On connection between values of Riemann zeta function at integers and generalized harmonic numbers

0.14 MB·English

by  Paweł J. Szabłowski

ON CONNECTION BETWEEN VALUES OF RIEMANN ZETA
FUNCTION AT INTEGERS AND GENERALIZED HARMONIC
NUMBERS
5
1
0
2 PAWEL J.SZABL OWSKI

n
a Abstract. Using Euler transformation of series we relate values of Hurwitz
J zeta function at integer and rational values of arguments to certain rapidly

2 convergingserieswheresomegeneralizedharmonicnumbersappear. Theform
1 of these generalized harmonic numbers carries information about the values
of the arguments of Hurwitz function. In particular we prove: ∀k ∈ N :
T] ζ(k,1) = 2k2−k−1−11P∞n=1 Hnn(k2−n1), where Hn(k) are defined below generalized
harmonicnumbers. Furtherwefindgeneratingfunctionofthenumbersˆζ(k)=
N
P∞ (−1)j−1/jk.
j=1
.
h
t
a
m
[ 1. Introduction

1 First let us recall basic notions and definitions that we will work with. By the
v Hurwitz function ζ(s,α) we will mean:
5
∞
0 1
6 ζ(s,α)= ,
(j+α)s
2 Xj=0
0
considered for Res > 1, Reα ∈ (0,1]. Function ζ(s,1) is called Riemann zeta
.
1 function. We will denote it also by ζ(s) if it will not cause misunderstanding. It
0
turns out that both these functions can be extended to holomorphic functions of s
5
on the whole complex plane except s=1 where a single pole exists. Of great help
1
: in doing so is the formula
v
i 2s−1 ∞ (−1)j−1
X (1.1) ζ(s)= ,
2s−1−1 js
r Xj=1
a
that enables to extend Riemann zeta function to the whole half plane Res>0.
We will consider numbers:
∞ (−1)j
M(m,i) = ,
k (mj+i)k
Xj=0
for m ∈ N and i ∈ {1,...,m−1}. Notice that M(1,1) = ∞ (−1)j−1/jk and
k j=1
M(2,1) =π/4. The number M(2,1) = ∞ (−1)j is called CaPtalan constant K.
1 2 j=0 (2j+1)2
P
Date:December,2014.
2010 Mathematics Subject Classification. Primary11M35,40G05; Secondary05A15,05E05.
Keywordsandphrases. RiemannzetafunctionHurwitzzetafunction,Eulersummation,Har-
monicnumbers,generalizedHarmonicnumbers,Catalanconstant.
1
2 PAWEL J.SZABL OWSKI
It is elementary to notice that
1
M(m,i) = (ζ(k,i/(2m))−ζ(k,1/2+i/(2m)).
k (2m)k
The main idea of this paper is to apply the so called Euler transformation that
was nicely recalled in [5]. As pointed out there we have:
∞ ∞
(−1)k−1a = ∆na /2n+1,
k 1
Xk=1 nX=0
where{a } isasequenceofcomplexnumbersandthe sequence∆na isdefined
k k≥1 k
recursively: ∆0a = a ∆na = ∆n−1a −∆n−1a = n (−1)m n a .
k k, k k k+1 m=0 m m+k
Sondow in [5] presented general idea of applying Euler transPformationto(cid:0)Ri(cid:1)emann
function. He however stopped half way in the sense that he calculated finite dif-
ferences ∆n applied to (j +1)−s only for s being negative integers. We are going
to make a few steps further and calculate these differences pointing out the rˆole of
the generalized harmonic numbers in those calculations.
The paper is organizedas follows. In the next section 2 we present an auxiliary
resultthat enablesapplicationof Eulertransformationto the analyzedseries. Fur-
therwe presenttransformedseriesapproximatingnumbersM(m,i). InSection3 we
k
calculategeneratingfunctionsofcertainseriesofnumbersandfunctions. Morepre-
cisely we calculate generating functions of the generalized harmonic numbers that
we have defined in the previous section. We also calculate generating function of
theseriesofthegeneratingfunctionsthatweredefinedpreviously. Itturnsoutthat
this calculation enables to obtain the generating function of the series sums that
appear on the right hand side of (1.1). Finally in the last Section 4 are collected
cases when exact values of numbers M are known.
2. Euler transformation
To proceed further we need the following result.
