ANALYTIC EXTENSION OF HYPERHARMONIC
NUMBERS
ISTVÁN MEZŐ
Abstract. We define the analytic extension of hyperharmonic
numbers involving the Pochhammer symbol, gamma and digamma
functions. In addition, some sum of hyperharmonic series have
been calculated. Surprisingly, the Lerch transcendent appears in
the closed form of the sums.
1. Introduction
Although the harmonic numbers and their connections with the
transcendental functions mentioned in the abstract are well known,
this is not true for hyperharmonic numbers because of their novelty.
The author points out that there is a close relationship between hyperharmonic numbers and transcendental functions, too. In 1996, J. H.
Conway and R. K. Guy in [3] have defined the notion of hyperharmonic
numbers.
The n-th harmonic number is the n-th partial sum of the harmonic
series:
n
X
1
Hn =
.
k
k=1
(1)
Hn := Hn , and for all r > 1 let
n
X
(r−1)
(r)
Hn =
Hk
k=1
be the n-th hyperharmonic number of order r. These numbers can be
expressed by binomial coefficients and ordinary harmonic numbers:
n+r−1
(r)
(1)
Hn =
(Hn+r−1 − Hr−1 ).
r−1
We use this form later to construct the analytic extension of these
numbers.
The hyperharmonic numbers have combinatorial connections. To
present this fact, we need to introduce the notion of r-Stirling numbers.
Getting deeper insight, see [1] and the references given there.
2000 Mathematics Subject Classification. 33E20.
Key words and phrases. harmonic numbers, hyperharmonic numbers, hyperharmonic function, analytic extension, Lerch transcendent.
1
2
ISTVÁN MEZŐ
Definition 1. nk r is the number of permutations of the set {1, . . . , n}
having k disjoint, non-empty cycles, in which the elements 1 through
r are restricted to appear in different cycles.
The following identity integrates the hyperharmonic- and the rStirling numbers.
n+r
r+1 r
= Hn(r) .
n!
Now we turn our attention to formulate the results of this paper. Let
us introduce the necessary notions. The gamma function for <(z) > 0
is
Z ∞
Γ(z) =
tz−1 e−t dt.
0
It can be extended to the whole complex plane, except the non-positive
integers. It is well known that
Γ(n) = (n − 1)! n = 1, 2, . . . .
(2)
The logarithmic derivative of the gamma function is called digamma
function:
d
Γ0 (z)
Ψ(z) =
log(Γ(z)) =
.
dz
Γ(z)
For integers,
Ψ(n) = Hn−1 − γ.
(3)
(γ is the Euler-Mascheroni constant.) Ψ’s derivatives are called polygamma functions.
The Pochhammer symbol is
(x)n =
Γ(x + n)
= x(x + 1) · · · (x + n − 1).
Γ(x)
It’s derivatives are
d
(4)
(x)n = (x)n (Ψ(x + n) − Ψ(x)),
dx
d
(5)
(x)n = (x)n Ψ(x + n).
dn
2. The hyperharmonic function
According to (1), (2) and (3) one can write
Hn(r) =
(n + 1)(n + 2) · · · (n + r − 1)
(Ψ(n + r) − Ψ(r)) =
(r − 1)!
(n + 1)r−1
(n)r
(Ψ(n + r) − Ψ(r)) =
(Ψ(n + r) − Ψ(r)).
Γ(r)
nΓ(r)
Since these functions are analytic except Γ and Ψ have poles at z ∈
Z− = {0, −1, −2, . . . } the following definition is correct.
=
ANALYTIC EXTENSION OF HYPERHARMONIC NUMBERS
3
Definition 2. The function
(z)w
Hz(w) =
(Ψ(z + w) − Ψ(w))
zΓ(w)
is called hyperharmonic function (w, z + w ∈ C \ Z− ).
Some interesting values are computed:
√
3
3 3Γ 23
4 ln(2)
( 13 )
H1 =
, H1 =
2
π √ 3
√ 4π
1
2
π
π
3
3
−
3
( )
3
H 23 =
,
3
3
3Γ 23
√
2 π 2 + ln(2) − 12 π
( 34 )
H3 =
,
2
4
3Γ 34
√ √
3Γ 23 − 32 ln(3) + 12 π 3
3Γ 65
( 21 )
√
,
H1 =
3
2 π3
√
√
(2)
H√2 = (1 + 2)(Ψ(2 + 2) − 1 + γ),
( 21 )
(2+i)
H1+i ≈ 0.355 + 0.965i.
