考研高等数学公式总结（三）

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无穷级数收敛性判断

定义

$$\begin{array}{rl}& \sum _{n=1}^{\infty }{u}_n=u_1+u_2+...+u_n+...叫无穷级数\\ & S_n=u_1+u_2+...+u_n叫级数的前n项和\\ & \sum _{n=1}^{\infty }{u}_n收敛⟺\underset{n\to +\infty }{\lim }{S}_n存在\\ & \sum _{n=1}^{\infty }{u}_n收敛\to \underset{n\to \infty }{\lim }{u}_n\underset{n\to \infty }{\lim }{u}_n存在=0(必要条件)\\ & \sum _{n=1}^{\infty }(u_{n+1}-u_n)收敛⟺\underset{n\to \infty }{\lim }{u}_n\underset{n\to \infty }{\lim }{u}_n存在\end{array}$$

正项级数

名称	具体
收敛原则	${\sum }_{n=1}^{\infty }{u}_n$收敛$\iff$部分和$S_n$有界
比较判别法	大的收敛$\Rightarrow$小的收敛，小的发散$\Rightarrow$大的发散
比较判别法的极限形式	$\lim_{n\rightarrow \infty}\cfrac{u_n}{v_n}=A= \begin{cases} 0 & v_n收敛\Rightarrow u_n也收敛\ +\infty& v_n发散\Rightarrow u_n也发散 \ C& v_n与u_n有相同的敛散性 \end{cases} \$
比值判别法	$\lim_{n\rightarrow \infty}\cfrac{u_{n+1}}{u_n}=\rho= \begin{cases} \rho<1 & 收敛\ \rho>1& 发散 \\rho=1& 失效\ \end{cases} \$
根植判别法	$\lim_{n\rightarrow \infty}\sqrt[n]{u_n}=\rho= \begin{cases} \rho<1 & 收敛\ \rho>1& 发散 \\rho=1& 失效 \ \end{cases} \$
积分判别法	若存在$[a,+\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=a}^{\infty }{u}_n$与反常积分${\int }_a^{+\infty }f(x)dx$的收敛性相同

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

$$|u_n|单调不增且\underset{n\to \infty }{\lim }{u}_n=0\to 级数收敛(不满足方法失效，可尝试拆项)\text{\hspace{1em}(莱布尼茨判别法)}$$

任意项级数${\sum }_{n=1}^{\infty }{u}_n$

$$\begin{array}{rl}& \sum _{n=1}^{\infty }|u_n|收敛\Rightarrow \sum _{n=1}^{\infty }{u}_n绝对收敛\\ & \sum _{n=1}^{\infty }|u_n|发散,\sum _{n=1}^{\infty }{u}_n收敛\Rightarrow \sum _{n=1}^{\infty }{u}_n条件收敛\end{array}$$

常用级数收敛性

$$\begin{array}{rlr}& 等比级数\sum _{n=1}^{\infty }a{q}^{n-1}\{\begin{array}{ll}=\frac{a}{1-q},& |q|<1\\ 发散,& |q|⩾1\end{array}& p级数\sum _{n=1}^{\infty }\frac{1}{n^p}\{\begin{array}{ll}收敛,& p>1\\ 发散,& p⩽1\end{array}\\ & 广义p级数\sum _{n=1}^{\infty }\frac{1}{n(\ln n{)}^p}\{\begin{array}{ll}收敛,& p>1\\ 发散,& p⩽1\end{array}& 交错p级数\sum _{n=1}^{\infty }(-1{)}^{n-1}\frac{1}{n^p}\{\begin{array}{ll}绝对收敛,& p>1\\ 条件收敛,& 0<p⩽1\end{array}\end{array}$$