Proposition 1. Let us denote A(m,i) = n (−1)j n /(mj +i)k, n = 0,1,...,
n,k j=0 j
and the family of sequences defined recursPively: B(m(cid:0),i)(cid:1)= 1, B = 1 , k ≥ 1,
n,0 0,k ik−1
∀n,k≥0:B(m,i) = n 1 B(m,i). We have then:
n,k j=0 (mj+i) j,k−1
∀ m∈N : A(m,i) =1P, A(m,i) =0, A(m,i) = n! , where
0,0 n,0 n,1 m(i/m)n+1
(a) =a(a+1)...(a+n−1) is the so called ’rising factorial’.
n
∀ n≥0,k≥1 we get:
n!
A(m,i) = B(m,i) .
n,k m(i/m) n,k−1
n+1
HURWITZ FUNCTION 3
Proof. i) The fact that A =0 follows immediately properties of binomial coeffi-
n,0
cients. Notice that we have
m(n+1) n n+1 (−1)n+1
A(m,i) − A(m,i) = (−1)j /(mj+i)k+
n+1,k m(n+1)+i n,k Xj=0 (cid:18) j (cid:19) (m(n+1)+i)k
m(n+1) n n (−1)n+1
− (−1)j /(mj+i)k =
m(n++1)+iXj=0 (cid:18)j(cid:19) (m(n+1)+i)k
n
n+1 m(n+1)!
+ (−1)j( − )/(mj+i)k
(cid:18) j (cid:19) j!(n−j)!(m(n+1)+i)
Xj=0
(−1)n+1 1 n n+1
= + (−1)j /(mj+i)k−1
(m(n+1)+i)k m(n+1)+j Xj=0 (cid:18) j (cid:19)
n+1
1 n+1 1
= (−1)j /(mj+i)k−1 = A(m,i)
m(n+1)+i (cid:18) j (cid:19) m(n+1)+i n+1,k−1
Xj=0
since(1− m(n+1−j))= jm+i . NownoticethatwehaveA(m.i) − m(n+1) A(m,i)
(mn+m+i) m+mn+i n+1,1 m(n+1)+i n,1
=0fromwhichimmediatelyfollowsthatA(m,i) = n! sinceA(m,i) = 1. Now
n,1 m(i/m)n+1 0,1 i
divide both sides of the identity A(m,i) − m(n+1) A(m,i) = 1 A(m,i) by
n+1,k mn+m+i n,k mn+m+i n+1,k−1
A(m,i) anddenoteB(m,i) =A(m,i)/A(m,i).WegetB(m,i) −B(m,i) = 1 B(m,i)
n+1,1 n,k n,k n,1 n+1,k n,k m(n+1)+i n+1,k−1
since A(m.i) = m(n+1) A(m,i). Hence B(m,i) = n 1 B(m,i) since ∀k ≥ 1 :
n+1,1 mn+m+i n,1 n,k j=0 mj+i j,k−1
B(n,i) =1/ik−1. P (cid:3)
0,k
Remark 1. In the literature (compare e.g. [1], [2], [4]) there function notions of
harmonic and generalized harmonic numbers defined by h(k) = n 1/jk, n ≥ 1.
n j=1
Numbers h(1) are called simply (ordinary) harmonic numbers. P
n
We are going to define differently generalized harmonic numbers.
Definition1. Foreveryk ∈Nnumbers H(k) definedrecursivelybyH(0)
n n
n on≥1,k≥0
=1, H(k) = n H(k−1)/j, n≥ 1 will be called generalized harmonic numbers of
n j=1 j
order k. P
Remark 2. It is easy to see that B(1,1) =H(k) and that H(1) is an ordinary n−th
n,k n+1 n
harmonic number.
Remark 3. Notice that H(k) is a symmetric function of order k of the numbers
n
{1,1/2,...,1/n} hence it can be expressed as a linear combination of some other
symmetric functions of order less or equal k. For example we have: H(1) = h(1)
n n
= H (the ordinary harmonic number), H(2) = H2/2+h(2)/2, H(3) = H3/6+
n n n n n n
H h(2)/2+h(3)/3 and so on.
n n n
Remark 4. Notice also that recursive equation that was obtained in the proof of
Proposition 1 i.e.
m(n+1) 1
A(m,i) − A(m,i) = A ,
n+1,k m(n+1)+i n,k m(n+1)+i n+1,k−1
4 PAWEL J.SZABL OWSKI
is valid also for k = 0,−1,−2,.... . Of course then we apply it in the following
form:
A =(m(n+1)+i)A −m(n+1)A(m,i),
n+1,k−1 n+1,k n,k
getting for example : A(m,i) = 1, A(m,i) = −m, A(m,i) = 0, A(m,i) = 1, A(m,i) =
0,−1 1,−1 n,−1 0,−2 1,−2
−m(m+i+1), A(m,i) = 2m2 , A(m,i) = 0 for n = 3,4,... . The fact that A(m,i)
2,−2 n,−2 n,−k
=0 for n≥k+1 was already noticed, justified and applied by Sondow in [5].
