SỰ HỘI TỤ CỦA MỘT CHUỖI CON CỦA CHUỖI ĐIỀU HÒA

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THE CONVERGENCE OF A SUB-SERIES OF HARMONIC SERIES

Dinh Van Tiep*, Pham Thi Thu Hang

College of Technology-TNU

ABSTRACT

When studying the convergence of a numeric series, one important technique we often use is to
compare that series with a series whose each term is a power of the reciprocal of an integer. We
sometimes call this technique p-test. In general, we often estimate that series with one of sub-series
(these are series whose each term is an integer) of the harmonic series. Besides, to test the
convergence of Riemann Zeta function at a given point, by comparing this series with a such
sub-series, since then we may know whether the function is defined at this point or not. Therefore,
finding the conditions in which such sub-series converges becomes a meaningful work. This
article aims to present new result for this problem.

Keyword: sub-series of the harmonic series, Riemann Zeta function, convergent, numeric series,

the reciprocal of an integer.

INTRODUCTION*

We know that the harmonic series

1
1

n n






is
divergent. However, there are its convergent
sub-series, such as 2 3

1 1

, .

k k k k



 

We have a

general and useful result which we often
consider to use first to test the convergence of
a given series, namely, the series

1
1

p
k k



converges if and only if

p



1

and diverges if

1 p



. This method is called the p-test. Now,
consider a sub-series

1
1

k nk



of the harmonic
series. If this series converges, whether the
sequence

 

n

k k1 increases at a higher speed
than some sequence

 

, (

1)

k





 

does.
This question is answered in Proposition 1
below. Besides, if we look at the distance of
two consecutive denominators,

:

k

n

k

n

k

 



, we also want to know how
different it is between this distance with that
distance of the series

1
1

k k





, the main result
is stated in Theorem 1.

THE RATE OF THE INCREASING OF
TERMS FOR A DIVERGENT SUB
-SERIES

We first consider

 

k k

n

 to be an increasing
sequence of positive integers. It naturally
establishes the series

k nk



, we index this by
(1). To find out the rate at which each term 
increases, we compare this series with
convergent series in the form

1
1

k k





(





1)

, and we get the following statement.

Proposition 1. If the series

1

k

n

k



diverges,
then

lim inf

k

0

k

n

k







for all



1.

This statement is easy to verify. Indeed, we
suppose by contradiction that there exist



 and



1 such that

lim inf

k

0

k

n

k







 

. This means, there has an
index k0 1 such that

1 ,

k

n





k

 for all

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However, this statement only includes one

direction, i.e if

lim inf

k

0

k

n

k







(

 



1)

, we
do not know whether (1) diverges or not. For 
example, consider the series (1) with

: (ln ) , ( 1)

k

n k k  



 . We have

1
(ln )

lim k lim 0, 1.

k k

n k

k k



 



     

However, by using the integral test, we can
see that the series

1
1
(ln )

k k k





converges,
and so does (1).

In the above counter-example, we use the
floor function

 

 

.

and apply the following
fact, which said that for an increasing

sequence

 

a

k k1 of real number, either both 
series

1
1

k ak



and

1

k

 

 

a

k



converge or

diverge. This fact can be implied easily by

applying the comparison test from the
observation that

1

k

2, (

1).

k

a

k

a





 

 

 

For another counter-example, let us consider
the series 1

1
1

n n

n




. It is clear that

1
1

1 lim

0

n

n n

n

 







  



for all



1.
However, this series diverges because

1
1

ln

n





n

for all n5 and the series

2
1
ln

n n n



diverges, which can be seen by
applying the integral test.

Proposition 2 (Tiep’s Test). Let

:

(0, )







be an increasing function
defined on the set of all natural numbers .
Then, the series (n)

2
1

n n





converges if

[ (n) 1] ln

lim inf

1 ln ln

n





  



 

, and it diverges if

[ (n) 1] ln

lim sup

1. ln ln

n





  



 

Proof. Firstly, if

[ (n) 1] ln

lim inf

1 ln ln

n







  

 

 

, we have

ln ln

(n) 1

ln

n



 





for all n large enough.
So,

n

(n)



n

ln



n

. By the integral test, we
can see that the series

2
1
ln

n n n





converges.
Therefore, (n)

2
1

n n





converges, too.
Finally, if

lim sup

[ (n) 1] ln

1 ln ln

n





  



 

, by

comparing the series (n)

2
1

n n





with the series

2
1
ln

n n n



, which diverges, we can see that
our series diverges, too.
We now illustrate the use of this test to some
following examples.

