高數(shù)下冊(cè)12十二章1節(jié)

1、用(比較、比值、根值)判別法判別正項(xiàng)級(jí)數(shù)的斂散性.2、用Leibniz判別法、絕對(duì)收斂法判別任意項(xiàng)級(jí)數(shù)的斂散性.3、絕對(duì)收斂與條件收斂的概念.4、求冪級(jí)數(shù)的收斂半徑與收斂域.5、利用逐項(xiàng)積分或求導(dǎo)求冪級(jí)數(shù)的和函數(shù).6、利用常見(jiàn)展式將簡(jiǎn)單函數(shù)展開成冪級(jí)數(shù).第十二章知識(shí)點(diǎn)復(fù)習(xí)：b=

a0

n

<mn

=m￥

n

>mxfi

￥

bxm

+axn

+1、lim2、x

fi

0時(shí),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均為常數(shù).￥(2)un為常數(shù)項(xiàng)級(jí)數(shù)n=1￥12.1無(wú)窮級(jí)數(shù)概念和性質(zhì)一、概念：1、無(wú)窮級(jí)數(shù)un

=

u1

+u2

++un

+n=1(1)un稱為un的通項(xiàng)(一般項(xiàng)).n=1un為x的函數(shù).￥(3)un為函數(shù)項(xiàng)級(jí)數(shù)n=1通項(xiàng)：un

=Sn

-Sn-1.2、無(wú)窮級(jí)數(shù)的斂散性：(1)部分和：Sn

=u1

+u2

++un.nfi

￥S

=

limSn.￥(2)un收斂于Sn=1￥余項(xiàng)：rn

=ui

=un+1

+un+2

+=S

-Sn.i=n+1nfi

￥例1、用定義判斷斂散性：un

fi

Sn

fi

limSn

fi

結(jié)論￥

1

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

=

1

=

1

-

1n(n+1)

n

n+1n+1Sn

==1-

1

nfi

￥limSn

==1收斂于1(1)解：un

=

1

-

1

n

n+1n+1nfi

￥

nfi

￥\

S

=limSn

=lim(1-

1

)

=1\原級(jí)數(shù)收斂于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

￥\原級(jí)數(shù)發(fā)散解：un

=

1

=

n+1-

nn+1-1k=1n\

Sn

=uk

==￥n=0(2)aqn

(a,q均為常數(shù),a

?0)(幾何級(jí)數(shù))a(1-qn

)1-qnkn+1aq

=解：S

=k=0\

S

=limS

=1-qan+1nfi

￥|

q

|<1不$

|

q

|?1￥\

aq

|

q

|<1發(fā)散|

q

|?1n

收斂n=0發(fā)散例2、證明￥1n(調(diào)和級(jí)數(shù))收斂于S,則有證明：反設(shè)1nn

=1￥n=1nfi

￥

nfi

￥nfi

￥limSn

=limS2n

=S

lim(S2n

-Sn

)

=0n+kn2n而S

-S

=1lim(S2n

-Sn

)?0

(與假設(shè)矛盾)發(fā)散nfi

￥￥n

=1\1n1?

=

n

nk=1

k=11n+n

2n=1

n=1二、級(jí)數(shù)收斂的性質(zhì)：￥

￥1、若un

=

A,則"

k

?

R,

kun

=

kAn=1

n=1￥

￥

￥2、若un

=

A,

vn

=

B,則

(un

–vn)

=

A–Bn=1￥3、un去(添)有限項(xiàng)后不改變斂散性；n=1收斂+發(fā)散

發(fā)散發(fā)散+發(fā)散

不定￥n=1￥n=1'n4、nuu

任意加括號(hào)后得新級(jí)數(shù)=A

(反之不一定)u

=

A

u

n

n￥

￥'n=1

n=1則：nfi

￥如：(1-1)+(1-1)++(1-1)+收斂于0而1-1+1-1++1-1+發(fā)散Sn

=

0

1

lim

Sn

$￥nfi

￥5、性質(zhì)5：un收斂

lim

un

=

0n=1(1)反之不一定(如：調(diào)和級(jí)數(shù))n=1￥nfi

￥(2) limun

?0

un發(fā)散.(1)

3

25nnnu

=

[(

)

+

]

￥

￥n=1

n=1例3、判斷斂散性：(幾何級(jí)數(shù))且￥￥￥

n=1

n=1+3n=11

25(

)nnnu

=解：(調(diào)和級(jí)數(shù))發(fā)散

￥

￥1n

25,(

)

收斂n

=1

n

=1n￥\原級(jí)數(shù)un發(fā)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(-)均收斂

25(

)

及nnn=1

n=1(幾何級(jí)數(shù))￥\原級(jí)數(shù)un收斂n=1n=1n=1n

n

+

32n

-1￥

￥(3)u

=

=lim

=

?0n+3

1nfi

￥

2n-1

2nnfi

￥\

limu2n

-1n解：u

=

n

+3￥\原級(jí)數(shù)un發(fā)散(性質(zhì)5)n=1小結(jié)：無(wú)窮級(jí)數(shù)1、相關(guān)概念2、斂散性的判斷方法3、性質(zhì)及其應(yīng)用4、常用結(jié)論的記用作業(yè)12--1：254頁(yè)3(1,2),4(2,3,5)發(fā)散2、調(diào)和級(jí)數(shù)：n=11nn=1￥常用結(jié)論：(