Chapter 7
Unit 1
数项级数
New Words
series of numbers 数项级数
partial sums 部分和
divergence 发散
convergence 收敛
the harmonic series 调和级数
alternate series 交错级数
positive series 正项级数
absolute convergence 绝对收敛
conditional convergence 条件收敛
convergence radius 收敛半径
convergence domain 收敛域
convergence interval 收敛区间
Purpose of teaching
1 To understand the concept of series、positive series、
alternate series、power series；
2 To understand the concept of convergence 、absolutely
convergence、conditionally convergence and
divergence for series；
3 To know the properties of convergence of series;
4 To master all tests for positive series: comparison test、
ratio test、root test；
5 To master Leibniz test for alternate series;
In Chapter 2, we studied the convergence of sequences
of numbers. In particular, we proved the important
result: “A monotone sequence of numbers converges if
and only if it is bounded”.
In this section we discuss the convergence of series of
numbers. The above result together with some limit
formulas are exactly what is needed to establish the
convergence of the series of numbers
1. The concept of series of numbers
At first we give the concept of series of numbers
Definition 1
Suppose that u n is a sequence of numbers,
then the following form

u
n 1
n
 u1  u 2  u 3    u n  
is called the series of numbers, where u n is
called the n  th term or the general term of
the series.
We define
n
sn   ui for every natural number n
i 1
and obtain a new sequence of numbers sn is called

the sequence of partial sums for the series
u .
n 1
n
2.The convergence and divergence of series
Definition 2
An infinite series

u
n 1
n
is convergent , if lim S n  S ,
n 
and S is called the sum of this infinite series; we also
say this series converges to S . Otherwise, we say that
it is divergent.
Remarks

1 That is,  un is convergent divergent   lim
Sn
n 
n 1
exists does not exist 
2 The term

rn  S  S n  u n 1  u n  2     u n i
i 1
is called the remainder term of this series
The following examples will use the definition
Example 1
Examine the geometrica l series for convergenc e :

n
2
n 1
aq

a

aq

aq



aq
 a  0

n 0
Solution
If q  1, then the sequence of partial sums is
s n  a  aq  aq 2    aq n 1
a  aq n

1 q
a
aq n


,
1 q 1 q


 a
n
,
as
q

1

lim
q
0
1  q
n 
 lim s n  
n 
 , as q  1 lim q n  
n 


That is, the geometrica l series  aq n is convergent


n 0
a
to
as q  1, and it is divergent as q  1.
1 q
If q  1, then sn  na  , the series is divergent
If q  1, then the series is a  a  a  a  
 lim s n does not exist, it is divergent.
n 
a

is convergent to
, when q  1,
n
1 q
Thus,  aq 
n 0
is divergent, when q  1


Example 2
Determine the following series for converence
1
1
1



2n  1  2n  1
1 3 3  5
Solution
1
1
1
1
 un 
 (

),
( 2n  1)( 2n  1) 2 2n  1 2n  1
1
1
1
 sn 


1 3 3 5
(2n  1)  (2n  1)
1
1 1 1 1
1
1
1
 (1  )  (  )    (

)
2
3 2 3 5
2 2n  1 2n  1
1
1
 (1 
),
2
2n  1
1
1
1
 lim s n  lim (1 
) ,
n 
n  2
2n  1
2
1
 the series is convergent , and the sum is .
2
Example 3
Determine the following series for converence

1

n 1
n  n 1
Solution
Since the n - th term
1
un 
 n 1  n
n  n 1
Thus,
1
sn 

1
1

n  n 1
2 3
1 2
 2 1  3  2    n 1  n

 

 n  1  1   n   


That is the original series diverge
3. The basic properties of a series of numbers
The series of numbers has the following properties
used often
Property 1


 ku k is a constant  converge , if  u
n 1
n
n 1
n
converge
Property 2

If the two series

 u and  v are convergent ,
n 1
n
n 1

then so is the series
 u
n 1

 u
n 1
n
n
n
 v n  and


n 1
n 1
 vn    u n   vn
Property 1 and 2 of the series of numbers imply the
linearity property of convergent series.
Property 3

If
u
n 1

n
converge , then
 u k  1 converge
n 1
nk
Proof
Let s n  u1  u 2    u n , and
 n  u k 1  u k  2    u k  n  s n  k  s k ,
Thus lim  n  lim s n  k  lim s k  s  s k .
n 
n 
n 
Property 3 implies that the convergence of the series
does not change when adding or reducing finite terms
before the series.
Property 4
The series after adding the bracket of a convergent series
is still convergent to its original sum
Proof
Let S n  u1  u 2    u n is the partial sums of
the series, and add any brackets, for instance,
(u1  u 2 )  (u 3  u 4  u 5 )  
The sequence of its partial sums is  1  s 2 ,  2  s5 ,
 3  s9 ,,  m  s n ,
Thus, lim  m  lim s n  s.
m 
n 
Remark
(1) The series moved off brackets of a convergence
series is not surely convergent
For example,
the series (1  1)  (1  1)   converge to 0,
but after moving off its brackets, the series is
1 1  1 1  ,
it diverges.
(2) If the series after adding the bracket is divergent,
then the original series is also divergent
Example 4

