数组无限求和
$$\left\{ \begin{array}{l} \text{数项级数}\left\{ \begin{array}{l} \text{一般级数：绝对值 }or \ \ \ \text{定义}\\ \text{正项级数：比较}+\text{极限}\left( \text{等价} \right)\\ \text{交错级数：绝对值收敛}+\text{莱布尼茨}+\text{定义}\\ \end{array} \right.\\ \text{幂级数求和}\left\{ \begin{array}{l} \text{常规：拆配凑补}\\ \text{微分方程：代入、合并}\\ \text{混合型：求导积分}\\ \end{array} \right.\\ \text{展开}\left\{ \begin{array}{l} x=0\\ \text{求导积分}\\ \text{合并最简}\\ \end{array} \right.\\ \end{array} \right. $$常数项级数
$$
\sum_{n=1}^∞ u_n=u_1+u_2+…+u_n+…
$$
线性性质：$\sum_{n=1}^{∞}(au_n+bv_n)=a\sum_{n=1}^{∞}u_n+b\sum_{n=1}^{∞}v_n$
前有限项不改变收敛性：$\underset{n=1}{\overset{∞}{\sum}}u_n\Rightarrow\underset{n=m}{\overset{∞}{\sum}}u_n$收敛
级数收敛必要条件：
- $\underset{n=1}{\overset{∞}{\sum}}u_n$收敛$\Leftrightarrow \underset{n \rightarrow ∞}{\lim}S_n$存在
- $\underset{n=1}{\overset{∞}{\sum}}u_n$收敛$\begin{array}{c}
\Rightarrow\
\nLeftarrow\
\end{array}\underset{n \rightarrow ∞}{\lim}u_n=0$ - $\underset{n=1}{\overset{∞}{\sum}}(u_{n+1}-u_n)$收敛$\Leftrightarrow{\lim}u_n$存在
敛散性判别方法
正项级数
$$
\sum u_n， u_n>0
$$
$\sum_{n=1}^{\infty}{u_n}$收敛$\Leftrightarrow{S_n}$有界
$\sum u_n$收敛$\Rightarrow$$\underset{n\rightarrow \infty}{\lim}u_n=0$
极限推敛散而非敛散推极限
若选项说了正项级数，有的没说具体什么级数，没说具体级数的排除
比较判别法：若$u_n\le v_n$
$\underset{n=1}{\overset{∞}{\sum}}v_n$收敛$\Leftrightarrow\underset{n=1}{\overset{∞}{\sum}}u_n$收敛
$\underset{n=1}{\overset{∞}{\sum}}u_n$发散$\Leftrightarrow \underset{n=1}{\overset{∞}{\sum}}v_n$发散
$\underset{n\rightarrow \infty}{\lim}\dfrac{u_n}{v_n}=\left\{ \begin{array}{l} 0\Rightarrow u_n\text{是}v_n\text{高阶无穷小}\Rightarrow \left\{ \begin{array}{l} \sum_{n=1}^{\infty}{v_n}\text{收敛}\Rightarrow \sum_{n=1}^{\infty}{u_n}\text{收敛}\\ \\ \sum_{n=1}^{\infty}{u_n}\text{发散}\Rightarrow \sum_{n=1}^{\infty}{v_n}\text{发散}\\ \end{array} \right.\\ \\ \infty \Rightarrow v_n\text{是}u_n\text{高阶无穷小}\Rightarrow \left\{ \begin{array}{l} \sum_{n=1}^{\infty}{u_n}\text{收敛}\Rightarrow \sum_{n=1}^{\infty}{v_n}\text{收敛}\\ \\ \sum_{n=1}^{\infty}{v_n}\text{发散}\Rightarrow \sum_{n=1}^{\infty}{u_n}\text{发散}\\ \end{array} \right.\\ \\ A\ne 0\Rightarrow u_n\text{是}v_n\text{同阶无穷小}\Rightarrow \sum_{n=1}^{\infty}{u_n}\sum_{n=1}^{\infty}{v_n}\text{同敛散}\\ \end{array} \right. $比值判别法：（达朗贝尔判别法）
