Dinh Van Ti$p vd Dtg Tap chi KHOA HOC & CONG NGHE 162(02): 177-182
THE CONVERGENCE OF A SUB-SERIES OF HARMONIC SERIES
Dinh Van Tiep , Pham Thi Thu Hang College ofTechnoiogy-TNU ABSTRACT
When studymg 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 fimction 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 / ^ —
n
T H E R A T E OF T H E I N C R E A S I N G OF T E R M S F O R A D I V E R G E N T SUB SERIES
W e first consider {"A}^^, to be an increasing sequence of positive integers. It naturally establishes the series V — , we index this by ( I ) . To find out the rate at which each term increases, we compare this series with
^ 1 convergent series in the form 2 ^ —
( « > 1 ) , and we get the following statement.
P r o p o s i t i o n 1. I f t h e series ^ — diverges.
divergent. However, there are hs convergent sub-series, such as zl'T'ZllJ' ^^ ^^^^ ^
ASl ^ i a i ^
general and useful result which we often consider to use first to test the convergence of a given series, namely, the series / ^ - — converges if and only if p>l and diverges if P<1. This method is called the p-test. N o w ,
consider a sub-series / — of the harmonic series. If this series converges, whether the sequence {"ij^^, increases at a higher speed than some sequence \k"\ ,{a>\) does.
This question is answered in Proposhion 1 ,. • , ,.1.
below Besides, if we look at the distance of ^"PP"^^ by contradiction that there exist two consecutive denominators, « > " M d « > ' '''"=1' * ^ '
\-=n,u-n,, w e also want to k n o w how l i n i i n f , ^ > £ > 0 . This means, there has an different it is between this distance with that *"*" ^
distance of the series T — , the main result index k„>l such that — - ~ j T ' ^°' " "
is stated in T h e o r e m 1. k>k^. This leads to the conclusion that (1) must be convergent, we meet the then lim i n f - ^ = 0 for all a>\.
This statement is easy to verify. Indeed, we
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177
BinhYlkn Tiep vd Dtg Tgp chi KHOA HQC & CONG NGHE 162(02): 177-182^
However, this statement only includes one rffl(n)-lllnn
^ litn inf i^-^:^—i > 1, and it diverges if direction, i.e if lim i n f - ^ = 0 ( V a > 1), we ""*' 1" 1" "
* - " ^ ,. [ ^ ( n ) - l ] l n ^ do not know whether (1) diverges or not. For il^^^P i i ~ example, consider the series (1) with
« . : = [ * ( l n * ) ' J , ( ^ > l ) . We have ^^"-^ ^ ^ ^ ,^, « ' * ' '' (i„t\P htninfi^^^^^^— >fi>l, we have lini-!^ = l i m i H i i | - = 0 , V a > l . " " taln«
However, by using the integral test, we can ^ ( n ) > l + y9 for all n large enough.
I l n « see ftat the series Tj^J^ converges, ^^^ ^„„, ^ „ ^ , i „ gy the integral test, we and so does (1).
can see that the series X 3— converges.
In the above counter-example, we use the ^ " In'^ « floor function |_.J and apply the following \ r . . . . A .A. . r Therefore, > —j-r converges, too.
fact, which said that for an increasing ^ !,'•<"'
sequence \ct.\, , of real number, either both r , , ,-,,
*^' ,. [oj(n)-l]ln« , , 1 1 Finally, if lim sup'^^^ \ -* < 1 , by series 2 ^ — '^'^ / . ] T converge or "^^ I n m n
*ai ^A isi |_OiJ J
<//verge. This fact can be implied easily by comparing the series Yj~m with the series applying the comparison test from the "^^
observation that 1 < r - ^ < 2 , ( V ^ > 1). T ] — ; — , which diverges, we can see thai [a^J f^2nlnn For another counter-example, let us consider our series diverges, too.
E
l , We now illustrate the use of this test to some—T-. It IS clear that - „ .
^^, i+- following examples,
I *^ -c- 1 '+- . Example I, The series > j lim = lim j- = 0 for all a>\. " " »t'^iiiin"lnn
n " converges.
However, this series diverges because , '•^ ^ , f „ >. c J .L • • Indeed, set n'°'"' = n ''"'"" In n, then
n " < « I n « for all n>5 and the series
1 / \ _ i ^ Inlnfi 7 diverges, which can be seen by 9\^) — ^'^' j ^~ •
~^n\nn i n m « mn applying the integral test. . [^(n) - 1 ] h n Proposition 2 (Tiep's Test). Let ' «-»» lnln«
^:N—>(0,oo) be an increasing function
defined on the set of all natural numbers N . =1-1- lim 7 = 00. By the above test, 1 n-xo (In In n)
Then, the series J ] — converges if „ e imply that our series converges.D
Dinh Van Ti|p va Dtg T9p chl KHOA HOC & CONG NGHE 162(02): 177-182
lim
Therefore, Example 2. The series J ] 1 diverges.
