Mathematica Aeterna, Vol. 5, 2015, no. 4, 583-592
On connection between values
of Riemann zeta function at rationals
and generalized harmonic numbers
Pawe÷J. Szab÷
owski
Department of Mathematics and Information Sciences,
Warsaw University of Technology
ul. Koszykowa 75,
00-662 Warsaw, Poland
Abstract
Using Euler transformation of series, we relate values of Hurwitz zeta function (s; t) at integer and rational values of arguments to certain rapidly converging series, where some generalized harmonic numbers appear. Most of
the results of the paper can be derived from the recent, more advanced results, on the properties of Arakawa-Kaneko zeta functions. We derive our
results directly, by solving simple recursions. The form of mentioned above
generalized harmonic numbers carries information, about the values of the arguments of Hurwitz function. In particular we prove: 8k 2 N : (k; 1) =
(k 1)
P
k 1
(k)
Hn
(k) = 2k2 1 1 1
n=1 n2n ; where Hn are de…ned below generalized harmonic
P
n!(H2n+1 Hn =2)
numbers, or that K = 1
; where K denotes Calatan constant
n=0
2(2n+1)!!
and Hn denotes n th (ordinary) harmonicPnumber. Further we show that
j 1 k
generating function of the numbers ^(k) = 1
=j , k 2 N and ^(0)
j=1 ( 1)
= 1=2 is equal to B(1=2; 1 y; 1 + y) where B(x; a; b) denotes incomplete beta
function.
Mathematics Subject Classi…cation: Primary 111M35, 40G05, ; Secondary 05A15, 05E05
Keywords: Riemann zeta function Hurwitz zeta function, Euler summation, Harmonic numbers, generalized Harmonic numbers, Catalan constant.
584
1
Pawe÷J. Szab÷owski
Introduction
First let us recall basic notions and de…nitions that we will work with. By the
Hurwitz function (s; ) we will mean:
(s; ) =
1
X
j=0
1
;
(j + )s
considered for Re s > 1; Re 2 (0; 1]: Function (s; 1) is called Riemann zeta
function. We will denote it also by (s); if it will not cause misunderstanding.
It turns out that both these functions can be extended to holomorphic functions of s on the whole complex plane except s = 1 where a single pole exists.
Of great help in doing so is the formula
2s
(s) =
2s
1
1
1
X
( 1)j
1 j=1
js
1
;
(1)
that enables to extend Riemann zeta function to the whole half plane Re s > 0.
We will consider numbers:
(m;i)
Mk
=
1
X
j=0
( 1)j
;
(mj + i)k
P
(1;1)
j 1 k
=j and
1g : Notice that Mk = 1
j=1 ( 1)
P
j
(2;1)
(2;1)
1
( 1)
M1 = =4: The number M2 = j=0 (2j+1)2 is called Catalan constant K:
It is elementary to notice that
for m 2 N and i 2 f1; : : : ; m
(m;i)
Mk
=
1
( (k; i=(2m))
(2m)k
(k; 1=2 + i=(2m)):
The main idea of this paper is to apply the so called Euler transformation,
that was nicely recalled by Sondow in [12]. As pointed out there we have:
1
X
k=1
k 1
( 1)
ak =
1
X
n
a1 =2n+1 ;
n=0
where fak gk 1 is a sequence of complex numbers and the sequence n ak is deP
n
n 1
…ned recursively: 0 ak = ak; n ak = n 1 ak
ak+1 = nm=0 ( 1)m m
am+k :
Sondow in [12] presented general idea of applying Euler transformation to
Riemann function. He however stopped half way in the sense that he calculated
…nite di¤erences n applied to (j + 1) s only for s being negative integers. We
are going to make a few steps further and calculate these di¤erences pointing
out the rôle of the generalized harmonic numbers in those calculations.
585
Hurwitz function
As stated in the abstract most of the results of this paper can be derived
from recent more advanced results concerning Lerch and Arakawa-Koneko zeta
functions that were presented in the series of papers [3], [4], [5], [6], [7].
We present here an alternative, simple way of obtaining them by solving
simple recursion.