If the series
u
n 1

n
converges to S , then
 u
n 1
n
 u n 1 
is  C ?
 A convergent to 2S ;
B  convergent to 2S  u1 ;
C  convergent to 2S  u1 ; D  divergent
4. The necessary condition of convergence of
series

Theorem 1 If
u
n 0
n
converges , then lim u n  0
n 
Proof

n
n 1
k 1
Define s   u n and s n   u k , then u n  s n  s n 1 ,
By the difference property of convergent sequences,
 lim u n  lim s n  lim s n 1  s  s  0
n 
n 
n 
Remarks
1. The necessary condition tells us : " if the general

term u n does not approach 0, then
For example, since lim (1)
n 
n 1
u
n 1
n
diverges"
n
 0,
n 1
1 2 3
n
n 1
      (1)
  is divergent
2 3 4
n 1
2 But this condition is not sufficient, it is only a
necessary condition.
1
For example, the general term u n   0 n    of
n
the harmonic series
1 1
1
1   
2 3
n
however, the series is divergent
Proof
We do this grouping the terms of the harmonic series
in a particular way:
1 1 1 1
1     
2 3 4 5
2 terms
2 terms
4 terms
8 terms
1
1 1
1 1 1 1
1 1
1
(1  )  (  )  (    )  (     )
16
9 10
5 6 7 8
3 4
2
1
1
1
   m 1 )  
 m
  ( m
2
2 1 2  2
2 mterms
1 1 1 1
1    
2 2 2 2
The right-hand side of this inequality is clearly divergent
so the harmonic series is divergent
5 The series of positive terms and its properties
There are many tests that can be used to be determine
the convergent properties of infinite series. In this
block we examine the convergence of series of positive
terms.
Definition 3

If all u n  0 of  u n , then
n 1
series of positive terms

u
n 1
n
is said to be a
Remark

1 If  u n is a series of positive
terms, then the
n 1
sequence {s n } of its partial sum is increasing ,
that is, s1  s 2    s n  
Since the sequence of its partial sums is monotonically
increasing. The Monotone Convergence Theorem
asserts that the sequence of partial sums converges if
and only if the sequence of partial sums is bounded.
Hence we can obtain the following test:
Theorem 2

A series
u
n 1
n
of positive terms is convergent
 the sequence {s n } of its partial sum is bounded
This test is important to obtain the following tests
The Comparison Test


n 1
n 1
Let  u n and  v n be series of positive terms,
and u n  v n n  1,2,, then, it is possible to
show that :


n 1

n 1

1 if  vn is convergent , then  u n is also convergent ;
2 if  u n is divergent,
n 1
then
v
n 1
n
is also divergent
n
n

k 1
k 1
n 1
Proof Let S n   u k ,  n   v k ,    v n ,

1 if  vn converges,  u n  vn ,
n 1
 S n  u1  u 2    u n  v1  v 2    v n   n   ,
that is, the sequence {S n }of partial sums is bounded,


u
n 1
n
is convergent

(2) if
u
n 1
n
diverges, then S n   (n  ),
 u n  v n   n  S n  , thus,

v
n 1
n
is divergent
This test, which is one of the most useful and widely
used convergence tests, applies only to series of
positive terms
Corollary of the comparison test

If a series of positive terms
u
n 1
n
is convergent
divergent , and vn  kun n  N or kun  vn ,

then,
v
n 1
n
is convergent divergent

Example 5
To examine the convergenc e of the P  series
1
1
1
1  p  p    p    p  0
2
3
n
Solution
1
1
If p  1, since for every natural number n, p  ,
n
n

1
and the harmonic series  diverges, it follows
n 1 n

1
from the Comparison Test that the P - series  p
n 1 n
is divergent
n dx
1
If p  1, see figure, we have p  
p
n

1
n
x
2 dx
n dx
1
1
1
 Sn  1  p  p    p  1   p    
1 x
n 1 x p
2
3
n
dx
 1  p
1 x
1
1
 1
(1  p 1 )
p 1
n
n
y
y
o
1
2
1
( p  1)
xp
3
4
x
1
 1
p 1
that is, S n is bounded, thus the P  series converges
1 convergent , as p  1;
 the P  series  p is 
n 1 n
divergent, as p  1

Example 6
Discussing the convergenc e of the following series

 n 



n 1  3n  1 
Solution
n
n
n

n
 n 
1
1

    , and the series    converges,
 3n  1 
 3
n 1  3 
it follows the Comparison Test that the original series
converges.
The important reference series are:
the geometric series, the P-series, the harmonic series.

