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

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

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

  k k1, (    1) does.

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

:

1

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

  n

k k_1 to be an increasing sequence of positive integers. It naturally establishes the series

1

1

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

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



0 and



1 such that

lim inf

^k

0

k

n k

^





 

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

1 1

,

n

k

  k

^ for all

k  k

0. This leads to the conclusion that (1)

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)

nk 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

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

1

n^ n^n



. It is clear that

1 1

1 1

lim lim 1 0

n

n n

n

n

n n

 







  



for all



1.

However, this series diverges because

1 1

ln

n

^n

 n n

for all n5 and the series

2

1

n_ nlnn



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

n n





  

  

, and it diverges if

[ (n) 1] ln

lim sup 1.

ln ln

n

n n





  

  

Proof. Firstly, if

[ (n) 1] ln

lim inf 1

ln ln

n

n n

 



    

 

^{, we have}

ln ln (n) 1

ln n

    n 



for all n large enough.

So,

n

^(n)

 n ln

^

n

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

2

1

n_ nln^n



converges.

Therefore, _(n)

2

1

n_ n^



converges, too.

Finally, if

[ (n) 1] ln

lim sup 1

ln ln

n

n n





   

 

^{, by}

comparing the series _(n)

2

1

n_ n^



with the series

2

1

n_ nln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

2 1 ln ln

1

n^ n^ nlnn



converges.

• Indeed, set

1 1 (n) ln lnⁿ

ln

n

^

 n

^

n

, then

1 ln ln

( ) 1

ln ln ln n n

n n

   

.

Hence,

[ ( ) 1] ln lim ln ln

n

n n

n





  

2

1 lim ln .

(ln ln )

n

n

 n

    By the above test, we imply that our series converges.

Example 2. The series ₁

2 1 ln

1

n^ n^ nlnn



diverges.

• Indeed, by taking

1 1

(n) lnⁿ

ln

n

^

 n

^

n

, we

have

1 ln ln

( ) 1

ln ln

n n

n n

   

, and

[ ( ) 1] ln lim

ⁿ

ln ln

n n

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

n

^{( )n}

 (ln ) n

^, we get

ln ln

( ) ln

n n

   n

. Therefore,

[ ( ) 1] ln lim

ⁿ

ln ln

n n

n





  

lnln ln

lim 1

ln ln ln

n

n n

n n





 

     . So, the

series diverges. □ Example 4. Consider the series

2 ln

1 (ln )

ⁿ

n

n

^



^{, where}

  0.

Case 1:

0    1,

if

n

^{( )n}

 (ln ) n

^lnn, then

1

ln ln ( ) ln

n n



n

 

_ , so

[ ( ) 1] ln

lim

ⁿ

ln ln

n n

n





  

 

. Hence, the series diverges.

Case 2:

1  

, if

n

^{( )n}

 (ln ) n

^lnn, we have ( )n (ln^ 1n)(ln ln )n



 ^ , so

[ ( ) 1] ln lim

ⁿ

ln ln

n n

n





    

. Therefore, the series converges.

The limit series for sequence of increasing series

Let

    a

k k_1

, b

k k_1 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 ln1x: ln , x ln₂x: ln ln ,... x , ln_mx: ln(ln _m1x). Using these notations, we have an order sequence of sequences as follows:

 ln1   (ln1 )(ln2 ) 

n n

n

^

n n n

^

n



^{1 2}



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

_m

) ... (*) n n n

^

n

n





1 2



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

_m

)

n n n

^

n

n



1 2



... (ln )(ln ) n n

^

n

n



1



... (ln )

n

^

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

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 { } ,{ }a_{n n} b_{n n} such that

{ }a_{n n} { }k_{n n} { }b_{n n}, the series ¹

1 n n

a ^





diverges, but ¹

1 n 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

_ n

n



diverges.

Indeed, set

( )

1 2 ( )

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

n

n

^

 n n n

_ n

n

, then

3 ( )

2 ln ln

( ) 1 ln ... .

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

2

,

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

( ( ) 2)ln ln n n



n

 

2 ( ) 2

2 2

ln(ln ) ( ( ) 2)

n ln

n n

e n







 

with _{( ) 2}

2

( ( ) 2)

lim _n 0

n

n e





 

  and

ln(ln2 )

lim 0.

ln

n

n

 n  Therefore, since the

previous arguments, we conclude that

2

[ ( ) 1] ln

lim 1.

ln

n

n n

n





  

 This shows that the series diverges.

We also can verify that

 (ln1 )(ln2 )...(ln ( )n ) 

2

n n n

_

n

n

 n lnn 

_n2

  n n1 for all    ,   1.

Indeed, we have

( ) 1

1 2 (n) 2

1

1 1

ln ln ...ln ln

0 ln ln

n n n n n

n n

 

 



   . By

setting x_n: ln ₂n, we have

( ) 1 ( ) 1

2

( 1) 1

1

limln lim 0.

ln ⁿ

n n

n

n n x

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

2

1

pk prime

p

k

 



diverges. Hence, by Proposition 1, we have

lim inf

^k

0

k

p k

^





for all



1. 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

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

2

1 ,

k

2

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

To study how different each pair of consecutive elements of the above sequence are, we consider the sequence whose elements are in form  _k: n_k1n k_k, 1, 2,... . We

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

1 k k

q

S q

^



 

^with



^{0, then}

there exists a constant



0 such that

1

Sk 



k^ for all k large enough.

Indeed, because

1

1 1

1

(k 1) 1

lim lim

( 1)

( 1) 1

k k

k k

S S

k k

k k



 



 



 

  

   

   

, so by Stolz-Cesàro theorem, we have

1

lim 1

1

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



0 and an integer k₀ 1 such that

, .

k

 k k

   

¹  ¹

for all

k  k

₀

.

So,

0 0

1

1 1

1 1

,

k k k

n _ ^ n 



S _ 



k^ for all

k  k

₀

.

However, the series

0

1

1

k k





 k

^ converges and therefore so does the series

0 0 0

1 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

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

1 k k

a



 converges. Set

1

1 1

k:

k k

a a





  , then the series (1) converges

if lim inf ^k 0.

k^



k

 

Proof. Assume that lim inf ^k 0

k 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

(

1

).

k k k

n  a

^_

a

^

  

So,

0 0 0 0

1

1 1 1

1 1 1

1 1

( ) ( )

k

k k k k k k k

a

n_ ^n 



a^_ a^{ }a_ n ^



a^ 





index

k

₁

 1

such that

0 0

1

1

( )

k k k

2

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

1 k k

a



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

1

1 1

k:

k k

a a





  , then the series

(1) also diverges if lim inf ^k 0

k



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 hòa.

Ngoài 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 không. 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