【考研数学一·高数(8)】无穷级数

1.数项级数的判敛

1.1.定义

1. 无穷级数：$\sum _{n=1}^{\infty }{u}_n=u_1+u_2+\cdots +u_n+\cdots$

2. 级数的前$n$项和：$S_n=u_1+u_2+\cdots +u_n$

3. $\sum _{n=1}^{\infty }{u}_n$收敛$\iff \underset{n\to \infty }{\lim }{S}_n$存在

4. $\sum _{n=1}^{\infty }{u}_n$收敛$\begin{array}{c}\Rightarrow \\ \nLeftarrow \end{array}\underset{n\to \infty }{\lim }{u}_n=0$

5. $\sum _{n=1}^{\infty }(u_{n+1}-u_n)$收敛$\iff \underset{n\to \infty }{\lim }{u}_n$存在

   • 证：$S_n=(u_2-u_1)+(u_3-u_2)+\cdots +(u_{n+1}-u_n)=u_{n+1}-u_1$

     故$\sum _{n=1}^{\infty }(u_{n+1}-u_n)$收敛$\iff \underset{n\to \infty }{\lim }{S}_n$存在$\iff \underset{n\to \infty }{\lim }{u}_n$存在

1.2.判敛法

1.2.1.正项级数$\sum _{n=1}^{\infty }{u}_n,u_n⩾0$

1. 基本定理
   • $\sum _{n=1}^{\infty }{u}_n$收敛 ⇔ { S n } \Leftrightarrow\{S_n\} ⇔{Sn​}有界
2. 比较判别法
   • 正项级数$\sum _{n=1}^{\infty }{u}_n$和$\sum _{n=1}^{\infty }{v}_n$，从某项起有$u_n⩽v_n$，则$\{\begin{array}{l}\sum _{n=1}^{\infty }{v}_n收敛\Rightarrow \sum _{n=1}^{\infty }{u}_n收敛\\ \sum _{n=1}^{\infty }{u}_n发散\Rightarrow \sum _{n=1}^{\infty }{v}_n发散\end{array}$
3. 比较判别法的极限形式
   • $\underset{n\to \infty }{\lim }\frac{u_n}{v_n}=\{\begin{array}{l}0\Rightarrow \{\begin{array}{l}\sum _{n=1}^{\infty }{v}_n收敛\Rightarrow \sum _{n=1}^{\infty }{u}_n收敛\\ \sum _{n=1}^{\infty }{u}_n发散\Rightarrow \sum _{n=1}^{\infty }{v}_n发散\end{array}\\ +\infty \Rightarrow \{\begin{array}{l}\sum _{n=1}^{\infty }{u}_n收敛\Rightarrow \sum _{n=1}^{\infty }{v}_n收敛\\ \sum _{n=1}^{\infty }{v}_n发散\Rightarrow \sum _{n=1}^{\infty }{u}_n发散\end{array}\\ A\ne 0\Rightarrow \sum _{n=1}^{\infty }{u}_n与\sum _{n=1}^{\infty }{v}_n同敛散\end{array}$
4. 比值判别法（达朗贝尔）
   • $\underset{n\to \infty }{\lim }\frac{u_{n+1}}{u_n}=\rho \{\begin{array}{l}<1,收敛\\ >1,发散\\ =1,失效\end{array}$
5. 根值判别法（柯西）
   • $\underset{n\to \infty }{\lim }\sqrt[n]{u_n}=\rho \{\begin{array}{l}<1,收敛\\ >1,发散\\ =1,失效\end{array}$
6. 积分判别法（柯西）
   • 设$\sum _{n=1}^{\infty }{u}_n$为正项级数，若存在$[1,+\infty )$上单调减少的非负连续函数$f(x)$，使得$u_n=f(n)$，则级数$\sum _{n=1}^{\infty }{u}_n$与反常积分${\int }_1^{+\infty }f(x)\mathrm{d}x$的敛散性相同.
7. 对数判别法
   • $\underset{n\to \infty }{\lim }\frac{\ln \frac{1}{u_n}}{\ln n}=\rho \{\begin{array}{l}>1,收敛\\ <1,发散\\ =1,失效\end{array}适用:u_n=\frac{1}{{[f(n)]}^{g(n)}}$

1.2.2.交错级数$\sum _{n=1}^{\infty }(-1{)}^{n-1}{u}_n,u_n>0$

• 莱布尼茨判别法

  1. $\underset{n\to \infty }{\lim }{u}_n=0$
  2. $u_n⩾u_{n+1}$

• 莱布尼茨判别法只是充分条件，不是必要条件，若失效，可以将级数拆成多个级数之和.

