數(shù)分相關(guān)大一下第九章

作業(yè)

(數(shù)學(xué)分析習(xí)題集)習(xí)題7.4

一般項(xiàng)級(jí)數(shù)的斂散性1、1),

4);

5、2),

3);6;

9;

102010／03／05§9.4一般級(jí)數(shù)一、柯西收斂準(zhǔn)則n⒈

a

收斂￥n=1"e

>0,

$N

?

N*

,當(dāng)n

>N時(shí)，<

e.n+

pk

=n+1k"p

?

N*

,

恒有

a已證.￥例1.

{an

}單調(diào)減,

an

>

0,證明若

an收斂,n=1nfi

￥則lim

nan

=

0.22nn+1+

+

a

<

e

.證明："e

>0,

$N

>0,

n

>N時(shí),aan

fl,

\

2na2n

￡

2

an+1

+

+

a2n

<

e.\

lim

2na2n

=

0nfi

￥(2n

+

1)a2n+1

￡

2na2n

+

a2n+1

fi

0.

(n

fi

￥

)nfi

￥\

lim

nan

=

0nn

ln

n1反例:

a

=若對任意的e

>0和任意的正整數(shù)p,存在N

s.t.<

e.an+1

++

an+

p對"n

>N有￥問級(jí)數(shù)an是否收斂？n=1￥

e

p，取N

=

p

"e

>0,和任意正數(shù)1

,n=1

n<

en

+1pn+

pn+1++

a

<當(dāng)n

>N時(shí)，a二、萊布尼茨（Leibniz）判別法⒈￥n=1n-1(-1)

an

,

an

>

0,設(shè)交錯(cuò)級(jí)數(shù)則￥n=1nn-1(-1)

a

收斂.knn+

pS

-

S

=n+

pk

=n+1(-1)k

-1

a若{an

}遞減趨于0,稱此類交錯(cuò)級(jí)數(shù)為

Leibniz

級(jí)數(shù)證明："n,p

>0,

p為偶數(shù)時(shí)：=

(an+1

-

an+2

+

an+3

-

an+4

+

+

an+

p-1

-

an+

p

)=

an+1

-(an+2

-

an+3

)

-

-

(an+

p-2

-

an+

p-1

)

-

an+

pn+1￡

ap為奇數(shù)時(shí)，Sn+

p

-

Sn

=

an+1

-

(an+2

-

an+3

)

-

-

(an+

p-1

-

an+

p

)n+1￡

a￥n=1nn-1n\lim

S

存在,nfi

￥2<

e2(-1)

a

收斂.<

eRn

=S

-Sn

￡

an+1

(令p

fi+￥

即得)1￥n例1.

⑴

n=1p(-1)n-1(p

>0)

收斂特別+ -

+1

-

收斂⑵nln

n1

1

1

1

12

3

4 5

-

61￥(-1)n=2收斂￥⑶

n=1(-1)n-1(-1)n-1

(n

+

1

+

n￥n

+

1

-

n)

=

n=1⒉Leibniz級(jí)數(shù)副產(chǎn)品(誤差估計(jì))：lim￥nnfi

￥nna

=

0(4)

(-1)n-1

an=1(a

?0)且￥￥=

(-1)1n=1

nn=2(-1)n-

nnn

+1n

+(-1)n=

( 2

-1)

-

( 3

+1)

+21

說明：Leibniz判別法中單調(diào)性條件不可少三、Abel和Dirichlet判別法￥n=1——

anbn的判斂1.分部求和公式：此時(shí)

Sk

=

a1

+

a2

+

+

ak

,

S0

=

0.

ak

bkn

n-1k

=1

k

=1=

Sk

(bk

-

bk

+1)

+

Snbn設(shè){an

},{bn

}是兩列實(shí)數(shù),則對任意正整數(shù)n有注：

DSk

=

Sk

-

Sk

-1

=

ak

,

Dbk

=

bk

+1

-

bkk

k

k

k k

0

可視為nb

DS=

S

bn

-

n-1Sk

Dbk

=1k

=1類似于分部積分公式n

n

n

nk

=1k

=1k

=1k

=1

ak

bk

=

(

Sk

-

Sk

-1

)bk

=

Sk

bk

-

Sk

-1bkn

n-1

n-1=

Sk

bk

-

Sk

bk

+1

=

Sk

(bk

-

bk

+1

)

+

Snbnk

=1

k

=0

k

=1證明：2.