Using a formula given for the digamma function, the complex hyperharmonic function can be written as follows.
Proposition 3. If w, z + w ∈ C \ Z− , then
∞
(z)w X
1
(w)
Hz =
.
Γ(w) n=0 (z + w + n)(w + n)
Proof. According to [7], the digamma function satisfies the equality
∞ X
1
1
Ψ(z) = −γ +
−
(z ∈ C \ Z− ),
n
z
+
n
−
1
n=1
whence
∞ X
1
1
1
1
−
− +
Ψ(z + w) − Ψ(w) =
n z+w+n−1 n w+n−1
n=1
∞ X
1
1
=
+
=
−
z+w+n−1 w+n−1
n=1
∞ X
−w − n + 1 + z + w + n − 1
=
=
(z
+
w
+
n
−
1)(w
+
n
−
1)
n=1
=z
∞
X
n=1
=
1
,
(z + w + n − 1)(w + n − 1)
changing the range of summation, the result follows.
4
ISTVÁN MEZŐ
The derivatives of the hyperharmonic function can be easily computed:
Proposition 4.
(z)w 0
d (w)
Hz = Hz(w) (Ψ(z + w) − Ψ(z + 1)) +
Ψ (z + w),
dz
zΓ(w)
(z)w Ψ(w) 0
d (w)
Hz = Hz(w) (Ψ(z + w) − Ψ(w)) +
(Ψ (z + w) − Ψ0 (z)).
dw
zΓ(w)
Proof. The formulas (4), (5) and the definition of the digamma function
gives
(z)w
(z + 1)w−1
(Ψ(z + w) − Ψ(w)) =
(Ψ(z + w) − Ψ(w)) =
zΓ(w)
Γ(w)
(z + 1)w−1
(Ψ(z + w) − Ψ(z + 1))(Ψ(z + w) − Ψ(w))+
=
Γ(w)
(z + 1)w−1 0
Ψ (z + w) =
+
Γ(w)
(z)w
(z)w 0
=
(Ψ(z + w) − Ψ(w))(Ψ(z + w) − Ψ(z + 1)) +
Ψ (z + w) =
zΓ(w)
zΓ(w)
(z)w 0
= Hz(w) (Ψ(z + w) − Ψ(z + 1)) +
Ψ (z + w).
zΓ(w)
The second derivative can be calculated in the same way.
3. Series involving hyperharmonic numbers
Some interesting sums with respect to harmonic numbers are known,
see [2] and [4]. For example,
∞
X
Hn
= 2ζ(3),
2
n
n=1
∞
X
Hn
1
= π2.
n
n2
12
n=1
Our goal is to derive similar equalities for hyperharmonic series.
The main tool is the generating function. Since (see [6])
∞
ln(1 − x) X (r) n
(6)
−
=
Hn x ,
(1 − x)r
n=1
the following results come immediately:
∞
∞
(2)
(2)
X
X
Hn
9
2
Hn
= 4 ln(2),
= − ln
,
n
n
2
3
4
3
n=1
n=1
∞
∞
(3)
(3)
X Hn
X
Hn
27
2
=
8
ln(2),
=
−
ln
,
n
n
2
3
8
3
n=1
n=1
ANALYTIC EXTENSION OF HYPERHARMONIC NUMBERS
5
and so on. The following proposition helps us to calculate more diffcult
sums.
Proposition 5. We have
Z
1
ln(t)
dt = Li2 (t) + ln2 (t),
(1 − t)t
2
and for all 2 ≤ r ∈ N
Z
Z
ln(t)
ln(t)
1
ln(t)
,
dt =
dt −
−
r
r−1
r−1
(1 − t)t
(1 − t)t
(r − 1)t
(r − 1)2 tr−1
or, equivalently,
Z
r−1 X
ln(t)
1 2
ln(t)
1
dt = Li2 (t) + ln (t) −
+ 2k .
(1 − t)tr
2
ktk
k t
k=1
up to additive constants, where
Z
x
Li2 (x) =
1
ln(t)
dt
1−t
is the dilogarithm function.
Proof. The definition of Li2 gives that
Li02 (t) =
ln(t)
.
1−t
Moreover,
0
1 2
ln(t)
ln (t) =
,
2
t
whence
Li02 (t)
0
1 2
t ln(t) + (1 − t) ln(t)
ln(t)
+
ln (t) =
=
.
