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Trang 1THE 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
k k k k
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 n k
of the harmonic series If this series converges, whether the
sequence nk k1 increases at a higher speed
than some sequence 1, ( 1)
k
k
does
This question is answered in Proposition 1
below Besides, if we look at the distance of
two consecutive denominators,
1
:
k nk nk
, 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 nk k1 to be an increasing sequence of positive integers It naturally establishes the series
1
1
k n k
, 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 nk
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
k
n
k
This means, there has an index k0 1 such that 1 1
,
k
n k for all
0
k k This leads to the conclusion that (1)
must be convergent, we meet the
Trang 2However, 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
k
n k k We have
1
(ln )
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 ak k1 of real number, either both
series
1
1
k a k
1
1
k ak
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
It is clear that
1
1
1 1
1
n
n
n
n
n
for all 1
However, this series diverges because
1
1
ln
n
n n n for all n5 and the series
2
1
ln
n n n
diverges, which can be seen by
applying the integral test
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
ln ln
n
n n
, and it diverges if
[ (n) 1] ln
ln ln
n
n n
[ (n) 1] ln
ln ln
n
n n
ln ln (n) 1
ln
n n
for all n large enough
So, n(n) n lnn 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 [ (n) 1] ln
ln ln
n
n n
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
1
2 ln ln
1 ln
converges
• Indeed, set
1 1 (n) ln ln
ln
n
n n n, then
1 ln ln ( ) 1
ln ln ln
n n
Hence, [ ( ) 1] ln
lim
ln ln
n
n
2
ln
(ln ln )
n
n n
By the above test,
we imply that our series converges
Trang 3Example 2 The series 1
1
2 ln
1 ln
n n
diverges
• Indeed, by taking
1 1 (n) ln
ln
n
n n n, we
( ) 1
n n
, and
[ ( ) 1] ln
lim
ln ln
n
n
ln ln
n n
efore, the series diverges
Example 3 The series
2
1 (ln )
n n
diverges
• Indeed, by setting ( )
(ln )
n
n n , we get
ln ln
( )
ln
n n
n
[ ( ) 1] ln
lim
ln ln
n
n
n
series diverges □
ln
2
1
(ln ) n
n n
, where 0.
Case 1: 0 1, if n( )n (ln ) n lnn, then
1
ln ln
( )
ln
n
n
n
, so [ ( ) 1] ln
lim
ln ln
n
n
Hence, the series diverges
Case 2: 1 , if n( )n (ln ) n lnn, we have
1
( )n (ln n)(ln ln )n
[ ( ) 1] ln
lim
ln ln
n
n
Therefore, the series converges
The limit series for sequence of increasing
series
Let k 1, k 1
a b be two sequences of positive numbers We denote
ak k1 bk k1 to mean that the sequence
ak k1 is an infinitesimal of a higher order than bk k1. That is, lim k 0
k k
a b
Denote
1
ln x: ln , x ln2x: ln ln , x ,
1
lnm x: ln(ln m x) Using these notations,
we have an order sequence of sequences as
follows:
ln1 (ln1 )(ln2 )
n n n n n
(ln )(ln ) (lnm ) (*)
n
n n n n
(ln )(ln ) (lnm )
n
1 2
(ln )(ln )
n
n n n
1
(ln )
n
n n n
for all 1 , , and m1
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 { } kn n for the sequence of sequences on the left of (*) (this means that such sequence satisfies
(ln )(ln ) (ln )
m n
for all m1, and for every sequence
{ } ,{ }a n n b n n such that
Trang 4{ }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 k n
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 1ln( )n ne for
1 (ln )(ln ) (ln )
n n n n n n
( )
(ln )(ln ) (ln )
n
n
( ) 3
ln
n n n
n n
Therefore,
( ) 3
ln ln
[ ( ) 1] ln
n
n n
We now prove that this limit is 1 In fact,
because 1ln( )n ne, we have
( ) 1
e n e and so on,
2
e n e where we denote
k k
e e e e e e We can prove
by induction that
k k
e e for all integers k
Hence,
( ) 3
ln
ln
n n n
2
ln ( ( ) 2)
ln
n n
n
2 ( ) 2
2 2
ln(ln )
( ( ) 2)
ln
n
n n
n e
2
( ( ) 2)
n
n
e
2
2
ln(ln )
ln
n
n
n
Therefore, since the
previous arguments, we conclude that
2
[ ( ) 1] ln
ln
n
n
This shows that the series diverges
We also can verify that
(ln1 )(ln2 ) (ln ( )n ) 2
n
ln 2 1
n n n
for all , 1.
( ) 1
1
0
n
setting x n: ln 2n, we have
( ) 1 ( ) 1
2
( 1) 1
1
ln
n n
n x
x n
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
k
p prime pk
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 qk
where q kk( 1, 2, ) are all composite numbers, is divergent Indeed, to see this point, we only take the sum of the type
2
1 , 2
k
p prime pk
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 k1n k k, 1, 2, We have very nice results as follows
Trang 5Theorem 1 (Tiep) If there exists a real
(k 1)
k
k k
converges
In this theorem, because
1
(k 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
k
S k for all k large enough
1
1
( 1)
S S
k k
, so by Stolz-Cesàro theorem, we have
1
1
lim
1
k
k
S
k
Therefore, we only need to
take to be a positive number less than this
limit, for example, 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
Hence, 1 1
1
,
n n S k
for all k k0.
However, the series
0
1 1
k k k
converges and therefore so does the series
1 1
1
k k n k S k k
(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 { } ak k1 is decreasing, and the series
1
k k
a
converges Set
1
:
k
, then the series (1) converges
if lim inf k 0
k
k
k
k
, so there exists an integer k0 1 such that
k k
for all k k0. Therefore,
n n a a n a a
1
n a a
1
k
a
n n a a a n a
But, since 0lima , we can find an
Trang 6index k1 1 such that
1
2
a n a
for all kk1 Hence, 1
1
2
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
:
k
, 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
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