1.2.3.任意项级数

• $\sum _{n=1}^{\infty }|u_n|$收敛，称$\sum _{n=1}^{\infty }{u}_n$绝对收敛
• $\sum _{n=1}^{\infty }|u_n|$发散，$\sum _{n=1}^{\infty }{u}_n$收敛，则称$\sum _{n=1}^{\infty }{u}_n$条件收敛

1. $\sum _{n=1}^{\infty }|u_n|$收敛$\Rightarrow \sum _{n=1}^{\infty }{u}_n$收敛
2. $\sum _{n=1}^{\infty }{u}_n,\sum _{n=1}^{\infty }{v}_n$均绝对收敛$\Rightarrow \sum _{n=1}^{\infty }(u_n\pm v_n)$绝对收敛
3. $\sum _{n=1}^{\infty }{u}_n$绝对收敛，$\sum _{n=1}^{\infty }{v}_n$条件收敛$\Rightarrow \sum _{n=1}^{\infty }(u_n\pm v_n)$条件收敛
4. $\sum _{n=1}^{\infty }{u}_n,\sum _{n=1}^{\infty }{v}_n$均条件收敛$\Rightarrow \sum _{n=1}^{\infty }(u_n\pm v_n)$收敛（可能绝对收敛，可能条件收敛）
5. 交错$p$级数$\sum _{n=1}^{\infty }(-1{)}^{n-1}\frac{1}{n^p}\{\begin{array}{l}p>1,绝对收敛\\ 0<p⩽1,条件收敛\end{array}$

1.3.常用结论

1. $\sum _{n=1}^{\infty }{u}_n$收敛$\Rightarrow \sum _{n=1}^{\infty }|u_n|$不定

2. $\sum _{n=1}^{\infty }{u}_n$收敛$\{\begin{array}{l}{u}_n⩾0\Rightarrow \sum _{n=1}^{\infty }{u}_n^2收敛(\underset{n\to \infty }{\lim }{u}_n=0,从某项起,u_n<1,u_n^2<u_n)\\ u_n任意\Rightarrow \sum _{n=1}^{\infty }{u}_n^2不定\end{array}$

3. $\sum _{n=1}^{\infty }{u}_n$收敛$\{\begin{array}{l}{u}_n⩾0\Rightarrow \sum _{n=1}^{\infty }{u}_n{u}_{n+1}收敛(u_n{u}_{n+1}⩽\frac{u_n^2+u_{n+1}^2}{2})\\ u_n任意\Rightarrow \sum _{n=1}^{\infty }{u}_n{u}_{n+1}不定\end{array}$

4. $\sum _{n=1}^{\infty }{u}_n$收敛$\Rightarrow \sum _{n=1}^{\infty }(-1{)}^n{u}_n$不定

5. $\sum _{n=1}^{\infty }{u}_n$收敛$\Rightarrow \sum _{n=1}^{\infty }(-1{)}^n\frac{u_n}{n}$不定

6. $\sum _{n=1}^{\infty }{u}_n$收敛$\{\begin{array}{l}{u}_n⩾0\Rightarrow \sum _{n=1}^{\infty }{u}_{2n},\sum _{n=1}^{\infty }{u}_{2n-1}收敛\\ u_n任意\Rightarrow \sum _{n=1}^{\infty }{u}_{2n},\sum _{n=1}^{\infty }{u}_{2n-1}不定\end{array}$

7. $\sum _{n=1}^{\infty }{u}_n$收敛$\Rightarrow \sum _{n=1}^{\infty }(u_{2n-1}+u_{2n})$收敛

   $\sum _{n=1}^{\infty }(u_{2n-1}+u_{2n})$收敛且$\underset{n\to \infty }{\lim }{u}_n=0\Rightarrow \sum _{n=1}^{\infty }{u}_n$收敛

8. $\sum _{n=1}^{\infty }{u}_n$收敛$\Rightarrow \sum _{n=1}^{\infty }(u_{2n-1}-u_{2n})$不定