1
• Indeed, by taking «*•'"'=« '""Inn, we , , , 1 bikirt have g}{n) = l-\ h , and
InH Inn ,. [ ^ ( « ) - l ] l n n , ,. 1 , limiTLJ—J = l - f l i m =l.Ther
"-»" In In K "-*" In In w
efore, the series diverges.
D
Example 3. The series V diverges.
• Indeed, by setting M*^"' = ( l n n ) ' ' , we get in Inn
ln«
[ ^ ( K ) - l ] l n « l n l n « ,. r InlnK ,1 Inn
= lim a 1 = -00. So, the
"-»*[ in« J l n l n n
series diverges. n Example 4. Consider the series
/ —,where y>0.
t?(ln«y""
Case 1: 0 < ; ' < ! , if n*"*"* =(lnn)''''", then , ^ In bin ,. [c>(«)-l]lnw
^(«) = — , so lim^^-^^^^— = -co.
]n'^ n "-*" In In«
Hence, the series diverges.
Case 2: 1 < / , if «"*"*= (Inn)'"'", we have
^(n) = (In^"' n)(In In « ) , so Iim ^^^-^^^— = CO. Therefore, the series
"^" \n\nn converges.
The limit series for sequence of increasing series
Let {^i};t5^r{^*}Aai ^^ *wo sequences of positive numbers. We denote
1*^*1*21 **• t^t/tai *° mean that the sequence {'^*}ASI '^ ^" infinitesimal of a higher order than {b/,} . That is, lim-^ = 0. Denote
In, x:=ln;c, bi2;c:=lnlnx,..., ln^ji::=hi(ln„_, x) . Using these notations, we have an order sequence of sequences as follows:
{/7inf«} <e {n(ln, H)(ln^«)}
<s:... <K {^(Ini n)(lni2 «)...(In^ «)} <£...«: (*)
«...«{«(hi,«)(ln,«)...(lnfn)}^
« . . . « { « ( l n , « ) ( l n f « ) }
«:...^{w(lnf«)} « { « " } for all y<l<a,p, 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 Proposhion 1, that is lim—^ = 0. It is natural to ask that whether exists the limit sequence {ft„}„ for the sequence of sequences on the left of (*) (this means that such sequence satisfies
{«(hii«)(ln2w)...(ln^,«)} « :
for all m>\, and for every sequence {«„}„.{*„}„ such that
Dinh van Ti^p vo Dtg T?ip chi KHOA HQC & C6NG N G H $ 162(02): 177 - 18:
{^Jfl^iJ^Jn ^ { ^ J « ' *he series J ] a„"' previous arguments, we conclude that
"-»" hi; n series diverges, D
,. [©(«)-!] Inn , . . ,
• ^ , _, . J J lim^^t-^^-^—^ = 1. This shows that the diverges, but 2 j ^» converges), and does "-»» In^«
the series ^ — still diverge? We also can verify that
""*" {"(tai«Xta2«)...(ln„„«)}_^^
Consider another sequence lying somewhere
at the position of (*) in the following <s:{«bi^n} <e:|n''} for all p,a>\.
, , _ , . . . . , . ^ J flS2 V / n&l
examp e. Then, since this, we will T ^ ^^ ,„ u •j. ^ , ~ . xL . .L J * Indeed, we have immediately figure out that there does not , , ,
exist such limit sequence as expected. Q ^ ' " i " ' " ^ " - % ( " ) " < ^ 2 " " g Example 5 [2]. Let ^ n ) be the smallest Inf" Inf"'n positive integer such that 1 <ln^(„jn<.e for setting x„ : - h i j « , we have each 11. The series ,. ln?'''^'n ,. xf"'"' „
, lim-- „ •— = lim v"n n, • = 0- This
„a2 «(l"i ")(ln2«)—(Inp(n)") completes the proof of the observation, n Indeed set "^^^ series of the reciprocals of primes
n*""' =«(ln,rt)(ln2n)...(hi ,„j«), then We have a famous result in [1], which said that the sum of reciprocals of all prime hijK ln3« ^ f w " 1 '^^''^-^^~^^~^^-^'^^' - " ' b e r s Y. - diverges. Hence, by
Therefore, 2^.*,^"-^*
,. [^(«)-l]lnn r hj/j ' " K " ) " ] Proposition 1, we have lim i n f " = 0 for all lim = lim IH 1-...-I . *-»ai ^°
x-^" In.K "-•• ln,n In,/? 1 T^L- • • 1 n
^ L i i ^ a>\. This is a simple corollary.
We now prove that this limit is 1. In fact, . ,, .. . ,, ^. . ,. . r ,u
*^ , , , Another corollary is directly implied from the because 1 < ln^,„, n<e, we have f^^^ ^^^^ ^^e sum of reciprocals of all primes e < kL„-„, , K < e% and so on, ,. . . . „,,»„. • V ^
*''"' ' diverges, which says that the series ? —>
-*,(«)-2 .^«)-l , J , *^i 9*
e < I n , « < e , where we denote
,[ ^2 At+i -* where g4(A: = l,2,...) are all composite e \=e,e •.= e\...,e •.= e' We can prove numbers, is divergent. Indeed, to see this
. , . , " * * . „ . , point, we only take the sum of the by induction that e >e for all integers k. . Hence, type ^ , this sum is obviously less
^j^ „ In w ^^V''."^ ^) .^^ ^ than the sum we want to study, It diverges, so
^ ^ ^ the first series diverges, too.