常用结论

$$\begin{array}{rl}1.& 若\sum _{n=1}^{\infty }{u}_n收敛,则\sum _{n=1}^{\infty }|u_n|不定(反例:\sum _{n=1}^{\infty }(-1{)}^n\frac{1}{n}收敛,但\sum _{n=1}^{\infty }\frac{1}{n}发散)\\ 2.& 设\sum _{n=1}^{\infty }{u}_n收敛,则\{\begin{array}{l}{u}_n⩾0时,{\sum }_{n=1}^{\infty }{u}_n^2收敛({\lim }_{n\to \infty }{u}_n=0,从某项起,u_n<1,u_n^2<u_n)\\ u_n任意时,{\sum }_{n=1}^{\infty }{u}_n^2不定(反例:{\sum }_{n=1}^{\infty }(-1{)}^n\frac{1}{\sqrt{n}}收敛,但{\sum }_{n=1}^{\infty }\frac{1}{n}发散)\end{array}\\ 3.& 设\sum _{n=1}^{\infty }{u}_n收敛,则\{\begin{array}{l}{u}_n⩾0时,{\sum }_{n=1}^{\infty }{u}_n{u}_{n+1}收敛(u_n\cdot u_{n+1}⩽\frac{u_n^2+u_{n+1}^2}{2})\\ u_n任意时,{\sum }_{n=1}^{\infty }{u}_n{u}_{n+1}不定(反例:u_n=(-1{)}^n\frac{1}{\sqrt{n}},u_n{u}_{n+1}=(-1{)}^n\frac{1}{\sqrt{n}}(-1{)}^{n+1}\frac{1}{\sqrt{n+1}}=-\frac{1}{\sqrt{n(n+1)}},级数发散)\end{array}\\ 4.& 设\sum _{n=1}^{\infty }{u}_n收敛,则\sum _{n=1}^{\infty }(-1{)}^n{u}_n不定(反例:\sum _{n=1}^{\infty }(-1{)}^n\frac{1}{n}收敛,但\sum _{n=1}^{\infty }\frac{1}{n}发散)\\ 5.& 设\sum _{n=1}^{\infty }{u}_n收敛,则\sum _{n=1}^{\infty }(-1{)}^n\frac{u_n}{n}不定(反例:\sum _{n=2}^{\infty }(-1{)}^n\frac{1}{\ln n}收敛,但\sum _{n=2}^{\infty }\frac{1}{n\ln n}发散)\\ 6.& 设\sum _{n=1}^{\infty }{u}_n收敛,则\{\begin{array}{l}{u}_n⩾0时,{\sum }_{n=1}^{\infty }{u}_{2n},{\sum }_{n=1}^{\infty }{u}_{2n-1}均收敛\\ u_n任意时,{\sum }_{n=1}^{\infty }{u}_{2n},{\sum }_{n=1}^{\infty }{u}_{2n-1}不定(反例:1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots \\ ={\sum }_{n=1}^{\infty }(-1{)}^{n-1}\frac{1}{n}收敛，但是其奇数项和与偶数项和都发散)\end{array}\\ 7.& 设\sum _{n=1}^{\infty }{u}_n收敛,则\sum _{n=1}^{\infty }(u_{2n-1}+u_{2n})收敛.\\ & (收敛级数任意加括号所得的新级数仍收敛,且和不变,但反过来推要增加\underset{n\to \infty }{\lim }=0的条件,即\sum _{n=1}^{\infty }(u_{2n-1}+u_{2n})收敛且\underset{n\to \infty }{\lim }=0,则\sum _{n=1}^{\infty }{u}_n收敛,\\ & 因S_{2n}=(u_1+u_2)+(u_3+u_4)+\cdots +(u_{2n-1}+u_{2n}),\underset{n\to \infty }{\lim }{S}_{2n}=S存在,\\ & S_{2n+1}=S_{2n}+u_{2n+1},\underset{n\to \infty }{\lim }{S}_{2n+1}=S+\underset{n\to \infty }{\lim }{u}_{2n+1}=S,即可得\sum _{n=1}^{\infty }{u}_n收敛）\\ 8.