Example 7 Prove that

n 1
Proof

1
1
is divergent
n(n  1)
1

, since
n(n  1) n  1

1
is divergent,

n 1 n  1
it follows the Comparison Test that the series

1
is also divergent .

n(n  1)
n 1
Example 8
1
1 
Suppose that a1  2, a n 1   a n  n  1,2,,
2
an 
prove that 1 lim a n exists;
n 
 an

2 The seires  
 1 converges
n 1  a n 1

Proof

1 an 1
1
1 
1
  a n    a n 
 1,
2
an 
an
1  a n2
1
1 
a n 1  a n   a n    a n 
0
2
an 
2a n
 a n is decreasing and bounded, thus lim a n exists
n 
2 Apply 1, we have
an
a n  a n 1
0
1 
 a n  a n 1 ,
a n 1
a n 1
n
let S n   a k  a k 1   a1  a n 1 , lim a n 1 exists,
k 1
n 
 lim S n exists, thus
n 

 a
n 1
n
 a n 1  converges,
it follows the Comparison Test that that the seires
 an


 1 converges

n 1  a n 1


The Limit Form of The Comparison Test


n 1
n 1
Let  un and  vn be series of positive terms, and
un
lim
 l , then, it is possible to show that :
n  v
n
1 if
0  l  , then the convergenc e of

u
n 1
n

v
n 1
n
is the same;

2 l  0, if  vn converges,

then
n 1
n 1


3 l  , if  vn diverges,
n 1
u
then
n
u
n 1
n
converges
diverges.
Proof
un
l
(1)  lim
 l , for    0,  N , when n  N ,
n v
2
n
and
l un
l
we have l  
l
2 vn
2
l
3l
that is v n  u n  vn (n  N ), it follows the
2
2
corollary of the Comparison Test that this
conclusion is correct.
Similarly, we can obtain others results
Example 9
To examine the convergenc e of the following series

1
1  sin ;
n
n 1

1
2   n
n 1 3  n
Solution
1
sin

1
1
n
1 lim
n sin  lim
 1,  sin is divergent
n 
n n  1
n
n 1
n
1

n
1
1
3

n
2 lim
 lim
 1, and  n is convergent
n 
n 
1
n
n 1 3
1 n
n
3
3

1
 n
is convergent
n 1 3  n
Example 10
To examine the convergenc e of the following series
1
 1 

 n  ln 1  n ;


n 1 

Solution
By applying the second Tayor formula of ln 1  x 
at x  0, we obtain
1
1
 1  1 1
 1
 ln 1       2  o 2
n
 n  n  n 2n
n
1
 1 
 2  o 2 
2n
n 



1
1
 1
 1 
 ln 1  
 o 2 
2
n
n
2n
n  1


Thus, lim
 lim
 .
n 
n 
1
1
2
n2
n2

1
And the series  2 converegs, thus the original
n 1 n
series converges
The Ratio Test (D’Alembert Test)

Let  un be a series of positive terms, suppose that
n 1
un 1
lim
  , then, it is possible to show that :
n  u
n

1 if   1, then  un is convergent
n 1

2 if   1 or   , then  un is divergent
3 if

n 1
  1, then  u n may be convergent or
n 1
divergent
Proof
un 1
(1)  lim
   1, for   0,  N , when n  N ,
n  u
n
un 1
un 1
we have
    , that is,    
  
un
un
let   1   , such that r      1,
u N 2  ru N 1 , u N 3  ru N 2  r 2 u N 1 , , u N m  r m1u N 1 ,

and the series
m 1
r
 u N 1 is convergent ,
m 1

  u N m 
m 1

u
n  N 1

u
is convergent ,   uu is convergent
n 1
Example 11
To examine the convergenc e of the following series

1
1  ;
n 1 n!

n!
2   n ;
n 1 10

1
3 
n 1 2n  1  2n
Solution
1
u n 1 (n  1)!
1
1


 0 (n  ),
1
un
n 1
n!

1
thus,  is convergent
n 1 n!
u n 1 (n  1)! 10 n n  1
2



  (n  ),
n 1
un
n!
10
10

n!
thus,  n is divergent
n 1 10
u n 1
(2n  1)  2n
3 lim
 lim
 1, in
n  u
n  ( 2n  1)  ( 2n  2)
n
this caes, we can not use the Ratio Test.

