高数下册12十二章1节

1、用(比较、比值、根值)判别法判别正项级数的敛散性.2、用Leibniz判别法、绝对收敛法判别任意项级数的敛散性.3、绝对收敛与条件收敛的概念.4、求幂级数的收敛半径与收敛域.5、利用逐项积分或求导求幂级数的和函数.6、利用常见展式将简单函数展开成幂级数.第十二章知识点复习：b=

a0

n

<mn

=m¥

n

>mxfi

¥

bxm

+axn

+1、lim2、x

fi

0时,x

~

sin

x

~

tan

x

~

arcsinx~

arctanx

~

ln(1+

x)

~

ex

-1；21-cosx

~

1

x2；(1+

x)a

-1~

ax.=

0limlnalna1nfi

¥n1n1n

a

=

a

=

en1lnn1nfi

¥

nlnnlim

=lim

=0nfi

¥n1n1nn

=

n

=

en4、a

=elna

limn

a

=limn

n

=1

(a

>0)nfi

¥

nfi

¥n

n+1nfi

¥

nfi

¥3、lim(1+

1

)n

=

e

lim(

n

)n

=

e-1-(n+1)

-nn+1

-(n+1)(

n

)n

=[1+

1

]nfi

¥

n+1n+1且lim

-n

=

-1¥un均为常数.¥(2)un为常数项级数n=1¥12.1无穷级数概念和性质一、概念：1、无穷级数un

=

u1

+u2

++un

+n=1(1)un称为un的通项(一般项).n=1un为x的函数.¥(3)un为函数项级数n=1通项：un

=Sn

-Sn-1.2、无穷级数的敛散性：(1)部分和：Sn

=u1

+u2

++un.nfi

¥S

=

limSn.¥(2)un收敛于Sn=1¥余项：rn

=ui

=un+1

+un+2

+=S

-Sn.i=n+1nfi

¥例1、用定义判断敛散性：un

fi

Sn

fi

limSn

fi

结论¥

1

n(n+1)(1)n=1un

=

1

=

1

-

1n(n+1)

n

n+1n+1Sn

==1-

1

nfi

¥limSn

==1收敛于1(1)解：un

=

1

-

1

n

n+1n+1nfi

¥

nfi

¥\

S

=limSn

=lim(1-

1

)

=1\原级数收敛于1

1

n+1nk=1k\

Sn

=u

==1-¥n=1

1

n+1+

nex：n+1+

n\

S

=

limSn

=

lim(

n

+1

-1)

=¥nfi

¥

nfi

¥\原级数发散解：un

=

1

=

n+1-

nn+1-1k=1n\

Sn

=uk

==¥n=0(2)aqn

(a,q均为常数,a

„0)(几何级数)a(1-qn

)1-qnkn+1aq

=解：S

=k=0\

S

=limS

=1-qan+1nfi

¥|

q

|<1不$

|

q

|‡1¥\

aq

|

q

|<1发散|

q

|‡1n

收敛n=0发散例2、证明¥1n(调和级数)收敛于S,则有证明：反设1nn

=1¥n=1nfi

¥

nfi

¥nfi

¥limSn

=limS2n

=S

lim(S2n

-Sn

)

=0n+kn2n而S

-S

=1lim(S2n

-Sn

)„0

(与假设矛盾)发散nfi

¥¥n

=1\1n1‡

=

n

nk=1

k=11n+n

2n=1

n=1二、级数收敛的性质：¥

¥1、若un

=

A,则"

k

˛

R,

kun

=

kAn=1

n=1¥

¥

¥2、若un

=

A,

vn

=

B,则

(un

–vn)

=

A–Bn=1¥3、un去(添)有限项后不改变敛散性；n=1收敛+发散

发散发散+发散

不定¥n=1¥n=1'n4、nuu

任意加括号后得新级数=A

(反之不一定)u

=

A

u

n

n¥

¥'n=1

n=1则：nfi

¥如：(1-1)+(1-1)++(1-1)+收敛于0而1-1+1-1++1-1+发散Sn

=

0

1

lim

Sn

$¥nfi

¥5、性质5：un收敛

lim

un

=

0n=1(1)反之不一定(如：调和级数)n=1¥nfi

¥(2) limun

„0

un发散.(1)

3

25nnnu

=

[(

)

+

]

¥

¥n=1

n=1例3、判断敛散性：(几何级数)且¥¥¥

n=1

n=1+3n=11

25(

)nnnu

=解：(调和级数)发散

¥

¥1n

25,(

)

收敛n

=1

n

=1n¥\原级数un发散n=1(2)52nn¥n=1[(2)n

-(-1)n-1

3

]¥n=1u

=¥¥¥21(-

)+3n=125(

)nnn=1n=1n

u

=解：且

¥

¥21(-)均收敛

25(

)

及nnn=1

n=1(几何级数)¥\原级数un收敛n=1n=1n=1n

n

+

32n

-1¥

¥(3)u

=

=lim

=

„0n+3

1nfi

¥

2n-1

2nnfi

¥\

limu2n

-1n解：u

=

n

+3¥\原级数un发散(性质5)n=1小结：无穷级数1、相关概念2、敛散性的判断方法3、性质及其应用4、常用结论的记用作业12--1：254页3(1,2),4(2,3,5)发散2、调和级数：n=11nn=1¥常用结论：(