& 设\sum _{n=1}^{\infty }{u}_n收敛,则\sum _{n=1}^{\infty }(u_{2n-1}-u_{2n})不定\\ & (反例：(u_1+u_2)+(u_3+u_4)+(u_5+u_6)+\cdots =1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots 收敛，\\ & 但1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\cdots 发散)\\ 9.& 设\sum _{n=1}^{\infty }{u}_n收敛,则\{\begin{array}{l}{\sum }_{n=1}^{\infty }(u_n+u_{n+1})收敛,{\sum }_{n=1}^{\infty }{u}_n+{\sum }_{n=1}^{\infty }{u}_{n+1}收敛\\ {\sum }_{n=1}^{\infty }(u_n-u_{n+1})收敛,{\sum }_{n=1}^{\infty }{u}_n-{\sum }_{n=1}^{\infty }{u}_{n+1}收敛\end{array}\\ 10.& 若\sum _{n=1}^{\infty }|u_n|收敛,则\sum _{n=1}^{\infty }{u}_n收敛;若\sum _{n=1}^{\infty }{u}_n发散,则\sum _{n=1}^{\infty }|u_n|发散\\ 11.& 若\sum _{n=1}^{\infty }{u}_n^2收敛,则\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.& 设a,b,c为非零常数,且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中只要有两个级数是收敛的,另一个必收敛\\ 13.& 若\sum _{n=1}^{\infty }{u}_n收敛,\sum _{n=1}^{\infty }{v}_n收敛,则\sum _{n=1}^{\infty }(u_n\pm v_n)收敛\\ 14.& 若\sum _{n=1}^{\infty }{u}_n收敛,\sum _{n=1}^{\infty }{v}_n发散,则\sum _{n=1}^{\infty }(u_n\pm v_n)发散\\ 15.& 若\sum _{n=1}^{\infty }{u}_n收敛,\sum _{n=1}^{\infty }{v}_n发散,则\{\begin{array}{l}{u}_n⩾0,v_n⩾0时,{\sum }_{n=1}^{\infty }(u_n+v_n)发散\\ u_n,v_n任意时,{\sum }_{n=1}^{\infty }(u_n\pm v_n)不定\end{array}\\ 16.& 若\sum _{n=1}^{\infty }{u}_n收敛,\sum _{n=1}^{\infty }{v}_n收敛,则\{\begin{array}{l}{u}_n⩾0,v_n⩾0时,{\sum }_{n=1}^{\infty }{u}_n{v}_n收敛(u_n\cdot v_n⩽\frac{u_n^2+v_n^2}{2})\\ u_n任意,v_n⩾0时,{\sum }_{n=1}^{\infty }|u_n|\cdot v_n收敛({\lim }_{n\to \infty }\frac{|u_n|\cdot v_n}{v_n}={\lim }_{n\to \infty }|u_n|=0)\\ u_n任意,v_n任意时,{\sum }_{n=1}^{\infty }{u}_n{v}_n不定\end{array}\end{array}$$

傅里叶级数

$$\begin{array}{rl}f(x)\sim S(x)=& \frac{a_0}{2}+\sum _{n=1}^{\infty }(a_n\cos \frac{n\pi }{l}x+b_n\sin \frac{n\pi }{l}x)\\ & a_n=\frac{1}{l}{\int }_{-l}^{l}f(x)\cos \frac{n\pi }{l}xdx\\ & b_n=\frac{1}{l}{\int }_{-l}^{l}f(x)\sin \frac{n\pi }{l}xdx\end{array}$$

$$S(x)=\{\begin{array}{ll}f(x)& x为连续点\\ \frac{f(x^{-})+f(x^{+})}{2}& x为间断点\\ \frac{f(-l^{+})+f(l^{-})}{2}& x=\pm l\end{array}\text{\hspace{1em}(狄利克雷收敛定理)}$$