Abel引理：設(shè){bn

}單調(diào)，Sk

=

a1

+

a2

+

+

ak

,nk

=1若Sk

￡

M

,

k

=

1,2,,

n,則

ak

bk

￡

M

(

b1

+

2

bn

).n

n-1k

=1

k

=1=

Sk

(bk

-

bk

+1

)

+

Snbn證明：

ak

bkk

=1￡

Sk

bk

-

bk

+1

+

Snbn

￡

M

(

bk

-

bk

+1

+

bn

)n-1

n-1k

=1=

M

(

b1

-

bn

+

bn

)

￡

M

(

b1

+

2

bn

).bn單調(diào)⒉Sk

有界；３.Dirichlet判別法設(shè){an

},{bn

}是兩個(gè)數(shù)列,

Sk

=

a1

+

a2

+

+

ak

,￥如果它們滿足：⒈{bn

}單調(diào)fi

0;則

ak

bk收斂.k

=1注:

取a

=

(-1)k

-1

,

|S

|￡

1,

b

fl

0,k

k

k則￥kk

-1(-1)

b

收斂.k

=1Dirichlet判別法是Leibniz判別法的推廣=

Sn+

p

-

Sn

￡

2M證明：an+1

+an+2

+

+an+p￡

2M

(

bn+1

+

2

bn+

p

)nfi

￥

lim

bn

=

08Me

,n+1\

"

e

>

0,

$N

?

N*

,

n

>

N時(shí),

b

<8Me

.b

<n+

p￡

2M

(

e

+

2e

)

<

e,8M

8M"

p,n+

pk

=n+1

ak

bk￥\

ak

bk收斂.k

=1n+

p

ak

bkk

=n+1４.阿貝爾（Ａbel）判別法￥k

=1設(shè){ak

},{bk

}滿足

1.

{bk

}單調(diào)有界

2.

ak收斂，￥則ak

bk收斂.k

=1證明：{bk

}單調(diào)有界,\{bk

-b}單調(diào)fi

0￥k

=1￥由Dirichlet法,

ak(bk

-

b)收斂.k

=1￥

￥

￥k

=1k

=1k

=1

ak

bk

=

ak

(bk

-

b)

+

b

ak收斂\lim

bk

=b存在,kfi￥

ak收斂,

\

{Sk

}有界.例2.￥nsin

nxn=1p(0

<

x

<

p

,

p

>

0)

"

x

?

(0,p

)np

1

單調(diào)fi

0;nnk

=1k

=12xx

sin

kx

sin2sin1

sin

kx

=2sin1x￡由Dirichlet判別法,￥n=1pnsin

nx收斂.同理可證:￥n=1pncos

nx收斂.x

?

2kp例3.￥

cos

3n(1

+

1

)nn=1

n

n解：￥⒈

n=1ncos

3n收斂n⒉

(1

+

1

)n

fi

e)單調(diào)有界.(1

+\n1n由Abel，級(jí)數(shù)收斂.nn

sin2

n￥n=1例4.

(-1)解：sin2

n

=1

-cos

2n￥￥￥1=

(-1)n=1-

(-1)n=1n\

(-1)n=12nn+1

cos

2nn

2n2n

sin2

n收斂（萊法）￥1由

(-1)n=1nn￥￥(-1)=

n=1n=1nnncos

2ncos(2n

+np

)

收斂（例2）故原級(jí)數(shù)收斂.例5.n1

(1

+

1

)n

(5

-

arctan

n)ln

n

n(-1)￥n=21

收斂(萊法)ln

nn￥n=2(-1)解：

(1

+1

)n

單調(diào)有界n收斂(Abel

)1

(1

+

1

)nln

n

n\￥(-1)n=2n{5

-arctan

n}單調(diào)有界,故原級(jí)數(shù)收斂（Abel）.n=2

n=2￥

￥例6.若{nxn

}收斂，

n(xn

-xn-1

)收斂，則級(jí)數(shù)

xn收斂k解：

令an

=

xn

,

bn

=1,則Bk

=

bi

=

k,由Abel變換i=1n-1n=

nxn

-

k

(xk

+1

-

xk

)