数分相关大一下第九章

作业

(数学分析习题集)习题7.4

一般项级数的敛散性1、1),

4);

5、2),

3);6;

9;

102010／03／05§9.4一般级数一、柯西收敛准则n⒈

a

收敛¥n=1"e

>0,

$N

˛

N*

,当n

>N时，<

e.n+

pk

=n+1k"p

˛

N*

,

恒有

a已证.¥例1.

{an

}单调减,

an

>

0,证明若

an收敛,n=1nfi

¥则lim

nan

=

0.22nn+1+

+

a

<

e

.证明："e

>0,

$N

>0,

n

>N时,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

=若对任意的e

>0和任意的正整数p,存在N

s.t.<

e.an+1

++

an+

p对"n

>N有¥问级数an是否收敛？n=1¥

e

p，取N

=

p

"e

>0,和任意正数1

,n=1

n<

en

+1pn+

pn+1++

a

<当n

>N时，a二、莱布尼茨（Leibniz）判别法⒈¥n=1n-1(-1)

an

,

an

>

0,设交错级数则¥n=1nn-1(-1)

a

收敛.knn+

pS

-

S

=n+

pk

=n+1(-1)k

-1

a若{an

}递减趋于0,称此类交错级数为

Leibniz

级数证明："n,p

>0,

p为偶数时：=

(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为奇数时，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

收敛.<

eRn

=S

-Sn

£

an+1

(令p

fi+¥

即得)1¥n例1.

⑴

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

>0)

收敛特别+ -

+1

-

收敛⑵nln

n1

1

1

1

12

3

4 5

-

61¥(-1)n=2收敛¥⑶

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

(n

+

1

+

n¥n

+

1

-

n)

=

n=1⒉Leibniz级数副产品(误差估计)：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

说明：Leibniz判别法中单调性条件不可少三、Abel和Dirichlet判别法¥n=1——

anbn的判敛1.分部求和公式：此时

Sk

=

a1

+

a2

+

+

ak

,

S0

=

0.

ak

bkn

n-1k

=1

k

=1=

Sk

(bk

-

bk

+1)

+

Snbn设{an

},{bn

}是两列实数,则对任意正整数n有注：

DSk

=

Sk

-

Sk

-1

=

ak

,

Dbk

=

bk

+1

-

bkk

k

k

k k

0

可视为nb

DS=

S

bn

-

n-1Sk

Dbk

=1k

=1类似于分部积分公式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证明：2.

Abel引理：设{bn

}单调，Sk

=

a1

+

a2

+

+

ak

,nk

=1若Sk

£

M

,

k

=

1,2,,

n,则

ak

bk

£

M

(

b1

+

2

bn

).n

n-1k

=1

k

=1=

Sk

(bk

-

bk

+1

)

+

Snbn证明：

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单调⒉Sk

有界；３.Dirichlet判别法设{an

},{bn

}是两个数列,

Sk

=

a1

+

a2

+

+

ak

,¥如果它们满足：⒈{bn

}单调fi

0;则

ak

bk收敛.k

=1注:

取a

=

(-1)k

-1

,

|S

|£

1,

b

fl

0,k

k

k则¥kk

-1(-1)

b

收敛.k

=1Dirichlet判别法是Leibniz判别法的推广=

Sn+

p

-

Sn

£

2M证明：an+1

+an+2

+

+an+p£

2M

(

bn+1

+

2

bn+

p

)nfi

¥

lim

bn

=

08Me

,n+1\

"

e

>

0,

$N

˛

N*

,

n

>

N时,

b

<8Me

.b

<n+

p£

2M

(

e

+

2e

)

<

e,8M

8M"

p,n+

pk

=n+1

ak

bk¥\

ak

bk收敛.k

=1n+

p

ak

bkk

=n+1４.阿贝尔（Ａbel）判别法¥k

=1设{ak

},{bk

}满足

1.

{bk

}单调有界

2.

ak收敛，¥则ak

bk收敛.k

=1证明：{bk

}单调有界,\{bk

-b}单调fi

0¥k

=1¥由Dirichlet法,

ak(bk

-

b)收敛.k

=1¥

¥

¥k

=1k

=1k

=1

ak

bk

=

ak

(bk

-

b)

+

b

ak收敛\lim

bk

=b存在,kfi¥

ak收敛,

\

{Sk

}有界.例2.¥nsin

nxn=1p(0

<

x

<

p

,

p

>

0)

"

x

˛

(0,p

)np

1

单调fi

0;nnk

=1k

=12xx

sin

kx

sin2sin1

sin

kx

=2sin1x£由Dirichlet判别法,¥n=1pnsin

nx收敛.同理可证:¥n=1pncos

nx收敛.x

„

2kp例3.¥

cos

3n(1

+

1

)nn=1

n

n解：¥⒈

n=1ncos

3n收敛n⒉

(1

+

1

)n

fi

e)单调有界.(1

+\n1n由Abel，级数收敛.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收敛（莱法）¥1由

(-1)n=1nn¥¥(-1)=

n=1n=1nnncos

2ncos(2n

+np

)

收敛（例2）故原级数收敛.例5.n1

(1

+

1

)n

(5

-

arctan

n)ln

n

n(-1)¥n=21

收敛(莱法)ln

nn¥n=2(-1)解：

(1

+1

)n

单调有界n收敛(Abel

)1

(1

+

1

)nln

n

n\¥(-1)n=2n{5

-arctan

n}单调有界,故原级数收敛（Abel）.n=2

n=2¥

¥例6.若{nxn

}收敛，

n(xn

-xn-1

)收敛，则级数

xn收敛k解：

令an

=

xn

,

bn

=1,则Bk

=

bi

=

k,由Abel变换i=1n-1n=

nxn

-

k

(xk

+1

-

xk

)