}{2^{2}} + \cdots = 3 + 1 = 4$
     • Case 2: $\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots = 1 - \frac{1}{2} = \frac{1}{2}$
     • Case 3: $2 + \frac{1}{2^{2}} + \frac{1}{2^{3}} + \cdots = 2-\frac{1}{2}+1 = \frac{5}{2}$
4. 级数内添括号，提升收敛性
   1. $\displaystyle \sum_{n=1}^{\infty}a_{n}$ 收敛 => 添加括号后收敛
   2. 添加括号后发散 => $\displaystyle \sum_{n=1}^{\infty}a_{n}$ 发散
   3. 例题 设 $\displaystyle \sum_{n=1}^{\infty}a_{n}$ 收敛，$\displaystyle \sum_{n=1}^{\infty}(a_{2n-1}+a_{2n})$ 的敛散性如何？ $$\begin{array}{ll} \because & \displaystyle \sum_{n=1}^{\infty}(a_{2n-1}+a_{2n}) = (a_{1}+a_{2}) + (a_{3}+a_{4}) + \cdots \\ \therefore & \displaystyle \sum_{n=1}^{\infty}(a_{2n-1}+a_{2n}) \text{收敛}\\ \end{array}$$
5. (必要条件) 若 $\displaystyle \sum_{n=1}^{\infty}a_{n}$ 收敛 $\nLeftarrow\Rightarrow$ $\displaystyle \lim_{n\to \infty}a_{n} = 0$

   Proof

   $\Rightarrow$ $$\begin{array}{ll} & S_{n} = a_{1} + a_{2} + \cdots + a_{n} \\ \because & \displaystyle \sum_{n=1}^{\infty}a_{n} \text{收敛} \\ \therefore & \displaystyle \lim_{n\to \infty}S_{n}\ \exists \text{, 令}\lim_{n\to \infty} = S \\ & \text{而} a_{n} = S_{n} - S_{n-1} \\ \therefore & \displaystyle \lim_{n\to \infty}a_{n} = \lim_{n\to \infty}S_{n} - \lim_{n\to \infty}S_{n-1} = S-S = 0 \\ \end{array}$$ $\nLeftarrow$: 反例 $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}$

Notes 若 $\displaystyle \sum_{n=1}^{\infty}a_{n}$ 收敛 $\Rightarrow \begin{cases} \displaystyle \lim_{n\to \infty} S_{n} \ \exists \\ \displaystyle \lim_{n\to \infty} a_{n} = 0 \end{cases}$

两个重要级数

$P$-级数

$$\sum_{n=1}^{\infty}\frac{1}{n^{p}} \begin{cases} \text{收敛}& p >1 \\ \text{发散}& p <1 \\ \text{发散} & p = 1\\ \end{cases}$$

• 其中 $p=1$ 时，$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}$ 被称为调和级数，发散

几何级数

$$\sum_{n=1}^{\infty}a\cdot q^{n}\quad (a\ne 0)$$

1. 若 $\vert q \vert \ge 1$，则 $\displaystyle \sum_{n=1}^{\infty}a\cdot q^{n}$ 发散
2. 若 $\vert q \vert < 1$，则 $\displaystyle \sum_{n=1}^{\infty}a\cdot q^{n} = \frac{\text{首项}}{1-q}$

   Proof $$\begin{array}{ll} & a_{n} = aq^{n} \\ & S_{n} = aq + aq^{2} + \cdots +aq^{n} \\ & qS_{n} = aq^{2} + aq^{3} + \cdots + aq^{n+1} \\ \therefore & (1-q)S_{n} = aq - aq^{n+1}\\ & S_{n} = \frac{aq - aq^{n+1}}{1-q} \\ \because & q<1 \\ \therefore & \displaystyle \lim_{n\to \infty}S_{n} = \frac{aq}{1-q} = \sum_{n=1}^{\infty}a\cdot q^{n} \\ \end{array}$$