( 0 ( n ) - 2 ) h i ( b i , « ) ^ , , . . ' .
^ (n)-2 —: The conditions on distance between
^ y l n j « consecutive terms
(m(n)-2) ^ ° ^^"'^y '^°w different each pair of witnlim ^(^p; ~ " ^""^ consecutive elements of the above sequence e ^ are, we consider the sequence whose elements l i m ^ i ^ ^ " ^ = 0 . Therefore, since the are in form A, •.= n,^,-n„k = \,2,... . We
V 2 " have very nice results as follows.
180
Dinh van Ti^p vd Dtg T?ip chi KHOA HOC & CONG NGHE 162(02): 177-182 Theorem 1 (Tiep). If there exists a real
number a > 1 such that
1 lim inf- A^
k>k^. So, for all k'^K.
- > 0 , then (1)
(k-HO^-Jfc' converges.
In this theorem, because lim •^ — ; = 1 , so the theorem is
*-»- ak"
equivalent to the following theorem: If there exists a positive number y such that
lim i n f —^ > 0 , then (1) converges. T o prove this theorem, w e need the following lemma.
k
Lemma. Let 5^ ^^q^ with x > 0 , then there exists a constant £ > 0 such that
5j > ek''*^ for all k large enough.
Indeed, because lun . — r = lini ; 7r = 7
"^'{k+iy^'-k'*' * — ^ , r . i ( r + i ) - r + i k , so by Stolz-Cesaro theorem, we have lim —~ = . Therefore, we only need to S 1 take E to be a positive number less than this limit, for example, s = 1
then the 2(r+i)
lemma is proved, a
ne proof of Theorem I: W e are going to
prove the equivalent form of Theorem 1, i.e with the hypothesis that l i m i n f — 4 - > 0 for some y>(i.
Firstly, from this hypothesis, there exist 5>0 and an integer k^i.\ such that A, 1 1
-^>S,\/k>k.. Hence, < —:
k' «M "i+Sk' for
1
n,, " n . - 5 S . ,+Sek'*''
* + l *6 *0 '
1 .._..; ssk'*'
and therefore so does the series y - . This shows that (1) converges. a From Theorem 1. we have the following simple remark: if A^ is bounded, or even the ratio ^ is bounded for some positive
( h i )
number /3, 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^Jiai is decreasing, and the series ^ a ^ converges. Set
,5^ := , then the series (I) converges if lim inf • ^ > 0 .
Proof Assume that lim inf - f = £ > 0, so there exists an integer k^^l such that
^.^>s5^ for all k>kQ. Therefore, n^^i > n^ + s{all^ - a^') ^ w^-, + £ifC\ - ^t-^) S « ^ + ^ ( « * ! i - « ^ ) -
1 . 1
So,
But, since 0 = lima(^,, we can find an
Dinh v a n TiSp vd Dig T ^ chi KHOA HQC & CONG NGHE 162(02): 177-182"
index k^>l such that a^^i {n,^ - sa^) > —
for all k>k,. Hence, < — ^ for all k>max.{kQ,kf}. This shows that the series (1) converges by the comparison test.
By the same technique, w e also can prove the following result, which is quite interesting.
T h e o r e m 3. Let ^ a^ diverge, whose terms ki.[
establish a decreasing sequence of positive numbers. Set ^^ := , then the series (1) also diverges if lim i n f —^ = 0 .
k-*=c §^
R E F E R R E N C E
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.
T O M T A T
S t r H O I T U C U A M O T C H U O I C O N C U A C H U O I D I E U H O A
Dinh Van Ti|p*, Phgrn Thi Thu H^ng Trudng Dgi hpc Ky thudl Cong Nghiep - O H Thai Nguyin
Khi xem xet s\r hOi tu ciia mpt chu6i s5, m$t phuang phap quan U-png hay ducrc can nhac truac tiSn Ik so sanh chuoi do vdi mgt chu5i s5 vdi cSc s6 hang ik luy thira ciia nghjch dao cac s6 nguySn. M§t each tong quat, ta thucmg so sanh chuoi so voi mot chuoi con ciia chufii dieu hoa.
Ngoii ra, de xem x6t su tbn tai ciia ham Riemann Zeta tgi mgt diem cho tru6c, bang each so sanh gia tri chuSi tai diem do voi mot chuoi con nhu the, tii do ta co the biet ham so co xac djnh tgi diSm do hay khSng. Do v$y, vi$c thn dieu kien de biet chuoi con ciia chu5i di^u h6a h^i tu hay phSn ky trd thanh m$t diiu rit co y nghia. Bai bao sS dua ra mot so ket qua mdi cho van de nay.
Til' khda: Chuoi con cua chudi dieu hoa, hdm Riemann Zeta, hqi tu. chudi so, nghich ddo cua m^l s6 nguyen
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