As a corollary we have the following result:
Theorem 1.
∞
n!
(2.1) M(m,i) = B(m,i) ,
k 2n+1m(i/m) n,k−1
nX=0 n+1
where numbers B(m,i) are defined above.
n,k
i) In particular :
1 (−1)kE
(2.2) M(2m,m) = M(2,1) =π2k+1 2k ,
2k+1 m2k+1 2k+1 2(2m)2k+1(2k)!
∞
n!(H −H /2)
(2.3) M(2,1) =K= 2n+1 n ,
2 2(2n+1)!!
nX=0
where H denotes n−th (ordinary) harmonic number.
n
ii) for m=i=1, k∈N:
∞ (−1)j−1 ∞ H(k−1)
n
(2.4) = ,
jk n2n
Xj=1 nX=1
and consequently for k =2,3,...
2k−1 ∞ H(k−1)
n
(2.5) ζ(k)= .
2k−1−1 n2n
nX=1
Proof. Applying Euler transformation to the series M(m,i) we have
n,k
∞
M(m,i) = A(m,i)/2n+1.
n,k n,k
nX=0
Now it remains to apply Proposition 1. i) To see that (2.1) reduces to (2.3) when
k = 2, m = 2 and i = 1 notice that B(2,1) = 1 and consequently B(2,1) =
n,0 n,1
n 1/(2j+1)=H −2H .Furtherwehave(1/2) = n (j+1/2)=(2n+
j=0 2n+1 n n+1 j=0
1P)!!/2n+1. To justify (2.2) we have to observethat 2M(2m,m)Q=Sˆ(2k+1,2m,m)=
2k+1
∞ (−1)j = 1 ∞ (−1)j . The fact that ∞ (−1)j
j=−∞ (j2m+m)2k+1 m2k+1 j=−∞ (j2+1)2k+1 j=−∞ (j2+1)2k+1
P=π2k+1(−1)k E2k dates bPack to Euler and was recalled in [P6].
22k+1(2k)!
ii) If m=i=1 we have (1) =(n+1)!. Recallalso that then B(m,i) =H(k) .
n+1 n,k n+1
(2.5) follows additionally (1.1). (cid:3)
HURWITZ FUNCTION 5
Remark 5. Notice that when i=1 then the sequence B(m,1) is generated by the
n,k
n o
recursion: B(m,1) =1, B(m,1) = n B(m,1)/(mj+1). Now arguing by induction
n,0 n,k j=0 n.k−1
we see that ∀n ≥ 0: B(m,1) ≥ BP(m,1). Consequently we deduce that the sequence
n,k n,k−1
M(m,1) is increasing which is not so obvious when considering only definition
k
n ok≥1
of these numbers. It is also elementary to notice that
lim M(m,1) =1.
k
k−→∞
In particular we deduce that the sequence {ζ(k)(1−1/2k−1)} is increasing.