第一类-正项级数

定义

• $\displaystyle \sum_{n=1}^{\infty}a_{n}\quad(a_{n}\ge 0)$ 为正项级数

Notes

1. $S_{1}\le S_{2} \le S_{3}\le\cdots$，即 $\{S_{n}\}$ 单调递增
2. $\begin{cases} \{S_{n}\} \text{无上界}&\Rightarrow& \lim\limits_{n\to \infty}S_{n} = +\infty \text{, 即发散} \\ S_{n} \le M &\Rightarrow& \lim\limits_{n\to \infty}S_{n}\ \exists \text{, 即收敛} \end{cases}$

判别法

法一: 比较法

Th1 $a_{n}\ge 0,b_{n}\ge 0$

1. $a_{n}\le b_{n}$ 且 $\displaystyle \sum_{n=1}^{\infty}b_{n}$ 收敛 $\Rightarrow$ $\displaystyle \sum_{n=1}^{\infty}a_{n}$ 收敛
2. $a_{n}\ge b_{n}$ 且 $\displaystyle \sum_{n=1}^{\infty}b_{n}$ 发散 $\Rightarrow$ $\displaystyle \sum_{n=1}^{\infty}a_{n}$ 发散

1. $\displaystyle \sum_{n=1}^{\infty}\sin \frac{\pi}{2^{n}}$ $$\begin{array}{ll} \because & \sin x \le x\\ \therefore & 0\le\sin \frac{\pi}{2^{n}} \le \frac{\pi}{2^{n}} \\ \because & \displaystyle \sum_{n=1}^{\infty}\frac{\pi}{2^{n}} \text{收敛} \\ \therefore & \displaystyle \sum_{n=1}^{\infty}\sin \frac{\pi}{2^{n}} \text{收敛} \\ \end{array}$$
2. $\displaystyle \sum_{n=1}^{\infty}\frac{1}{\int_{0}^{n}\sqrt[4]{1+x^{4}}\cdot dx}$ $$\begin{array}{ll} \because & \int_{0}^{n}\sqrt[4]{1+x^{4}}\cdot dx > \int_{0}^{n}x\cdot dx = \frac{1}{2}n^{2} \ge 0\\ \therefore & 0 \le \frac{1}{\int_{0}^{n}\sqrt[4]{1+x^{4}}\cdot dx} < \frac{1}{\frac{1}{2}n^{2}}\\ \because & \displaystyle \sum_{n=1}^{\infty}\frac{2}{n^{2}} \text{收敛} \\ \therefore & \displaystyle \sum_{n=1}^{\infty}\frac{1}{\int_{0}^{n}\sqrt[4]{1+x^{4}}\cdot dx} \text{收敛} \\ \end{array}$$
3. $a_{n}\le b_{n}\le c_{n}$，$\displaystyle \sum_{n=1}^{\infty}a_{n}, \sum_{n=1}^{\infty}c_{n}$ 收敛，证 $\displaystyle \sum_{n=1}^{\infty}b_{n}$ 收敛 $$\begin{array}{ll} \because & a_{n} \le b_{n} \le c_{n} \\ \therefore & 0 \le b_{n} - a_{n} \le c_{n} - a_{n} \\ \because & \sum_{n=1}^{\infty}a_{n}, \sum_{n=1}^{\infty}c_{n} \text{收敛} \\ \therefore & \sum_{n=1}^{\infty}(c_{n} - a_{n}) \text{收敛} \\ \therefore & \sum_{n=1}^{\infty}(b_{1}- c_{n}) \text{收敛} \\ \because & \sum_{n=1}^{\infty}c_{n} \text{收敛} \\ \therefore & \sum_{n=1}^{\infty}b_{n} \text{收敛} \\ \end{array}$$

Th1' $a_{n}>0,b_{n}>0$

• 若 $\displaystyle \lim_{n\to \infty}\frac{b_{n}}{a_{n}} = \ell\quad (0<\ell<+\infty)\Rightarrow \sum_{n=1}^{\infty}a_{n}, \sum_{n=1}^{\infty}b_{n}$敛散性相同

