本页面记录判断数项级数敛散性的常用定理


约定部分和符号 S:=n=1anS:=\sum^\infty_{n=1}a_nSn:=k=1nakS_n:=\sum^n_{k=1}a_k.

特别地,S+S_+ 是正项级数.

此外,P,PnP,P_n 用来表示无穷乘积,在这里对于一般的无穷乘积,写成P=n=1pnP=\prod^\infty_{n=1}p_n;对于收敛的无穷乘积,写成P=n=1(1+an)P=\prod_{n=1}^\infty(1+a_n);以下符号也沿用.

约定收敛符号SS\downarrow,条件收敛符号SS|\downarrow,绝对收敛符号 S|S|\downarrow,发散是上箭头 \uparrow.

约定 Cauchy 乘积S(a)×S(b)=n=1i=1naibn+1iS(a)\times S(b)=\sum^\infty_{n=1}\sum_{i=1}^na_{i}b_{n+1-i}.

显然,nn 是足够大的,因为去掉有限项后级数敛散性不变.

# 一般级数判断准则

# Cauchy 收敛准则

Sϵ>0,N>0,m,n>N:SmSn<ϵS\downarrow \iff \forall \epsilon>0,\exists N>0,\forall m,n>N:|S_m-S_n|<\epsilon

# Lebnitz 交错判敛

an|a_n| 单调趋于零,$\text {sgn}(a_n)=(-1)^n \implies S\downarrow $

# 必要条件

Sliman=0S\downarrow\implies \lim a_n=0

# Dirichlet 判别

Sn(a)S_n(a) 有界,bnb_n 单调趋于 00 S(ab)\implies S(ab)\downarrow

# Abel 判别

S(a)S(a)\downarrowbnb_n 单调有界 S(ab)\implies S(ab)\downarrow

# 绝对收敛

SS|S|\downarrow \implies S\downarrow

# 收敛级数其他性质

# 线性叠加

S(a),S(b)α,βR,S(αa+βb)=αS(a)+βS(b)S (a)\downarrow,S(b)\downarrow\implies \forall \alpha,\beta\in\mathbb R, S(\alpha a+\beta b)\downarrow =\alpha S(a)+\beta S(b)

# 定序归组

S(a)bk=nk1nkai:S(bk),S(bk)=S(a)S(a)\downarrow \implies b_k=\sum^{n_{k}}_{n_{k-1}}a_i:S(b_k)\downarrow ,\ S(b_k)=S(a)

# 定序归组逆命题

S(bk)S(b_k)\downarrowbkb_k 归组的项符号相同,且顺序不变S(a)\implies S(a)\downarrowS(a)=S(bk)S(a)=S(b_k)

# Riemann 重排

S(a)αβRS(a)|\downarrow\implies \forall \alpha \leq \beta \in \mathbb R,存在求和次序使得排序后 lim infS=α,lim supS=β\liminf S'=\alpha,\limsup S'=\beta

# 绝对换序

S|S|\downarrow\implies 任意改变求和次序得到新数列 S|S'|\downarrow,且S=SS'=S

# 正项级数判别法

# Weierstrass 有界收敛

S+(a)S+(a)<+S_+(a)\downarrow \iff S_+(a)<+\infty

# 比较判别法

limanbn=r\lim\dfrac{a_{n}}{b_n}=r

r=0(S+(a)S+(b)),(S+(b)S+(a))r=0\implies (S_+(a)\uparrow\implies S _+(b)\uparrow),(S_+(b)\downarrow \implies S_+(a)\downarrow )

r=(S+(a)S+(b)),(S+(b)S+(a))r=\infty\implies (S_+(a)\downarrow\implies S _+(b)\downarrow),(S_+(b)\uparrow \implies S_+(a)\uparrow )

0<r<(S+(a)S+(b))0<r<\infty \implies (S_+(a)\downarrow \iff S_+(b)\downarrow)

# Cauchy 广义积分判别

x1,f(x)0,f(x)单调递减(S(f(n))1f(x)dx)x\geq 1,f(x)\geq 0,\ f(x)\text{单调递减}\implies (S(f(n))\downarrow \iff \displaystyle \int^\infty_1 f(x)\mathrm dx\downarrow )

# Cauchy 根式判敛

$\limsup \sqrtn=r<1 \implies S_+ \downarrow $

$\limsup \sqrtn=r>1 \implies S_+ \uparrow $

# D'Alembert 比值判敛

lim supan+1an=r<1S+\limsup \dfrac{a_{n+1}}{a_n}=r<1\implies S_+\downarrow

lim infan+1an=r>1S+\liminf \dfrac{a_{n+1}}{a_n}=r>1\implies S_+\uparrow

# Cauchy 根式蕴含 D'Alembert

\liminf \dfrac{a_{n+1}}{a_n}\leq \liminf\sqrt[n]{a_n}\leq \limsup\sqrt[n]{a_n}\leq \limsup \dfrac{a_{n+1}}