The paper is organized as follows. In the next section 2 we present an auxiliary result that enables application of Euler transformation to the analyzed
(m;i)
series. Further we present transformed series approximating numbers Mk .
In Section 3 we calculate generating functions of certain series of numbers and
functions. More precisely we calculate generating functions of the generalized
harmonic numbers that we have de…ned in the previous section. We also calculate generating function of the series of the generating functions that were
de…ned previously. It turns out that this calculation enables to obtain the
generating function of the series sums that appear on the right hand side of
(1). Finally in the last Section 4 there are collected cases when exact values
of numbers M are known.
2
Euler transformation
To proceed further we need the following result.
(m;i)
Proposition 1 Let us denote An;k =
Pn
j=0 (
n
=(mj + i)k , n = 0; 1; : : : ;
j
(m;i)
Bn;0 = 1; B0;k = ik1 1 ; k 1;
1)j
and the family of sequences de…ned recursively:
P
(m;i)
(m;i)
1
8n; k 0 : Bn;k = nj=0 (mj+i)
Bj;k 1 . We have then:
(m;i)
(m;i)
(m;i)
n!
8 m 2 N : A0;0 = 1, An;0 = 0; An;1 = m(i=m)
; where
n+1
(a)n = a(a + 1) : : : (a + n 1) is the so called ’rising factorial’.
8 n 0; k 1 we get:
(m;i)
An;k =
n!
(m;i)
Bn;k 1 :
m(i=m)n+1
Proof. i) The fact that An;0 = 0 follows immediately properties of binomial
coe¢ cients. Notice that we have
X
m(n + 1)
n+1
( 1)n+1
(m;i)
=(mj + i)k +
An;k =
( 1)j
m(n + 1) + i
j
(m(n + 1) + i)k
j=0
n
(m;i)
An+1;k
m(n + 1) X
n
( 1)j
=(mj + i)k =
m(n + +1) + i j=0
j
n
586
Pawe÷J. Szab÷owski
( 1)n+1
k
(m(n + 1) + i)
+
n
X
( 1)j (
j=0
n+1
j
j!(n
m(n + 1)!
)=(mj + i)k
j)!(m(n + 1) + i)
n
X
n+1
1
=(mj + i)k
=
+
( 1)j
k
m(n + 1) + j j=0
j
(m(n + 1) + i)
( 1)n+1
X
1
n+1
=(mj + i)k
( 1)j
m(n + 1) + i j=0
j
n+1
=
1
=
1
1
(m;i)
A
m(n + 1) + i n+1;k
1
(m;i)
m(n+1)
A
m(n+1)+i n;1
(m;i)
(m;i)
n!
= 0 from which immediately follows that An;1 = m(i=m)
since A0;1 = 1i .
n+1
(m;i)
(m;i)
(m;i)
m(n+1)
1
Now divide both sides of the identity An+1;k mn+m+i
An;k = mn+m+i
An+1;k 1
(m;i)
(m;i)
(m;i)
(m;i)
(m;i)
(m;i)
(m;i)
1
by An+1;1 and denote Bn;k = An;k =An;1 : We get Bn+1;k Bn;k = m(n+1)+i
Bn+1;k 1
P
(m:i)
(m;i)
(m;i)
(m;i)
m(n+1)
1
since An+1;1 = mn+m+i
An;1 : Hence Bn;k = nj=0 mj+i
Bj;k 1 since 8k
1:
(n;i)
k 1
B0;k = 1=i :
since (1
m(n+1 j)
)
(mn+m+i)
=
jm+i
.
m+mn+i
(m:i)
Now notice that we have An+1;1
Remark 1 In the literature (compare e.g. [2], [8], [10]) there function notions
P
(k)
of harmonic and generalized harmonic numbers de…ned by hn = nj=1 1=j k ,
(1)
n 1: Numbers hn are called simply (ordinary) harmonic numbers. Another
way to generalize the notion of harmonic numbers was presented by Coppo
and Candelpergher in their papers [5], [6], [7]. There the generalized harmonic
numbers were de…ned using Bell’s polynomials.