k>1
Remark 6. Notice that one can easily prove (by induction) that ∀n,k ∈ N : 1 ≤
H(k) ≤n. Hence, utilizing (2.5) we have:
n
ln2−1/2 2k−1 m H(k−1) 1
≤ ζ(k)− n+1 ≤ ,
2m+1(m+1) (cid:12)(cid:12)(cid:12) 2k−1−1nX=02n+1(n+1)(cid:12)(cid:12)(cid:12) 2m+1
(cid:12) (cid:12)
since 2k−1 ≤2 for k≤2 a(cid:12)nd further ζ(k)− 2k−1 m (cid:12) Hn(k+−11) ≤
2k−1−1 (cid:12) 2k−1−1 n=0 2n+1(n+1)(cid:12)
2k−1 ∞ 1/2n+1≤ 2k−1 /2m+2(cid:12)(cid:12)and m+1 ≥ P1 and ∞ (cid:12)(cid:12) 1
2k−1−1 n=m+1 2k−1−1 (cid:12) n+1 n−m+1 n=m(cid:12)+1 2n+1(n+1)
≥ 1P ∞ 1 = ln2−1/2 . P
2m+1(m+1) n=m+1 2n−m+1(n−m+1) 2m+1(m+1)
P
Remark 7. Formulae (2.1) and (2.5) can be considered as a series transformation
to speed up its convergence. Apery for ζ(3) in his breakthrough paper and later
Hessami Pilehrood et al. in [3] obtained series transformations to speedup series
appearinginthedefinitionsofRiemannorHurwitzzetafunctions. Asitisremarked
in [3] all these transformation give series more or less of the form c /4n where c
n n
= O(1), but for different arguments of ζ one gets very different series in a very
different, particular way. Apery’s one is one of the simplest. Formulae (2.1) and
(2.5) offer unified form of the transformed series and speed of convergence is only
slightly worse. Namely of the form c /2n.
n
Remark 8. Notice also that analyzing the proof of Proposition 1 we can formulate
the following observation. Let us denote A(m,l) = n (−1)j n /(mj + l)s for
n,s j=0 j
Re(s)>0. Then P (cid:0) (cid:1)
m(n+1) 1
A(m,l) − A(m,l) = A(m,l) .
n+1,s m(n+1)+l n,s m(n+1)+l n+1,s−1
Hence keeping in mind that A(m,l) = 1/ls and assuming that we know numbers
0,s
A(m,l) weareabletogetnumbers A(m,l) andconsequentlyfindζ(s,l/m).
n,s−1 n,s
n on≥0 n on≥0
In particular if m = l = 1 we get A − n+1A = 1 A where we
n+1,s n+2 n,s n+2 n+1,s−1
denoted A = A(1,1) to simplify notation. Consequently we deduce that A =
n,s n,s n,s
1 n A . Since we can iterate this relationship we see that the knowledge
n+1 j=1 j,s−1
of fuPnctions An,s for Re(s)∈(0,1] implies knowledge of these functions for s with
Re(s)>0.
6 PAWEL J.SZABL OWSKI
3. Generating functions and integral representation Riemann zeta
functions at integer values
∞
Letusdenotebyf (x)thegeneratingfunctionofnumbers H(n) i.e. f (x)
n j n
n oj=0
= ∞ xjH(n). We have the following simple observation:
j=0 j+1
P
Proposition 2. i) ∀x∈(−1,1): f (x)=1, f (x)=1/(1−x):
−1 0
1 x
(3.1) f (x)= f (y)dy,
n n−1
x(1−x)Z
0
n≥1.
ii) Let us denote Q(x,y) the generating function of function series {f } i.e.
n n≥0
Q(x,y)= ∞ yjf (x), for y ∈(−1,1). We have
j=0 j
P
B(x,1−y,1+y)
(3.2) Q(x,y)= ,
x1−y(1−x)1+y
where B(x,a,b) denotes incomplete beta function.
Proof. i) We have f (x)= ∞ xj−1H(n) = ∞ xj−1 j H(n−1)/k =
n j=1 j j=1 k=1 k
∞ H(n−1)/k ∞ xj−1P= 1 xk−1HP(n−1)/k=P
k=1 k j=k 1−x k=1 k
P1 ∞ H(nP−1) syk−1dy = P1 x ∞ yk−1H(n−1)dy
x(1−x) k=1 k 0 x(1−x) 0 k=1 k
= 1 P xf (y)dRy. R P
x(1−x) 0 n−1
ii)WeRhave: (1−x)xQ(x,y)= ∞ yj(1−x)xf (x)=x+ ∞ yj xf (z)dz
j=0 j j=1 0 j−1
= P P R
x+ x ∞ yjf (z)dz)=x+y xQ(z,y))dz.Differentiatingwithrespecttoxwe
0 j=1 j−1 0
get:R(1P−2x)Q(x,y)+x(1−x)Q′(Rx,y)=1+yQ(x,y). Now solvingthis differential
equation we get Q(x,y)= Beta(x,1−y,1+y)−C(y). Recalling that Q(0,y)=1/(1−y)
x1−y(1−x)1+y
we see C(y)=0. (cid:3)
Let us denote for simplicity ˆζ(s) =df ∞ (−1)j/js for Re(s) > 0. Notice that
j=1
following (2.4) we have P
1/2 1
(3.3) ζˆ(k)= f (x)dx= f (1/2),
k−1 k
Z 4
0
for k =1,2,... .