1. $\displaystyle \sum_{n=1}^{\infty}\bigg[\frac{1}{n} - \ln\bigg(1+\frac{1}{n}\bigg)\bigg]$ 的敛散性 $$\begin{array}{ll} \because & x>0 \text{时，} x > \ln(1+x) \\ \therefore & 0 < \frac{1}{n} - \ln(1+\frac{1}{n}) \\ \because & \lim\limits_{x\to 0} \frac{x - \ln(1+x)}{x^{2}} = \lim\limits_{x\to 0} \frac{1 - \frac{1}{1+x}}{2x} = \frac{1}{2} \\ \therefore & \lim\limits_{n\to \infty} \frac{\frac{1}{n} - \ln(1+\frac{1}{n})}{\frac{1}{n^{2}}} = \frac{1}{2} \\ \therefore & \sum_{n=1}^{\infty}[\frac{1}{n} - \ln(1+\frac{1}{n})] \sim \sum_{n=1}^{\infty}\frac{1}{n^{2}}\\ \because & \sum_{n=1}^{\infty}\frac{1}{n^{2}} \text{收敛} \\ \therefore & \sum_{n=1}^{\infty}[\frac{1}{n}- \ln(1+ \frac{1}{n})] \text{收敛} \\ \end{array}$$

法二: 比值法

Th 2 $\displaystyle a_{n} > 0, \lim_{n\to \infty}\frac{a_{n+1}}{a_{n}} = \rho$

$$\lim_{n\to \infty}\frac{a_{n+1}}{a_{n}} = \rho \begin{cases} \text{收敛} & \rho < 1 \\ \text{发散} & \rho > 1 \\ \text{?} & \rho = 1 \\ \end{cases}$$

Notes 适合$\color{#D0104C}{a_{n}}$中含有阶乘的情形

1. $\displaystyle \sum_{n=1}^{\infty}\frac{2^{n}\cdot n!}{n^{n}}$ $$\begin{array}{ll} \because & \lim\limits_{n\to \infty}\frac{a_{n+1}}{a_{n}} = \lim\limits_{n\to \infty}\frac{2^{n+1}(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^{n}}{2^{n}\cdot n!} = \lim\limits_{n\to \infty}2 (\frac{n}{n+1})^{n}\\ & = 2\lim\limits_{n\to \infty} \frac{1}{(1+\frac{1}{n})^{n}} = \frac{2}{e} < 1\\ \therefore & \sum_{n=1}^{\infty}\frac{2^{n}\cdot n!}{n^{n}} \text{收敛} \\ \end{array}$$

法三: 根值法

Th 3 $a_{n}>0$

$$\lim_{n\to \infty}\sqrt[n]{a_{n}} = \rho \begin{cases} \text{收敛} & \rho <1 \\ \text{发散} & \rho >1 \\ \text{?} & \rho =1 \end{cases}$$

1. $\displaystyle \sum_{n=1}^{\infty}\bigg(\frac{n}{2n+1}\bigg)^{n}$ $$\begin{array}{ll} \because & \lim\limits_{n\to \infty} \sqrt[n]{a_{n}} = \lim\limits_{n\to \infty} \frac{n}{2n+1} = \frac{1}{2} < 1 \\ \therefore & \sum_{n=1}^{\infty}(\frac{n}{2n+1})^{n} \text{收敛} \\ \end{array}$$

第二类-交错级数

定义

$$\begin{array}{c} \displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}a_{n} = a_{1} - a_{2} + a_{3} - a_{4} + \cdots\\ \text{OR}\\ \displaystyle \sum_{n=1}^{\infty}(-1)^{n}a_{n} = -a_{1} + a_{2} - a_{3} + a_{4} - \cdots \end{array}\quad (a_{n} >0)$$

• 称为交错级数

判别法

Th 对于$\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}a_{n}\quad (a_{n}>0)$，若

1. $\{a_{n}\}$ 单调递减
2. $\lim\limits_{n\to \infty} a_{n} = 0$

则$\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}a_{n}$ 收敛

Notes

1. $\displaystyle \sum_{n=1}^{\infty}a_{n}^{}$ 收敛 $\nRightarrow$ $\displaystyle \sum_{n=1}^{\infty}a_{n}^{2}$ 收敛