2
(1 − t)t
(1 − t)t
The first statement is proved. The second one also can be deduced by
differentiation. The derivative of the right-hand side has the form
ln(t)
(r − 1)tr−2 − (r − 1)2 ln(t)tr−2
−(r − 1)
−
−
=
(1 − t)tr−1
(r − 1)2 (tr−1 )2
(r − 1)2 tr
t ln(t)
tr − (r − 1) ln(t)tr
1
−
+
=
(1 − t)tr
(r − 1)(tr )2
(r − 1)tr
tr+1 ln(t)(r − 1) − tr (1 − t) + (1 − t)(r − 1) ln(t)tr + tr (1 − t)
=
=
(r − 1)t2r (1 − t)
t ln(t) + (1 − t) ln(t)
ln(t)
=
= r
,
r
t (1 − t)
t (1 − t)
as we want. Using this result, the verification of the summation formula
is straightforward.
=
6
ISTVÁN MEZŐ
Example 6. Let us see some examples. For instance, let the order
equal 2 and 3.
∞
(2)
X
Hn
1
= π 2 + 2 ln(2) − 1,
n
2 n
12
n=1
∞
(2)
X
2
Hn
1
2
2
2
= Li2
+
ln
− 3 ln
−3
3n n
3
2
3
3
n=1
∞
(3)
X
Hn
n=1
∞
X
n=1
2n n
=
7
1 2
π + 4 ln(2) − ,
12
4
(3)
Hn
21
21
=
ln(3) −
ln(2)+
n
3 n
8
8
1
1
13
2
+ ln2 (2) + ln2 (3) −
− ln(3) ln(2).
+ Li2
3
2
2
16
Proof. The formula (6) can be transformed into the form
∞
ln(1 − x) X (r) n−1
Hn x .
=
−
x(1 − x)r
n=1
Formal integration gives that
Z
∞
(r)
X
Hn n
ln(1 − x)
dx
=
x .
−
x(1 − x)r
n
n=1
It is known that – for a fixed x – the sum of a series equals to the integral
of the corresponding generating function from 0 to x. Therefore
Z x
∞
(r)
X
ln(1 − t)
Hn n
dt.
x =
−
r
n
t(1
−
t)
0
n=1
The substitution 1 − t = x gives that
Z a
Z 1−a
ln(1 − t)
ln(x)
−
dt
=
dx,
t(1 − t)r
(1 − x)xr
0
1
therefore the last proposition is applicable.
Let us fix the order: r = 2 and let x = 12 .
t= 12
∞
(2)
X
Hn
ln(t) 1
1 2
= Li2 (t) + ln (t) −
−
=
nn
2
2
t
t
t=1
n=1
1
1 2 1
1
1
= Li2
+ ln
− 2 ln
− 1 = π 2 + 2 ln(2) − 1,
2
2
2
2
12
2
1 2
since Li2 12 = 12
π − 21 ln (2). The next case is x = 13 . As above,
∞
(2)
X
Hn
2
1
2
2
2
= Li2
+
ln
− 3 ln
−3 .
nn
3
3
2
3
3
n=1
ANALYTIC EXTENSION OF HYPERHARMONIC NUMBERS
7
Since there is no known closed form of Li2 32 ≈ 0.3662, it is not
possible to simplify this sum.
Let r be 3. In this case
t= 12
∞
(3)
X
Hn
1 2
ln(t) 1
1
ln(t)
=
= Li2 (t) + ln (t) −
− − 2−
nn
2
2
2
t
t
4t
2t
t=1
n=1
1
1 2 1
1
1
1
Li2
+ ln
− 2 ln
− 2 − 1 − 2 ln
+1+ =
2
2
2
2
2
4
7
1
= π 2 + 4 ln(2) − .
12
4
The last identity can be proved as the former ones. The approach
described above gives a method to calculate an arbitrary sum which
has the form
∞
(r)
X
Hn
(r = 1, 2, . . . ).
nn
c
n=1
In the next section we extend this result to arbitrary r ∈ C \ Z.
4. Hyperharmonic series and the Lerch transcendent
Now, we extend our result to real, even complex orders. As we shall
prove, there is a general form to express the integral
Z
ln(1 − x)
−
dx
x(1 − x)r
except the cases r = 0, 1, 2, . . . described in the last section. We introduce the notion of the Lerch transcendent (see [5], for example).
∞
X
zn
Φ(z, s, a) =
(a + n)s
n=0
The next proposition helps us to calculate the sum of some type of
hyperharmonic series with non-integer order.