$\underset{n\rightarrow {\infty }}{\lim}\dfrac{u_{n+1}}{u_n}=\rho \left\{ \begin{array}{l}<1\ \ \text{收敛}\\>1\ \ \text{发散}\\ =1\ \ \text{失效}\\ \end{array} \right. $[$a^n;n!;n^n$]
根植判别法：（柯西）
$\underset{n\rightarrow {\infty }}{\lim}\sqrt[n]{u_n}=\rho \left\{ \begin{array}{l} <1\ \ \text{收敛}\\>1\ \ \text{发散}\\ =1\ \ \text{失效}\\ \end{array} \right. $[$a^n;n^n$]
积分判别法：(柯西)
$u_n$与$\int^{\infty}_1f(x)\text dx$敛散性相同
交错级数
$$
\sum(-1)^{n-1}u_n
$$
莱布尼茨判别法：$\left. \begin{array}{r} \underset{n\rightarrow \infty}{\lim}u_n=0\\ u_n\text{单调递减}\\ \end{array} \right\} \Rightarrow \text{收敛}$
任意项级数
绝对收敛：$\underset{n=1}{\overset{∞}{\sum}}|u_n|$收敛
条件收敛：$\underset{n=1}{\overset{∞}{\sum}}u_n$收敛但$\underset{n=1}{\overset{∞}{\sum}}|u_n|$发散
收敛性质
考虑凑$\sum (-1)^n \dfrac1n$，变换后为$\sum \dfrac1n$
涉及$u_nv_n$，可凑$\dfrac1{n^p}$
$\sum{u_n}$收敛$\nRightarrow$$\sum{|u_n|}$
- $\sum{u_n}$收敛$\left\{ \begin{array}{l} \sum{u_{n}^{2}}\text{收敛 }\ \ \ \ u_n\ge 0\\ \\ \sum{u_{n}^{2}}\text{不定 }\ \ \ \ u_n\text{任意}\\ \end{array} \right. $
- $\sum{u_n}$收敛$\left\{ \begin{array}{l} \sum{u_{n}^{}u_{n+1}}\text{收敛 }\ \ \ \ u_n\ge 0\\ \\ \sum{u_{n}^{}u_{n+1}}\text{不定 }\ \ \ \ u_n\text{任意}\\ \end{array} \right. $
$\sum{u_n}$收敛$\nRightarrow$$\sum{(-1)^nu_n}$
$\sum{u_n}$收敛$\nRightarrow$$\sum{(-1)^n\dfrac{u_n}{n}}$
$\sum{u_n}$收敛$\left\{ \begin{array}{l} \sum{u_{2n}\text{、}\sum{u_{2n-1}}}\text{收敛 }\ \ \ \ u_n\ge 0\\ \\ \sum{u_{2n}\text{、}\sum{u_{2n-1}}}\text{不定 }\ \ \ \ u_n\text{任意}\\ \end{array} \right. $
$\sum{u_n}$收敛$\Rightarrow \sum({u_{2n-1}+u_{2n}})$收敛
$\Rightarrow \sum({u_{2n-1}-u_{2n}})$不定
- $\sum{u_n}\Rightarrow $$\left\{ \begin{array}{l} \sum{\left( u_n+u_{n+1} \right)}\\ \\ \sum{\left( u_n-u_{n+1} \right)}\\ \\ \sum{u_n}+\sum{u_{n+1}}\\ \\ \sum{u_n}-\sum{u_{n+1}}\\ \end{array} \right. $收敛
$\sum{|u_n| }$收敛$\Rightarrow$$\sum{u_n}$收敛
$\sum{u_n}$发散$\Rightarrow$$\sum{|u_n| }$发散
$\sum{u^2_n}$收敛$\Rightarrow$$\sum \dfrac {u_n}n$绝对收敛
- $\left. \begin{array}{r} a,b,c\ne 0\\ au_n+bv_n+cw_n=0\\ u_n,v_n\text{收敛}\\ \end{array} \right\} \Rightarrow w_n\text{收敛}$