Example 1. The series 1

1
2 ln ln

1
ln

n n

n n





converges.
• Indeed, set

1
1
(n) ln lnn

ln

n





n



n

, then

1 ln ln

( ) 1

ln ln

ln

n



 



Hence,

lim

[ ( ) 1] ln

ln ln

n





  

2
ln

1 lim .

(ln ln )

n

n
n


    By the above test,

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Example 2. The series 1

1
2 ln

1
ln

n n

n n




diverges.

• Indeed, by taking

1
1
(n) lnn

ln

n





n



n

, we

have

( ) 1

1 ln ln

ln

n



 



, and

[ ( ) 1] ln

lim

ln ln

n





  

1 1 lim

1. ln ln

n

n

 



Ther

efore, the series diverges.

Example 3. The series

2
1
(ln )

n n






diverges.
• Indeed, by setting ( )

(ln )

n





n

, we get

ln ln

( )

ln

n







. Therefore,

[ ( ) 1] ln

lim

ln ln

n





  

lnln ln

lim 1

ln ln ln

n

n n





 

     

  . So, the

series diverges. □

Example 4. Consider the series

ln
2

1 (ln )

n







, where





0.

Case 1:

0  



1,

n

( )n



(ln )

n

lnn, then

ln ln

( )

ln

n







 , so

lim

[ ( ) 1] ln

ln ln

n





  

 

.
Hence, the series diverges.

Case 2:

1 



, if

n

( )n



(ln )

n

lnn, we have

( )n (ln n)(ln ln )n



 

, so

[ ( ) 1] ln

lim

ln ln

n





  

 

. Therefore, the series
converges.

The limit series for sequence of increasing 
series

Let

   

k 1

,

k 1

k k

a



b

 be two sequences of
positive numbers. We denote

 

a

k k1

 

b

k k1 to mean that the sequence

 

a

k k1 is an infinitesimal of a higher order

than

 

b

k k1

.

That is, lim k 0.

k
k
a
b

  Denote

ln x: ln , x ln2x: ln ln ,... x ,

lnmx: ln(ln m x). Using these notations,
we have an order sequence of sequences as
follows:



n

ln

n

 

n

n

(ln

n

)(ln

n

)



n

 



1 2



...

(ln

)(ln

)...(ln

m

)

...

(*)

n



n





1 2



...

(ln

)(ln

)...(ln

m

)

n



n



1 2



...

(ln

)(ln

)

n



n





...

(ln

)

n



n

 

n
for all



 

1  

, ,

and m1.

We see that for any sequence on the left of
(*), the series whose each term is the
reciprocal of its corresponding term is
divergent. However, for any sequence on the
right of (*), the series whose each term is the
reciprocal of its corresponding term is
convergent. All of these sequences satisfy the

necessary condition in Proposition 1, that is

lim

n

0.

n

u

n







It is natural to ask that whether
exists the limit sequence

{ }

k

n n for the
sequence of sequences on the left of (*) (this
means that such sequence satisfies









1 2

(ln )(ln )...(ln )

{ } (ln )(ln )...(ln )

m n

n n m n

n n n n

k n n n n







for all m1, and for every sequence

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{ }an n { }kn n { }bn n, the series 1

n
n

a 




diverges, but 1

n

b 




converges), and does
the series

1
1

n kn



still diverge?

Consider another sequence lying somewhere
at the position of (*) in the following
example. Then, since this, we will
immediately figure out that there does not
exist such limit sequence as expected.

Example 5 [2]. Let



( )

n

be the smallest
positive integer such that 1ln( )n ne for

each n. The series

2 1 2 ( )

1 (ln

)(ln

)...(ln

)

n

n

 n

n



diverges.