9. $\sum _{n=1}^{\infty }{u}_n$收敛$\Rightarrow \sum _{n=1}^{\infty }(u_n\pm u_{n+1})$收敛，$\sum _{n=1}^{\infty }{u}_n\pm \sum _{n=1}^{\infty }{u}_{n+1}$收敛

10. $\sum _{n=1}^{\infty }|u_n|$收敛$\Rightarrow \sum _{n=1}^{\infty }{u}_n$收敛

    $\sum _{n=1}^{\infty }{u}_n$发散$\Rightarrow \sum _{n=1}^{\infty }|u_n|$发散

11. $\sum _{n=1}^{\infty }{u}_n^2$收敛$\Rightarrow \sum _{n=1}^{\infty }\frac{u_n}{n}$绝对收敛$(|\frac{u_n}{n}|⩽\frac{1}{2}(u_n^2+\frac{1}{n^2}))$

12. $a,b,c$为非零常数，$a{u}_n+b{v}_n+c{w}_n=0$，则在$\sum _{n=1}^{\infty }{u}_n,\sum _{n=1}^{\infty }{v}_n,\sum _{n=1}^{\infty }{w}_n$中只要有两个级数收敛，另一个必收敛

13. $\sum _{n=1}^{\infty }{u}_n$收敛，$\sum _{n=1}^{\infty }{v}_n$收敛$\Rightarrow \sum _{n=1}^{\infty }(u_n\pm v_n)$收敛

14. $\sum _{n=1}^{\infty }{u}_n$收敛，$\sum _{n=1}^{\infty }{v}_n$发散$\Rightarrow \sum _{n=1}^{\infty }(u_n\pm v_n)$发散

15. $\sum _{n=1}^{\infty }{u}_n$发散，$\sum _{n=1}^{\infty }{v}_n$发散$\Rightarrow \{\begin{array}{l}{u}_n⩾0,v_n⩾0\Rightarrow \sum _{n=1}^{\infty }(u_n+v_n)发散\\ u_n,v_n任意\Rightarrow \sum _{n=1}^{\infty }(u_n\pm v_n)不定\end{array}$

16. $\sum _{n=1}^{\infty }{u}_n$收敛，$\sum _{n=1}^{\infty }{v}_n$收敛$\{\begin{array}{l}{u}_n⩾0,v_n⩾0\Rightarrow \sum _{n=1}^{\infty }{u}_n{v}_n收敛(u_n{v}_n⩽\frac{u_n^2+v_n^2}{2})\\ u_n任意,v_n⩾0\Rightarrow \sum _{n=1}^{\infty }|u_n|\cdot v_n收敛(\underset{n\to \infty }{\lim }\frac{|u_n|\cdot v_n}{v_n}=\underset{n\to \infty }{\lim }|u_n|=0)\\ u_n任意,v_n任意\Rightarrow \sum _{n=1}^{\infty }{u}_n{v}_n不定\end{array}$

17. 选择题常见反例$\frac{{(-1)}^n}{n},\frac{{(-1)}^n}{\sqrt{n}},\frac{{(-1)}^n}{\ln n}$

2.级数的收敛域

2.1.概念

1. 函数项级数
   • 设函数列 { u n ( x ) } \{u_n(x)\} {un​(x)}定义在区间$I$上，称$u_1(x)+u_2(x)+\cdots +u_n(x)+\cdots$为定义在区间$I$上的函数项级数，记为$\sum _{n=1}^{\infty }{u}_n(x)$，当$x$取确定的值$x_0$时，$\sum _{n=1}^{\infty }{u}_n(x)$成为常数项级数$\sum _{n=1}^{\infty }{u}_n(x_0)$.
2. 幂级数
   • $\sum _{n=1}^{\infty }{u}_n(x)$的一般项$u_n(x)$是$n$次幂函数
   • 一般形式$\sum _{n=0}^{\infty }{a}_n(x-x_0{)}^n=a_0+a_1(x-x_0)+a_2(x-x_0{)}^2+\cdots +a_n(x-x_0{)}^n+\cdots$
   • 标准形式$\sum _{n=0}^{\infty }{a}_n{x}^n=a_0+a_{1}x+a_2{x}^2+\cdots +a_n{x}^n+\cdots$
3. 收敛点与发散点
   • 若给定$x_0\in I$，有$\sum _{n=1}^{\infty }{u}_n(x_0)$收敛（发散），称$x_0$为$\sum _{n=1}^{\infty }{u}_n(x)$的收敛（发散）点
4. 收敛域
   • $\sum _{n=1}^{\infty }{u}_n(x)$的所有收敛点的集合