$\sum{u_n}$收敛$\left\{ \begin{array}{l} \sum{v_n}\text{发散}\Rightarrow \left( u_n+v_n \right) \text{发散}\\ \sum{v_n}\text{收敛}\Rightarrow \left( u_n+v_n \right) \text{收敛}\\ \end{array} \right. $
$\sum{u_n}，\sum v_n$收敛$\Rightarrow$$\left\{ \begin{array}{l} \left( u_n+v_n \right) \text{发散 }\ \ \ \ u_n\ge 0,v_n\ge 0\\ \left( u_n+v_n \right) \text{不定 }\ \ \ \ u_n,v_n\text{不定}\\ \end{array} \right. $
$\sum{u_n}，\sum v_n$收敛$\Rightarrow$$\left\{ \begin{array}{l} \sum{u_nv_n\text{收敛}}\ \ \ \ \ u_n\ge 0,\ v_n\ge 0\\ \sum{|u|v_n\text{收敛 }\ \ \ \ u_n\text{任意，}v_n\ge 0}\\ \end{array} \right. $
若$\underset{n=1}{\overset{∞}{\sum}}u_n$收敛，则其中任意项数的和组成的新级数也收敛
若$\underset{n=1}{\overset{∞}{\sum}}u_n$绝对收敛，则无论各项如何排序，新级数绝对收敛
常见级数
等比级数：$\sum{aq^{n-1}}=\left\{ \begin{array}{l} \dfrac{a}{1-q}\ \ |q|<1\\ \text{发散 }\ \ \ \ \ |q|\ge 1\\ \end{array} \right. $
$p$级数：$\sum{\dfrac{1}{n^p}}\left\{ \begin{array}{l} \text{收敛 }\ \ p>1\\ \text{发散 }\ \ p\le 1\\ \end{array} \right. $
广义$p$级数：$\sum{\dfrac{1}{n\left( \ln n \right) ^p}}\left\{ \begin{array}{l} \text{收敛}\ \ p>1\\ \text{发散}\ \ p\le 1\\ \end{array} \right. $
交错$p$级数：$\sum{\left( -1 \right) ^{n-1}\dfrac{1}{n^p}}\left\{ \begin{array}{l}
\text{绝对收敛 }\ \ \ \ \ \ \ \ \ \ \ p > 1\\
\text{条件收敛 }\ \ \ \ 0 < p\le 1\\
\end{array} \right. $
$\sum{\dfrac{a^n}{n^s}}=\left\{ \begin{array}{l} a<1\ \ \text{收敛}\\ a>1\ \ \text{发散}\\ a=1\left\{ \begin{array}{l} s>1\ \ \text{收敛}\\ s\le 1\ \ \text{发散}\\ \end{array} \right.\\ \end{array} \right. $
调和级数：$\sum \dfrac1n $ 发散
其他级数
- $u_n=\dfrac1{\sqrt{n(n+1)}}\left\{ \begin{array}{l} \ >\dfrac{1}{b_n}\ \ \ \ \dfrac{1}{b_n}\text{发散}\\ \ <\dfrac{1}{a_n}\ \ \ \ \dfrac{1}{a_n}\text{收敛}\\ \end{array} \right. $
- $\sum\dfrac1{n^2-k}$收敛【$(n^2-k)\ge(n-a)^2$】
- $\sum\dfrac1{a^n+kn}\rightarrow \lim \dfrac{\dfrac1{a^n+kn}}{\dfrac1{a^n}}=1$敛散一致
- $A\sim B\Rightarrow A、B$敛散性相同
幂级数收敛域
幂级数一般形式：$\sum_{n=0}^∞a_n(x-x_0)^n$
幂级数标准形式：$\sum_{n=0}^∞a_nx^n$
收敛点：$\underset{n=1}{\overset{∞}{\sum}}u_n(x_0)$收敛
具体型问题
$\sum a_nx^n$型、缺项幂级数$\sum u_n(x) $
$\dfrac1{1-x}=1+x+x^2+\cdots+x^n$
收敛域的求法