1
1
1

 2 , and  2 is convergent ,
(2n  1)  2n n
n 1 n
it follows the Comparison Test that

1
the series 
is convergent
n 1 2n  ( 2n  1)
The Root Test ( Cauchy’s Test )

Let  u n be a series of positive terms, if
n 1
lim n u n   , then, it is possible to show that :
n 

1 if   1, then  un is convergent
n 1

2 if   1 or   , then  un is divergent
n 1

3 if   1, then  un may be convergent
n 1
Example 12
To examine the convergenc e of the series

2
 n  1n
n 1
Solution
 lim u n  lim 2
n
n 

 2
n 1
n 
 n   1n

1n
1
n
1
 1
2
is convergent
or divergent
Example 13
To examine the convergenc e of the following series
n2

 1
1   ;
 n
1
1  n
n 1 2
2   1
2 
n
2
n 1

n
Solution
1 By the Root Test,
n
1 1
e
since lim u n  lim 1     1, thus the
n 
n  2
2
 n
original series diverges.
n
By the Comparison Test, since

2   1
3
3
0  un 
 n , and  n converges, thus
n
2
2
n 1 2
the original series converges.
n
6 The tests for convergence of the alternating
series
Definition 4
An alternating series is a special type of series in
which the sign changes from one term to the next.
They have the form

n 1
(

1
)
un or

n 1

n
(

1
)
 un un  0
n 1
For series of this type there is a simple criterion for
convergence.
The Alternating Series Test ( Leibniz’s Test )
The alternatin g series
u1  u 2  u 3  u 4  
in which each ui , i  1,2,, is a positive number 
is convergent , if
1 u n  u n1 n  1,2,;
2 lim
un  0
n 
Proof
Define
n
S n    1 u k for every natural number n.
k 1
k 1
In order to prove that the sequence of partial sums
S n converges, we will first show that the
subsequenc e S 2n  converges.
Indeed,  u n 1  u n  0, and
S 2 n  (u1  u 2 )  (u 3  u 4 )    (u 2 n 1  u 2 n )
We conclude that S 2 n is monotonica lly increasing ,
and
S 2 n  u1  (u 2  u 3 )    (u 2 n  2  u 2 n 1 )  u 2 n  u1
 S 2 n is bounded, By the Monotone Convergenc e
Theorem, the sequence S n  converges, define
lim S 2 n  S  u1 and lim u 2 n 1  0,
n 
n 
 lim S 2 n 1  lim ( S 2 n  u 2 n 1 )  S , thus the
n 
n 
original series converges.
Example 14
To examine the convergenc e of the series


n 1
 1
n
n
n 1
Solution
x
 (1  x)
(
) 
 0 ( x  2)
2
x 1
2 x ( x  1)
x

is decreasing , thus un  un 1 ,
x 1
n


n
 1 n
and lim un  lim
 0, 
converge
n 
n  n  1
n 1
n 1
6 Absolute and conditional convergence
The comparison test, ratio test and root test apply to
series of positive terms. However, not all series are
series of positive terms. To apply these tests such
series must first be converted into series of positive
terms. Here, we will introduce the concept of the
absolute convergence
Theorem 3 ( The Absolute Convergence Test)

If
u
n 1

n
is convergent , then
u
n 1
n
is convergent
Proof
1
Let vn  (un  un ) ( n  1,2, ), then, vn  0,
2

and vn  un ,   vn is convergent , also
n 1



n 1
n 1
n 1
  un   ( 2v n  un ),  un is convergent
Taking the absolute value
Series
Series of
positive terms
Definition 6

1 If  un

is convergent , then
n 1
u
n 1
n
is siad to be
absolutely convergent ;

2 If  un
n 1

is divergent , but  un is convergent , then
n 1

u
n 1
n
is siad to be conditiona lly convergent
Example 15
To examine the convergenc e of the series

sin n

2
n
n 1
Solution

sin n
1
1
 2  2 , and  2 converge,
n
n
n 1 n


n 1
sin n
convrege
2
n
Thus, the original series converges absolutely
Example 16
Suppose a scalar   0 and the series

2
a
 n converges
n 1

, then the series
  1
n
an


C
n2  
 A diverges;
B  converges conditiona lly;
C  concerges absolutely ;
D  the convergenc e is dependent on .
n 1
Solution
  1
n
an
1 2 
1



an  
2
2
n   2 
n






2





and  a n
n 1
2

  a n2 converges,

1



n 1 
n2  
n 1
2


1
  2
converges, it follows

n 1 n  


an
n
the Comparison Test that the series   1
n 1
n2  
converges, thus the original series converges absolutely .

The theory of convergent and divergent series is a
broad and deep subject. The present section gives
but a brief glimpse of some ways in which the
monotone Convergence Theorem for sequences of
numbers can be applied to obtain criteria that are
sufficient for a series to converge.