2.2.具体型问题

2.2.1.对于不缺项幂级数$\sum _{n=0}^{\infty }{a}_n{x}^n$

1. 收敛半径
   • $\underset{n\to \infty }{\lim }|\frac{a_n+1}{a_n}|=\rho$，收敛半径$R=\{\begin{array}{l}\frac{1}{\rho },\rho \ne 0,+\infty \\ +\infty ,\rho =0\\ 0,\rho =+\infty \end{array}$
2. 收敛区间与收敛域
   • 收敛区间$(-R,R)$
   • 单独考察$x=\pm R$处的敛散性，确定收敛域$(-R,R)$或$[-R,R]$或$[-R,R)$或$(-R,R]$

2.2.2.对于缺项幂级数（如$\sum _{n=0}^{\infty }{a}_n{x}^{2n+1}$）或一般函数项级数$\sum u_n(x)$

1. 加绝对值，即$\sum |u_n(x)|$

2. 用正项级数的比值（或根值）判别法

   • 令$\underset{n\to \infty }{\lim }\frac{|u_{n+1}(x)|}{|u_n(x)|}<1$，求出收敛区间$(a,b)$

3. 单独讨论$x=1,x=b$时的敛散性，确定收敛域

2.3.抽象型问题

1. 阿贝尔定理
   • 当幂级数$\sum _{n=0}^{\infty }{a}_n{x}^n$在点$x=x_1(x_1\ne 0)$处收敛时，对于满足$|x|<|x_1|$的一切$x$，幂级数绝对收敛
   • 当幂级数$\sum _{n=0}^{\infty }{a}_n{x}^n$在点$x=x_2(x_2\ne 0)$处发散时，对于满足$|x|>|x_2|$的一切$x$，幂级数发散
2. 结论1
   • 已知$\sum _{n=0}^{\infty }{a}_n(x-x_0{)}^n$在某点$x_1(x_1\ne x_0)$的敛散性$\{\begin{array}{l}收敛\Rightarrow R⩾|x_1-x_0|\\ 发散\Rightarrow R⩽|x_1-x_0|\\ 条件收敛\Rightarrow R=|x_1-x_0|\end{array}$
3. 结论2
   • 已知$\sum a_n(x-x_1{)}^n$的敛散性，讨论$\sum b_n(x-x_2{)}^m$的敛散性$\{\begin{array}{l}提出或乘以因式(x-x_0{)}^k或作平移等，收敛半径不变\\ 逐项求导,收敛半径不变,收敛域可能缩小\\ 逐项积分,收敛半径不变,收敛域可能扩大\end{array}$

3.重要幂级数展开式

1. $e^x=\sum _{n=0}^{\infty }\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+...+{(-1)}^{n-1}\frac{x^n}{n}+...(-\infty <x<+\infty )$

2. $\frac{1}{1+x}=\sum _{n=0}^{\infty }{(-1)}^n{x}^n=1-x+x^2-x^3+...+{(-1)}^n{x}^n+...(-1<x<1)$

3. $\frac{1}{1-x}=\sum _{n=0}^{\infty }{x}^n=1+x+x^2+...+x^n+...(-1<x<1)$

4. $\ln (1+x)=\sum _{n=1}^{\infty }{(-1)}^{n-1}\frac{x^n}{n}=x-\frac{x^2}{2}+\frac{x^3}{3}-...+{(-1)}^{n-1}\frac{x^n}{n}+...(-1<x⩽1)$

5. $\sin x=\sum _{n=0}^{\infty }{(-1)}^n\frac{x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...+{(-1)}^n\frac{x^{2n+1}}{(2n+1)!}+...(-\infty <x<+\infty )$

6. $\frac{e^x-e^{-x}}{2}=\sum _{n=0}^{\infty }\frac{x^{2n+1}}{(2n+1)!}$

7. $\cos x=\sum _{n=0}^{\infty }{(-1)}^n\frac{x^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-...+{(-1)}^n\frac{x^{2n}}{(2n)!}+...(-\infty <x<+\infty )$