- 若$\underset{n \rightarrow ∞}{\lim}\left. | \right.\dfrac{a_{n+1}}{a_n} \left. \right|=\rho$，收敛半径$R$为$R=\left\{ \begin{array}{l} \dfrac{1}{\rho}\ \ \ \ \ \ \ \ \ \rho \ne 0\\ +{\infty }\ \ \ \ \ \rho =0\\ 0\ \ \ \ \ \ \ \ \ \ \rho =+\textit{∞ }\\ \end{array} \right.$
- 单独计算$±R$的敛散性
统一方法求收敛域
- 记$|u_n(x)|$
- 用分值判别法或根植判别法求$\rho$
- 令$\rho<1$,解不等式
- 单独讨论端点
阿贝尔定理
$\underset{n=1}{\overset{∞}{\sum}}a_nx^n$在$x_0$处收敛，$\forall |x|<x_0$，幂级数绝对收敛
$\underset{n=1}{\overset{∞}{\sum}}a_nx^n$在$x_0$处发散，$\forall |x|>x_0$，幂级数发散
推论1：已知$\underset{n=0}{\overset{∞}{\sum}}a_n(x-x_0)^n$
- 若在$x_1$处收敛，则收敛半径$R\ge|x_1-x_0|$
- 若在$x_1$处发散，则收敛半径$R\le|x_1-x_0|$
- 若在$x_1$处条件收敛，则收敛半径$R=|x_1-x_0|$
推论2：已知$\underset{n=0}{\overset{∞}{\sum}}a_n(x-x_0)^n$、已知$\underset{n=0}{\overset{∞}{\sum}}b_n(x-x_0)^m$
- $(x-x_0)^n$到$(x-x_0)^m$转换通过平移或乘$(x-x_0)^k$
- $a_n$到$b_n$转化通过微积分变形：对级数求导或积分
- 以下情况收敛半径不变
- 平移或乘$(x-x_0)^k$
- 对级数求导，收敛域可能缩小
- 对级数积分，收敛域可能扩大
- $\left(\ ,\ \right]$、$\left[\ ,\ \right)$，积分后$[\ ,\ ]$，求导后$(\ ,\ )$
- $\sum \left(a_n+b_n\right)x^n$收敛半径：$R\ge\min\left\{R_1,R_2\right\}$
求和
套公式
$\sum \dfrac{x^n}n=-\ln(1-x),\ -1\le x<1$
$\sum nx^{n-1}=\dfrac 1{(1-x)^2},\ -1<x<1$
先求导积分后求和
- $\sum(an+b)x^{an}$先积后导
- $\sum \dfrac{x^{an}}{an+b}$先导后积
- $\sum\dfrac{cn^2+dn+e}{an+b}=\sum_1+\sum_2$
拆解法
子型级数
$\underset{n=0}{\sum}{x^n}=\dfrac{1}{1-x}$
$\underset{n=0}{\sum}{nx^{n-1}}=\dfrac{1}{\left( 1-x \right) ^2}$
$\underset{n=0}{\sum}{n\left( n-1 \right) x^{n-2}}=\dfrac{2}{\left( 1-x \right) ^3}$
母型级数
$\underset{n=1}{\sum}{\dfrac{x^n}{n}}=-\ln \left( 1-x \right) $
$\underset{n=1}{\sum}{\left( -1 \right) ^n\dfrac{x^n}{n}}=-\ln \left( 1+x \right) $
$\underset{n=1}{\sum}{\dfrac{x^{2n-1}}{2n-1}}=\dfrac{1}{2}\ln \dfrac{1+x}{1-x}=\dfrac{\underset{n=1}{\sum}{\dfrac{1}{n}x^n}-\underset{n=1}{\sum}{\dfrac{1}{n}\left( -x \right) ^n}}{2}$
$\underset{n=1}{\sum}{\left( -1 \right) ^{n-1}\dfrac{x^{2n-1}}{2n-1}}=\arctan x$
泰勒函数
$\sum x^n=\dfrac1{1-x}$
$\sum\dfrac{x^n}{n!}=\text e^x$
$\sum\dfrac{2x^{2n}}{(2n)!}=\text e^x+\text e^{-x}$
$\sum \dfrac{n(n-1)}{2!}x^2=(1+x)^n$
$\sum (-1)^{n-1}\dfrac{x^{2n-1}}{(2n-1)!}=\sin x$