We are going to de…ne di¤erently generalized harmonic numbers.
o
n
(k)
de…ned recursively by
De…nition 1 For every k 2 N numbers Hn
n 1;k 0
P
(k 1)
(0)
(k)
Hn = 1; Hn = nj=1 Hj
=j; n
1 will be called generalized harmonic
numbers of order k:
(1;1)
(k)
(1)
Remark 2 It is easy to see that Bn;k = Hn+1 and that Hn is an ordinary
n th harmonic number.
(k)
Remark 3 Notice that Hn is a symmetric function of order k of the numbers
f1; 1=2; : : : ; 1=ng hence it can be expressed as a linear combination of some
(1)
other symmetric functions of order less or equal k: For example we have: Hn
(1)
(2)
(2)
(3)
= hn = Hn (the ordinary harmonic number), Hn = Hn2 =2 + hn =2; Hn =
(2)
(3)
Hn3 =6 + Hn hn =2 + hn =3 and so on.
Remark 4 Notice also that recursive equation, that was obtained in the proof
of Proposition 1 i.e.
(m;i)
An+1;k
m(n + 1)
1
(m;i)
An;k =
An+1;k 1 ;
m(n + 1) + i
m(n + 1) + i
587
Hurwitz function
is valid also for k = 0; 1; 2; : : : : . Of course then we apply it in the following
form:
(m;i)
An+1;k 1 = (m(n + 1) + i)An+1;k m(n + 1)An;k ;
(m;i)
(m;i)
(m;i)
(m;i)
(m;i)
getting for example : A0; 1 = 1; A1; 1 = m; An; 1 = 0; A0; 2 = 1; A1; 2
(m;i)
(m;i)
= m(m + i + 1); A2; 2 = 2m2 , An; 2 = 0 for n = 3; 4; : : : . The fact that
(m;i)
An; k = 0 for n k + 1 was already noticed, justi…ed and applied by Sondow
in [12].
As a corollary we have the following result:
Theorem 2
(m;i)
Mk
=
1
X
n=0
(m;i)
Bn;k
n!
(m;i)
2n+1 m(i=m)
Bn;k 1 ;
n+1
(2)
where numbers
are de…ned above.
i) In particular :
(2m;m)
M2k+1
=
1
(2;1)
M
=
m2k+1 2k+1
(2;1)
M2
=K=
( 1)k E2k
;
2(2m)2k+1 (2k)!
2k+1
1
X
n!(H2n+1
Hn =2)
;
2(2n + 1)!!
n=0
(3)
(4)
where Hn denotes n th (ordinary) harmonic number.
ii) for m = i = 1; k 2 N:
1
X
( 1)j
jk
j=1
1
=
1
(k
X
Hn
n=1
1)
(5)
;
n2n
and consequently for k = 2; 3; : : :
(k) =
2k
2k
1
1
1
1
(k
X
Hn
n=1
1)
n2n
(6)
:
(m;i)
Proof. Applying Euler transformation to the series Mn;k we have
(m;i)
Mn;k =
1
X
(m;i)
An;k =2n+1 :
n=0
Now it remains to apply Proposition 1. i) To see that (2) reduces to (4)
(2;1)
(2;1)
when
Pnk = 2; m = 2 and i = 1 notice that Bn;0 = 1 and consequently
Qn Bn;1
=
2Hn : Further we have (1=2)n+1 = j=0 (j +
j=0 1=(2j + 1) = H2n+1
588
Pawe÷J. Szab÷owski
(2m;m)
1=2) = (2n + 1)!!=2n+1 : To justify (3) we have to observe that 2M2k+1 =
P
P1
( 1)j
( 1)j
1
^
S(2k+1;
2m; m) = 1
j= 1 (j2m+m)2k+1 = m2k+1
j= 1 (j2+1)2k+1 : The fact that
P1
( 1)j
E2k
2k+1
( 1)k 22k+1
dates back to Euler and was recalled
j= 1 (j2+1)2k+1 =
(2k)!
in [13].