8. $\frac{e^x+e^{-x}}{2}=\sum _{n=0}^{\infty }\frac{x^{2n}}{(2n)!}$

9. $\arctan x=\sum _{n=1}^{\infty }{(-1)}^{n-1}\frac{x^{2n-1}}{2n-1}=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+...+{(-1)}^{n-1}\frac{x^{2n-1}}{2n-1}+...$

10. ${(1+x)}^{\alpha }=1+\alpha x+\frac{\alpha (\alpha -1)}{2!}{x}^2+...+\frac{\alpha (\alpha -1)...(\alpha -n+1)}{n!}{x}^n+...\{\begin{array}{l}\alpha ⩽-1\to -1<x<1\\ -1<\alpha <1\to -1<x⩽1\\ \alpha >0,\alpha \notin N_{+}\to -1⩽x⩽1\\ \alpha \in N_{+}\to -\infty <x<+\infty \end{array}$

4.重要结论

4.1.正项级数

$(1)\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}=A\Rightarrow \underset{n\to \infty }{\lim }\sqrt[n]{a_n}=A$

$(2)正项级数，若通项同阶，则级数同敛散$

4.2.幂级数

$\sum _{n=1}^{\infty }\frac{x^n}{n}=-\ln (1-x)(-1⩽x<1)$

$\sum _{n=1}^{\infty }n{x}^{n-1}=\frac{1}{{(1-x)}^2}(-1<x<1)$

$\{\begin{array}{l}\sum _{n=0}^{\infty }{x}^n=\frac{1}{1-x}\\ \sum _{n=0}^{\infty }(n+1)x^n=\frac{1}{{(1-x)}^2}\\ \sum _{n=0}^{\infty }(n+2)(n+1)x^n=\frac{2}{{(1-x)}^3}\end{array}$

4.3.p-级数

$p-级数\sum _{n=1}^{\infty }\frac{1}{n^p}\{\begin{array}{l}p>1收敛\\ p⩽1发散\end{array}$

$广义p-级数\sum _{n=2}^{\infty }\frac{1}{n{\ln }^{p}n}\{\begin{array}{l}p>1收敛\\ p⩽1发散\end{array}$

$p-积分{\int }_1^{+\infty }\frac{1}{x^p}\mathrm{d}x\{\begin{array}{l}p>1收敛\\ p⩽1发散\end{array}$

$广义p-积分{\int }_2^{+\infty }\frac{1}{x{\ln }^{p}x}\mathrm{d}x\{\begin{array}{l}p>1收敛\\ p⩽1发散\end{array}$

5.傅里叶级数

1. $S(x)=\frac{a_0}{2}+\sum _{n=1}^{\infty }(a_n\cos \frac{n\pi }{l}x+b_n\sin \frac{n\pi }{l}x)\{\begin{array}{l}{a}_n=\frac{1}{l}{\int }_{-l}^{l}f(x)\cos \frac{n\pi }{l}x\mathrm{d}x,n=0,1,2,...\\ b_n=\frac{1}{l}{\int }_{-l}^{l}f(x)\sin \frac{n\pi }{l}x\mathrm{d}x,n=1,2,...\end{array}$

2. 狄利克雷收敛$S(x)=\{\begin{array}{l}f(x),x为连续点\\ \frac{f(x-0)+f(x+0)}{2},x为间断点\\ \frac{f(-l+0)+f(l-0)}{2},x=\pm l\end{array}$

3. $\{\begin{array}{l}f(x)为奇函数\{\begin{array}{l}{a}_n=0\\ b_n=\frac{2}{l}{\int }_0^{l}f(x)\sin \frac{n\pi }{l}x\mathrm{d}x,n=1,2,...\\ f(x)=\sum _{n=1}^{\infty }{b}_n\sin \frac{n\pi }{l}x(正弦级数)\end{array}\\ f(x)为偶函数\{\begin{array}{l}{a}_n=\frac{2}{l}{\int }_0^{l}f(x)\cos \frac{n\pi }{l}x\mathrm{d}x\\ b_n=0\\ f(x)=\frac{a_0}{2}+\sum _{n=1}^{\infty }{a}_n\cos \frac{n\pi }{l}x\mathrm{d}x(余弦级数)\end{array}\end{array}$