$\sum (-1)^n\dfrac{x^{2n}}{(2n)!}=\cos x$
$\sum (-1)^{n-1}\dfrac{x^n}n=\ln(1+x)$
拆解步骤
- 求收敛半径$R=\underset{n\rightarrow\infty}{\lim}\sqrt[n]{f(x^n)}<1$
- 求边界的敛散性
- 拆：变为单一幂函数
- 补：补为字母型（根据系数凑$x$的阶数）
- 凑：凑泰勒
- 无定义点补齐
- 抽象数列考虑微分方程求和
微分方程求和
- 验证$y,y’,y’’$满足微分方程
- 求通解
- 根据初始条件定$\text C$或求$x=x_0$时的和
展开
幂级数展开
函数展开：$f(x)=\sum a_nx^n$
积分展开：$\int_a^bf(x)\text dx=\sum a_n\dfrac{b^{n+1}-a^{n+1}}{n+1}$
导数展开：$\dfrac{\text df(x)}{\text dx}=\sum a_nx^{n-1}$
常见展开
$\dfrac1{x+c}$在$(x+x_0)$展开：$\dfrac1{x+c}=-\sum_{n=0}^{\infty}\dfrac{(x+x_0)^n}{(x_0-c)^{n+1}}$
傅里叶级数
迪利克雷收敛
$f(x)\in T_{2l}$，若在$[-l.l]$满足$\left\{ \begin{array}{l} \text{连续或有有限个第一类间断点}\\ \text{至多有限个极值点}\\ \end{array} \right. $，则$f(x)$在$[-l,l]$收敛 $$ S\left( x \right) =\left\{ \begin{array}{l} f\left( x \right) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x\text{连续}\\ \dfrac{f\left( x-0 \right) +f\left( x+0 \right)}{2}\ \ \ \ \ x\text{间断}\\ \dfrac{f\left( -l+0 \right) +f\left( l-0 \right)}{2}\ \ \ \ x=±l\\ \end{array} \right. $$傅里叶系数
$$
f(x)\sim \dfrac{a_0}2+\sum(a_n\cos\dfrac{n\pi x}l+b_n\sin \dfrac{n\pi x}l)
$$
- 当$f(x)$为奇函数，其展开式$f(x)$$\sim$$ b_n\sum \sin \dfrac{n\pi x}l$
- 当$f(x)$为偶函数，其展开式$f(x)$$\sim$$ \dfrac{a_0}2 +a_n\sum \cos \dfrac{n\pi x}l$
周期$2\pi$：$\left\{ \begin{array}{l} \text{奇函数}\left\{ \begin{array}{l} a_n=0\\ b_n=\dfrac{2}{\pi}\int_0^{\pi}{f\left( x \right) \sin \left( nx \right)}\text{d}x\\ \end{array} \right.\\ \text{偶函数}\left\{ \begin{array}{l} a_n=\dfrac{2}{\pi}\int_0^{\pi}{f\left( x \right) \cos \left( nx \right) \text{d}x}\\ b_n=0\\ \end{array} \right.\\ \end{array} \right.$
延拓
$f(x)\in\text C_{[0,\pi]}$$\left\{ \begin{array}{l} \text{奇延拓}\Rightarrow F\left( x \right) =\left\{ \begin{array}{l} f\left( x \right)\ \ \ \ \ \ \ \ \ \ \ \ \ 0 < x\le \pi\\ 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=0\\ -f\left( -x \right) \ \ -\pi < x\le 0\\ \end{array} \right.\\ \text{偶延拓}\Rightarrow F\left( x \right) =\left\{ \begin{array}{l} f\left( x \right)\ \ \ \ \ \ \ \ \ \ \ \,\,\,\,0\le x\le \pi\\ \\ -f\left( -x \right) \,\,\,\,-\pi \le x < 0\\ \end{array} \right.\\ \end{array} \right. $