   反例 $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}}{\sqrt{n}}$

2. $\displaystyle \sum_{n=1}^{\infty}a_{n}^{} \quad (a_{n}\ge 0)$ 收敛 $\Rightarrow$ $\displaystyle \sum_{n=1}^{\infty}a_{n}^{2}$ 收敛

   Proof $$\begin{array}{ll} & \text{取}\ \epsilon = 1 > 0, \exists\ N > 0, \text{当}n>N \text{时}, \vert a_{n} - 0 \vert<1 \\ \Rightarrow & 0 \le a_{n}^{2} \le a_{n} < 1 \\ \because & \sum_{n=1}^{\infty}a_{n} \text{收敛} \\ \therefore & \sum_{n=1}^{\infty}a^{2}_{n} \text{收敛} \\ \end{array}$$

1. $\displaystyle \sum_{n=1}^{\infty}\sin \sqrt{n^{2} + 1}\pi$ $$\begin{array}{ll} & \sin \sqrt{n^{2} + 1} \pi = \sin [n\pi + (\sqrt{n^{2}+ 1} -n)\pi] \\ & = (-1)^{n}\sin \frac{\pi}{\sqrt{n^{2}+1} + n} \\ \because & \sin \frac{\pi}{\sqrt{n^{2} + 1} + n} > 0 \\ \therefore & \sum_{n=1}^{\infty}\sin \sqrt{n^{2} + 1} \pi \text{为交错级数} \\ \because & \sin \frac{\pi}{\sqrt{1+n^{2}}+ n} \text{单调递减且} \lim\limits_{n\to \infty} \sin \frac{\pi}{\sqrt{1+n^{2}}+ n} = 0\\ \therefore & \sum_{n=1}^{\infty}\sin \sqrt{n^{2} + 1} \pi \text{收敛} \\ \end{array}$$

任意级数-条件收敛与绝对收敛

定义

1. 若 $\displaystyle \sum_{n=1}^{\infty}a_{n}$ 收敛，但 $\displaystyle \sum_{n=1}^{\infty}\vert a_{n} \vert$ 发散，称 $\displaystyle \sum_{n=1}^{\infty}a_{n}$ 条件收敛
2. 若 $\displaystyle \sum_{n=1}^{\infty}a_{n}$ 收敛，且 $\displaystyle \sum_{n=1}^{\infty}\vert a_{n} \vert$ 收敛，称 $\displaystyle \sum_{n=1}^{\infty}a_{n}$ 绝对收敛

关系

Th

• $\displaystyle \sum_{n=1}^{\infty}a_{n}$ 绝对收敛 $\nLeftarrow \Rightarrow$ $\displaystyle \sum_{n=1}^{\infty}a_{n}$ 收敛

1. $\displaystyle \sum_{n=1}^{\infty}\sin \sqrt{n^{2}+1}\pi$ 是绝对收敛还是条件收敛？ $$\begin{array}{ll} & \sin \sqrt{n^{2} + 1}\pi = (-1)^{n} \sin \frac{\pi}{\sqrt{n^{2}+1}+n} \\ \because & \sum_{n=1}^{\infty}\sin \sqrt{n^{2} + 1}\pi \text{为交错级数} \\ & \lim\limits_{n\to \infty} \sin \frac{\pi}{\sqrt{n^{2} + 1} + n} = 0 \text{且} \sin \frac{\pi}{\sqrt{n^{2}+ 1}+n} \text{单调递减} \\ \therefore & \sum_{n=1}^{\infty}\sin \sqrt{n^{2}+ 1} \pi \text{收敛} \\ \because & \vert a_{n} \vert = \sin\frac{\pi}{\sqrt{n^{2} +1}+n} \le \frac{\pi}{\sqrt{n^{2} + 1} + n} \le \frac{\pi}{2n} \\ \because & \sum_{n=1}^{\infty}\frac{\pi}{2n} \text{发散} \\ \therefore & \sum_{n=1}^{\infty}\vert a_{n} \vert \text{发散}\\ & \sum_{n=1}^{\infty}\sin \frac{\pi}{\sqrt{n^{2}+1}+n} \text{条件收敛} \end{array}$$