Indeed, set

( )

1 2 ( )

(ln

)(ln

)...(ln

)

n





n



n

, then

( )
3

2 ln ln

( ) 1 ... .

ln ln ln

n n

n

n
n

n n n



    

Therefore,

( )
3

2 2 2

ln
ln

[ ( ) 1] ln

lim lim 1 ... .

ln ln ln

n

n n

n
n

n n

n n n



 

 

      

 

 

We now prove that this limit is 1. In fact,
because 1ln( )n ne, we have

( ) 1

ln

n e

,

e



 

n



e

and so on,

( ) 2 ( ) 1

ln

,

n n

e

 



n



e

  where we denote

1 2 1

: , : ,..., : .

k
k

e e

e e e e e  e We can prove
by induction that

k
k

e



e

for all integers k.
Hence,

( )
3

2 2

ln
ln

...

ln ln

n n
n

n n



  3

2
ln
( ( ) 2)

n
n

n



 

2
( ) 2

2
2

ln(ln )
( ( ) 2)

n

n
n

n
e








with ( ) 2

2
( ( ) 2)

lim n 0

n

e









 and

2
ln(ln )

lim 0.

n

n
n

  Therefore, since the

previous arguments, we conclude that

2
[ ( ) 1] ln

lim 1.

n

n n

n





  

 This shows that the
series diverges.

We also can verify that



n

(ln

n

)(ln

n

)...(ln

( )n

n

)



n2



ln



2

 

1

n n

n



n



  for all

 

,

 

1.

Indeed, we have

( ) 1

1 2 (n) 2

1 1

ln ln ...ln ln

ln ln

n

n n n n

n n




 




  . By

setting xn: ln 2n, we have
( ) 1

( ) 1
2

( 1)
1

1
ln

lim lim 0.

ln n

n
n

n
x

n n

x
n

n e










    This

completes the proof of the observation.

The series of the reciprocals of primes

We have a famous result in [1], which said
that the sum of reciprocals of all prime

numbers

1

k

p prime

p

k
 



diverges. Hence, by

Proposition 1, we have

lim inf

k

0

k

p

k







for all



 This is a simple corollary.

Another corollary is directly implied from the
fact that the sum of reciprocals of all primes
diverges, which says that the series

1 ,

k

q

k



where

q k

k

(



1, 2,...)

are all composite
numbers, is divergent. Indeed, to see this
point, we only take the sum of the
type

1 ,

2

k

p prime

p

k
 



this sum is obviously less
than the sum we want to study. It diverges, so
the first series diverges, too.

The conditions on distance between 
consecutive terms

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Theorem 1 (Tiep). If there exists a real

number



1 such that

lim inf

0 (k 1)

k
k 

k











, then (1)

converges.

In this theorem, because

1
(k 1)

lim 1

k

k
k

 







  

, so the theorem is
equivalent to the following theorem: If there

exists a positive number



such that

lim inf

k

0

k

k







, then (1) converges. To prove

this theorem, we need the following lemma.

Lemma. Let

k
k

q

S

q







with



0, then
there exists a constant



0 such that

k

S 



k for all k large enough.

Indeed, because

1 1

(k 1)

1 lim

(

1)

(

1)

1

k k

S

k


 





 

  















, so by Stolz-Cesàro theorem, we have

1
1
lim

k
k

S
k



   . Therefore, we only need to
take



to be a positive number less than this
limit, for example, 1

2( 1)







 , then the

lemma is proved. □

The proof of Theorem 1: We are going to

prove the equivalent form of Theorem 1, i.e
with the hypothesis that

lim inf

k

0

k

k







for

some





0.

Firstly, from this hypothesis, there exist



 and an integer k0 1 such that

,

.

k





  

1  1

for all

k



k

0

.

So,

0 0

1 1

k k k

n   n 



S  



k for all

k



k

.

However, the series

1

k k

k










converges
and therefore so does the series

0 0 0

1
1

k k nk Sk k








   



. This shows that

(1) converges.

From Theorem 1, we have the following
simple remark: if



k is bounded, or even the
ratio

(ln )

k





is bounded for some positive
number



, then (1) diverges.

We now generalize Theorem 1 by using a
general series to compare with (1).