(m;i)
ii) If m = i = 1 we have (1)n+1 = (n + 1)!: Recall also that then Bn;k =
(k)
Hn+1 : (6) follows additionally (1).
n
(m;1)
Bn;k
o
Remark 5 Notice that when i = 1, then the sequence
is generated
Pn
(m;1)
(m;1)
(m;1)
by the recursion: Bn;0 = 1; Bn;k = j=0 Bn:k 1 =(mj + 1): Now arguing
(m;1)
by induction we see that 8n
n
o
(m;1)
that the sequence Mk
(m;1)
Bn;k 1 : Consequently we deduce
0: Bn;k
is increasing, which is not so obvious when
k 1
considering only de…nition of these numbers. It is also elementary to notice
that
(m;1)
lim Mk
= 1:
k !1
1=2k 1 )gk>1 is increasing.
In particular we deduce that the sequence f (k)(1
Remark 6 Notice that one can easily prove (by induction) that 8n; k 2 N :
(k)
1 Hn
n: Hence, utilizing (6) we have:
ln 2 1=2
m+1
2
(m + 1)
2k
since
2k 1
2k 1
2k
1
P1
1
2 for k
n+1
n=m+1 1=2
P
1
1
n=m+1 2n
2m+1 (m+1)
1
1
(k)
2k
1
2k 1
1
2 and further
2k 1
2k
1
1
m+1)
=
n=0
(k)
(k 1)
Hn+1
n+1
2 (n + 1)
2k
2k 1
m+1
n+1
ln 2 1=2
.
2m+1 (m+1)
=2m+2 and
1
m+1 (n
m
X
1
1
Pm
1
n m+1
n=0
1
2m+1
;
(k 1)
Hn+1
n+1
2
(n+1)
and
P1
1
n=m+1 2n+1 (n+1)
Remark 7 Formulae (2) and (6) can be considered as a series transformation
to speed up its convergence. Apery for (3) in his breakthrough paper and later
Hessami Pilehrood et al. in [9] obtained series transformations to speedup
series appearing in the de…nitions of Riemann or Hurwitz zeta functions. As
it is remarked in [9] all these transformation give series more or less of the
form cn =4n where cn = O(1); but for di¤erent arguments of one gets very
di¤erent series in a very di¤erent, particular way. Apery’s one is one of the
simplest. Formulae (2) and (6) o¤er uni…ed form of the transformed series
and speed of convergence is only slightly worse. Namely of the form cn =2n :
589
Hurwitz function
Remark 8 Notice also that analyzing the proof of Proposition
1 we can formuPn
(m;l)
late the following observation. Let us denote An;s = j=0 ( 1)j nj =(mj + l)s
for Re(s) > 0: Then
(m;l)
An+1;s
1
m(n + 1)
(m;l)
A(m;l)
A
:
n;s =
m(n + 1) + l
m(n + 1) + l n+1;s 1
(m;l)
Hence keeping
in mind that A0;s = 1=ls and
that we know numbers
n
o
n assuming
o
(m;l)
(m;l)
An;s 1
we are able to get numbers An;s
and consequently …nd
n 0
n 0
(s; l=m):
In particular if m = l = 1 we get An+1;s
n+1
A
n+2 n;s
=
1
A
n+2 n+1;s 1
where we
(1;1)
An;s
denoted
to simplify notation. Consequently we deduce that An;s =
Pn An;s =
1
A
:
Since
we can iterate this relationship we see that the knowlj=1 j;s 1
n+1
edge of functions An;s , for Re(s) 2 (0; 1] implies knowledge of these functions
for s with Re(s) > 0:
3
Generating functions and integral representation Riemann zeta functions at integer values
n
o1
(n)
Let us denote by fn (x) the generating function of numbers Hj
i.e. fn (x)
j=0
P1 j (n)
= j=0 x Hj+1 : We have the following simple observation:
Proposition 3 i) 8x 2 ( 1; 1) : f 1 (x) = 1; f0 (x) = 1=(1
Z x
1
fn (x) =
fn 1 (y)dy;
x(1 x) 0
x) :
(7)
n
1:
ii) Let us denote
y) the generating function of function series ffn gn
P1 Q(x;
j
i.e. Q(x; y) = j=0 y fj (x); for y 2 ( 1; 1): We have
Q(x; y) =
B(x; 1 y; 1 + y)
;
x1 y (1 x)1+y
where B(x; a; b) denotes incomplete beta function.