# 比值引理

an+1anbn+1bn(S+(a)S+(b)),(S+(b)S+(a))\dfrac{a_{n+1}}{a_n}\leq \dfrac{b_{n+1}}{b_n}\implies (S_+(a)\uparrow\implies S_+(b)\uparrow),(S_+(b)\downarrow \implies S_+(a)\downarrow)

# Raabe 判别

limn(anan+11)=r:(r<1S+),(r>1S+)\lim n\left(\dfrac{a_n}{a_{n+1}}-1\right)=r:(r<1\implies S_+\uparrow ),(r>1\implies S_+\downarrow)

anan+1=1+rn+o(1n),n:(r<1S+),(r>1S+)\dfrac{a_{n}}{a_{n+1}}=1+\dfrac rn+o(\dfrac 1n),\ n \to \infty :(r<1\implies S_+\uparrow ),(r>1\implies S_+\downarrow )

# Gauss 判别

anan+1=1+1n+rnlnn+o(1nlnn)1nlnrn/1(n+1)lnr(n+1),n:\dfrac{a_n}{a_{n+1}}=1+\dfrac1n+\dfrac r{n\ln n}+o(\dfrac1{n\ln n})\sim\dfrac1{n\ln^r n}{\large/}\dfrac{1}{(n+1)\ln^r(n+1)},n\to\infty:

r<1S+r<1\implies S_+\uparrow

r>1S+r>1\implies S_+\downarrow

# Kummer 判别

S+(a),S+(b):S_+(a),S_+(b):

1bnanan+11bn+1λ>0S+(a)\dfrac1{b_n}\dfrac{a_n}{a_{n+1}}-\dfrac1{b_{n+1}}\geq \lambda >0\implies S_+(a)\downarrow

1bnanan+11bn+10(S+(b)S+(a))\dfrac1{b_n}\dfrac{a_n}{a_{n+1}}-\dfrac1{b_{n+1}}\leq 0\implies (S_+(b)\uparrow\implies S_+(a)\uparrow)

# Cauchy 凝聚

a>0a>0 单调递减(S2ka2k)\implies (S\downarrow\iff \sum 2^ka_{2^k}\downarrow)

# 级数的乘法

# Cauchy

S(a),S(b)S(a)×S(b)|S(a)|\downarrow,|S(b)|\downarrow\implies|S(a)\times S(b)|\downarrow,且S(a)×S(b)=S(a)S(b)S(a)\times S(b)=S(a)S(b)

# Mertens

S(a),S(b)S(a)×S(b)|S(a)|\downarrow,S(b)\downarrow\implies S(a)\times S(b)\downarrow,且S(a)×S(b)=S(a)S(b)S(a)\times S(b)=S(a)S(b)

# Abel

S(a)×S(b)S(a)×S(b)=S(a)S(b)S(a)\times S(b)\downarrow\implies S(a)\times S(b)=S(a)S(b)

# 无穷乘积的敛散

# 必要条件

Plimnpn=1P\downarrow\implies\lim_{n\to\infty}p_n=1

# 收敛等价

P(a)S(ln(1+a))P(a)\downarrow\iff S(\ln(1+a))\downarrow

P(a)S(ln(1+a))S(a)P(|a|)\downarrow\iff S(\ln(1+|a|))\downarrow\iff S(|a|)\downarrow

S(a2)(S(a)P(a))S(a^2)\downarrow\implies(S(a)\downarrow\iff P(a)\downarrow)

# 发散推论

1<a<0(S(a)0P(ln(1+a)))-1<a<0\implies (S(a)\uparrow 0\iff P(\ln(1+a))\uparrow -\infty)

1<a<0,S(a)P(a)0-1<a<0,S(a)\uparrow\implies P(a)\uparrow0

S(a),S(a2)P(a)0S(a)\downarrow,S(a^2)\uparrow\implies P(a)\uparrow 0

# 绝对收敛

P(a)P(a)|P(a)|\downarrow\implies P(a)\downarrow

P(a)|P(a)|\downarrow\implies 任意改变乘积次序得到新乘积 P(b)|P(b)|\downarrow,且P(a)=P(b)P(a)=P(b)

# Riemann 重排

P(a)0αβ,lim supP(a)=β,lim infP(a)=αP(a)|\downarrow\implies \forall 0\leq\alpha\leq\beta\leq \infty,\limsup P(a)=\beta,\liminf P(a)=\alpha

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