Theorem 2 (Tiep). Suppose that the sequence

of positive numbers

{ }

a

k k1 is decreasing,

and the series

k
k

a




converges. Set

1 1

k

k k

a a





  , then the series (1) converges

if lim inf k 0.

k





 

Proof. Assume that lim inf k 0

k







  

, so
there exists an integer k0 1 such that

k



k

  for all

k



k

0

.

Therefore,

1 1 1 1

1 ( 1 ) 1 ( 1 1)

k k k k k k k

n n 



a a n  



a a

0 0

1 1

(

).

k k k

n



a



a









So,

0 0 0 0

1 1 1

1 1

( ) ( )

k

k k k k k k k

a

n n



a a a n



a





  

  

 

   

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index

k

1



1

such that

0 0

(

)

2

k k k

a



n





a









for all kk1. Hence, 1

1
2

1 k

k

a

n






 for all

0 1
max{ , }

k  k k . This shows that the series
(1) converges by the comparison test.

By the same technique, we also can prove the
following result, which is quite interesting.

Theorem 3. Let

k
k

a




diverge, whose terms
establish a decreasing sequence of positive
numbers. Set

1 1

k

k k

a a





  , then the series

(1) also diverges if lim inf k 0

k






 .

REFERRENCE

1. Roman J. Dwilewicz, Jan Minac, Values of the 
Riemann zeta function at integers, Materials 
Mathematics. ISSN: 1887-1097.

2. W. J. Kaczor, M. T. Nowak, Problems in 
Mathematical Analysis I, ISBN:
978-0-8218-2050-6.

3. T. Thanh Nguyen, Fundamental Theory of

Functions of a complex variable, Vietnam

National University Hanoi Publisher, 2006.
4. Elias M. Stein, Rami Shakarchi, Complex

Analysis, Princeton University Press,

New Jersey, 2003.

5. Nick Lork (2015), Quick proofs that certain 
sums of fractions are not integers, The 
Mathematical Gazette Vol. 99.

6. William Dunham (1999). Euler The Master of 
Us All, Vol. 22, MAA. ISBN 0-88385-328-0.

TÓM TẮT

SỰ HỘI TỤ CỦA MỘT CHUỖI CON CỦA CHUỖI ĐIỀU HÒA

Đinh Văn Tiệp*, Phạm Thị Thu Hằng 
Trường Đại học Kỹ thuật Công Nghiệp – ĐH Thái Nguyên 
Khi xem xét sự hội tụ của một chuỗi số, một phương pháp quan trọng hay được cân nhắc trước
tiên là so sánh chuỗi đó với một chuỗi số với các số hạng là lũy thừa của nghịch đảo các số
nguyên. Một cách tổng quát, ta thường so sánh chuỗi số với một chuỗi con của chuỗi điều hịa.
Ngồi ra, để xem xét sự tồn tại của hàm Riemann Zeta tại một điểm cho trước, bằng cách so sánh
giá trị chuỗi tại điểm đó với một chuỗi con như thế, từ đó ta có thể biết hàm số có xác định tại
điểm đó hay khơng. Do vậy, việc tìm điều kiện để biết chuỗi con của chuỗi điều hòa hội tụ hay
phân kỳ trở thành một điều rất có ý nghĩa. Bài báo sẽ đưa ra một số kết quả mới cho vấn đề này.

Từ khóa: Chuỗi con của chuỗi điều hòa, hàm Riemann Zeta, hội tụ, chuỗi số, nghịch đảo của một

số nguyên

</div>

SỰ HỘI TỤ CỦA MỘT CHUỖI CON CỦA CHUỖI ĐIỀU HÒA

<b>THE CONVERGENCE OF A SUB-SERIES OF HARMONIC SERIES </b>



 



<i>p</i>



1

1

<i>p</i>





 

<i>n</i>

 

, (

1)

<i>k</i>



 

:

<i>n</i>

<i>n</i>

 





 

<i>n</i>





(





1)

1

<i>n</i>



lim inf

0

<i>n</i>

<i>k</i>









lim inf

0

<i>n</i>

<i>k</i>



 

1

1

,

<i>n</i>





<i>k</i>

lim inf

0

<i>n</i>

<i>k</i>



(

 



1)







 

<sub> </sub>

.

 

<i>a</i>



1

 

 

<i>a</i>