P
P1 j
j 1 (n)
Proof. i) We have fn (x) = 1
x
H
=
j
j=1
j=1 x
P1
P1 j 1
P
(n 1)
1
k 1 (n 1)
=k j=k x
= 1 x k=1 x Hk
=k =
k=1 Hk
R
R
P
P
s
x
(n
1)
(n
1
1
1
Hk
y k 1 dy = x(11 x) 0
y k 1 Hk
k=1
k=1
x(1 x)
0
Rx
= x(11 x) 0 fn 1 (y)dy:
0
(8)
1
Pj
k=1
1)
dy
(n 1)
Hk
=k =
590
Pawe÷J. Szab÷owski
ii) We have: (1 x)xQ(x; y) =
P1
j=0
R
P
j x
f (z)dz
y j (1 x)xfj (x) = x+ 1
j=1 y
0 j 1
= R P
Rx
x
1
j
x+ 0
j=1 y fj 1 (z)dz) = x + y 0 Q(z; y))dz: Di¤erentiating with respect
to x we get: (1 2x)Q(x; y) + x(1 x)Q0 (x; y) = 1 + yQ(x; y): Now solving
y;1+y) C(y)
this di¤erential equation we get Q(x; y) = Beta(x;1
: Recalling that
x1 y (1 x)1+y
Q(0; y) = 1=(1 y) we see C(y) = 0.
df P
j
s
Let us denote for simplicity ^(s) = 1
j=1 ( 1) =j for Re(s) > 0: Notice
that following (5) we have
^(k) =
Z
1=2
0
for k = 1; 2; : : : .
We also have:
1
X
1
fk 1 (x)dx = fk (1=2);
4
y j ^(j) = B(1=2; 1
(9)
y; 1 + y);
j=0
for y 2 ( 1; 1) following
(8).
P
2j
(2j)t
=1
Recall that 1
j=1
1 after some algebra. Hence
1
X
t cot( t) hence
y 2j+1 ^(2j + 1) = B(1=2; 1
P1 ^
2j
=
j=1 (2j)t
y; 1 + y) + 1
j=0
t
sin( t)
y
;
sin( y)
since ^(0) = 1=2: Let us remark that there exist some expansions of incomplete
beta function. Applying one of them we have for example:
1
X
j^
y 1
y (j) = 2
j=0
1
X
j=0
( y)j
;
j!(j + 1 y)2j
for y 2 (0; 1):
4
Remarks on particular values
P
P1
( 1)j
1
^
In [13] the sums of the form S(n; k; l) = 1
j= 1 (jk+l)n ; S(n; k; l) =
j= 1 (jk+l)n
were analyzed and some of them were calculated. From the results of this paper
it follows that the following sums:
(m;i)
Mk
(m;m i)
+ ( 1)k+1 Mk
have values of the form k times some known, analytic number. Notice that
this statement is trivial for k odd, m even and i = m=2:
591
Hurwitz function
(m;i)
(m;m i)
In particular we get for k = 2l we have M2l
M2l
= m12l ( (2l; l=(2m))
(2l; (m+i)=(2m)) (2l; (m i)=(2m))+ (2l; (2m i)=(2m)) = m12l ( (2l; l=(2m))
+ (2l; 1 i=(2m))
(2l; (m + i)=(2m)
(2l; (m i)=(2m)):
Following [13] we also have for k 1:
S(2k; 4; 1) =
1
( (2k; 1=4) + (2k; 3=4)) =
42k
where B2k denotes 2k
2k (2
2k
1)
( 1)k+1 B2k ;
2(2k)!
th Bernoulli number. In particular we have
(2; 3=4)); ( (2; 1=4) + (2; 3=4)) = 2 2 :
16K =( (2; 1=4)
2l
)
: Using formula (6) we
Finally let us recall that (2l; 1) = ( 1)l+1 B2l (2
2(2l)!
get:
1
(2l 1)
22l 1 X Hn
(2 )2l
l+1
= 2l 1
;
( 1) B2l
2(2l)!
2
1 n=1 n2n
and consequently we obtain the following expansions of even powers of
2l
l+1
= ( 1)
(22l
:
1
(2l 1)
X
(2l)!
Hn
:
1
1)B2l n=1 n2n
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Received: August, 2015