We also have:
∞
yjˆζ(j)=B(1/2,1−y,1+y),
Xj=0
for y ∈(−1,1) following (3.2).
Recall that ∞ ζ(2j)t2j = 1−πtcot(πt) hence ∞ ζˆ(2j)t2j = πt −1
j=1 j=1 sin(πt)
after some algePbra. Hence P
∞
πy
y2j+1ˆζ(2j+1)=B(1/2,1−y,1+y)+1− ,
sin(πy)
Xj=0
HURWITZ FUNCTION 7
since ˆζ(0) = 1/2. Let us remark that there exist some expansions of incomplete
beta function. Applying one of them one can we have for example:
∞ ∞ (−y)
yjζˆ(j)=2y−1 j ,
j!(j+1−y)2j
Xj=0 Xj=0
for y ∈(0,1).
4. Remarks on particular values
In[6]thesumsoftheformS(n,k,l)= ∞ 1 , Sˆ(n,k,l)= ∞ (−1)j
j=−∞ (jk+l)n j=−∞ (jk+l)n
were analyzed and some of them were caPlculated. From the results ofPthis paper it
follows that the following sums:
M(m,i)+(−1)k+1M(m,m−i)
k k
have values of the form πk times some know, analytic number. Notice that this
statement is trivial for k odd, m even and i=m/2.
Inparticularwegetfork =2l wehaveM(m,i)−M(m,m−i) = 1 (ζ(2l,l/(2m))−
2l 2l m2l
ζ(2l,(m+i)/(2m))−ζ(2l,(m−i)/(2m))+ζ(2l,(2m−i)/(2m))= 1 (ζ(2l,l/(2m))
m2l
+ζ(2l,1−i/(2m))−ζ(2l,(m+i)/(2m)−ζ(2l,(m−i)/(2m)).
Following [6] we also have for k ≥1:
1 (22k−1)
S(2k,4,1)= (ζ(2k,1/4)+ζ(2k,3/4))=π2k (−1)k+1B ,
42k 2(2k)! 2k
where B denotes 2k−th Bernoulli number. In particular we have
2k
16K=(ζ(2,1/4)−ζ(2,3/4));(ζ(2,1/4)+ζ(2,3/4))=2π2.
Finally letus recallthat ζ(2l,1)=(−1)l+1B (2π)2l. Using formula(2.5)we get:
2l2(2l)!
(2π)2l 22l−1 ∞ H(2l−1)
(−1)l+1B = n ,
2l2(2l)! 22l−1−1 n2n
nX=1
and consequently we obtain the following expansions of even powers of π :
(2l)! ∞ H(2l−1)
π2l =(−1)l+1 n .
(22l−1−1)B n2n
2l nX=1
References
[1] Choi, Junesang. Certain summation formulas involving harmonic numbers and generalized
harmonicnumbers.Appl. Math. Comput.218(2011), no.3,734–740. MR2831299
[2] Choi,Junesang. Summationformulasinvolvingbinomialcoefficients, harmonicnumbers,and
generalized harmonicnumbers. Abstr. Appl. Anal. 2014,Art.ID 501906, 10pp.MR3246339
(Reviewed)
[3] Hessami Pilehrood,Kh.;HessamiPilehrood, T.Bivariateidentities forvalues of theHurwitz
zeta function andsupercongruences. Electron. J. Combin. 18(2011), no.2, Paper 35,30pp.
MR2900448(Reviewed)
[4] Kronenburg,M.,J.Somegeneralizedharmonicnumeridentities,arxiv: 1103.5430v2
[5] Sondow, Jonathan. Analytic continuation of Riemann’s zeta function and values at negative
integers viaEuler’stransformationofseries.Proc. Amer. Math. Soc. 120(1994), no.2,421–
424.MR1172954(94d:11066)
[6] Szab lowski,Pawel J.AfewremarksonvaluesofHurwitzZetafunctionatnaturalandrational
arguments,submitted, http://arxiv.org/abs/1405.6270
8 PAWEL J.SZABL OWSKI
DepartmentofMathematicsandInformationSciences,,WarsawUniversityofTech-
nology,ul. Koszykowa 75,00-662Warsaw, Poland
E-mail address: [email protected]