Proposition 7. If r ∈ C\Z, then the integral of the generating function
(r)
of Hn divided by x can be written as follows:
Z
ln(t)
r ln(t) + 1 + r2 ln(t)Φ(t, 1, −r) − r2 Φ(t, 2, −r)
dt
=
.
(1 − t)tr
tr r2
Proof. To prove our statement, let us differentiate the right hand side
of the formula given in the proposition. It equals to
r ln(t) + 1 + r2 ln(t)Φ(t, 1, −r) − r2 Φ(t, 2, −r)
+
tr+1 r
rΦ(t,1,−r)
r
1
2 Φ(t,1,−r)
2
+
r
+
r
ln(t)
+
t
t
t(t−1)
t
+
−
r
2
tr
−
8
ISTVÁN MEZŐ
r2
−
Φ(t,1,−r)
t
+ r Φ(t,2,−r)
t
=
tr r2
−r2 ln(t) − r − r3 ln(t)Φ(t, 1, −r) + r3 Φ(t, 2, −r) + r + r2 Φ(t, 1, −r)
=
+
tr+1 r2
1
r2 ln(t) t−1
+ rΦ(t, 1, −r) − r2 (Φ(t, 1, −r) + rΦ(t, 2, −r))
+
=
tr+1 r2
1
− ln(t) +
−r2 ln(t) + r2 ln(t) 1−t
=
=
r+1
2
t r
tr+1
as we expect. We used the facts
ln(t)
1−t
=
ln(t)
,
(1 − t)tr
d
1
aΦ(z, 1, a)
Φ(z, 1, a) =
−
,
dz
z(1 − z)
z
d
Φ(z, 1, a) aΦ(z, 2, a)
Φ(z, 2, a) =
−
.
dz
z
z
Since the Φ(z, s, a) function have singularities in the points a ∈ Z−
if <(s) > 0 and the hyperharmonic function has poles at negative
integers, we have to restrict us the cases r ∈ C \ Z.
Example 8. Let us see a concrete series. For instance, let r = 32 .
First of all, we make the substitution 1 − x = t. It yields that
Z
Z
ln(t)
ln(1 − x)
dx =
dt.
−
r
x(1 − x)
(1 − t)tr
Using this substitution to find the interval of the integration, we get
that
Z 1
( 3 ) Z 21
∞
X
2
ln(1 − x)
ln(t)
Hn 2
=
−
3 dt =
3 dt =
n
2
n
2
2
x(1
−
x)
(1
−
t)t
0
1
n=1
"
#t= 12
3
3 2
3
3 2
3
ln(t) + 1 + 2 ln(t)Φ t, 1, − 2 − 2 Φ t, 2, − 2
= 2
=
2
3
t 2 32
t=1
√
8
3
9
1
3
9
1
3
, 1, −
− Φ
, 2, −
+
2 − ln(2) + 1 − ln(2)Φ
9
2
4
2
2
4
2
2
1
+4 + π 2 ≈ 0.99222242.
2
An other exotic sum is computed with the same method.
√
Z 2
Z 1
∞
( 2)
X
3
3
ln(1 − x)
ln(t)
Hn
√ =
√ =
=
−
n
2
2
3
n
x(1
−
x)
(1
−
t)t
0
1
n=1
"√
=
√
√ #t= 23
2 ln(t) + 1 + 2 ln(t)Φ(t, 1, − 2) − 2Φ(t, 2, − 2)
√
=
2t 2
t=1
ANALYTIC EXTENSION OF HYPERHARMONIC NUMBERS
√
=
2 ln
2
3
9
√
√
Φ( 23 , 1, − 2) − 2Φ( 32 , 2, − 2)
−
√2
2 23
√
1 + 2Φ(1, 2, − 2)
≈ 0.48960626.
−
2
+ 1 + 2 ln
2
3
References
[1]
[2]
[3]
[4]
[5]
[6]
[7]
A. T. Benjamin, D. Gaebler and R. Gaebler A combinatorial approach
to hyperharmonic numbers, Integers 3, #A15 p. 1-9, 2003.
D. Borwein and J. M. Borwein, On an intriguing integral and some
series related to ζ(4), Proc. Amer. Math. Soc. 123, 1191-1198, 1995.
J. H. Conway and R. K. Guy, The book of numbers, Springer-Verlag,
New York, 1996.
P. J. de Doelder, On some series containing Ψ(x) − Ψ(y) and (Ψ(x) −
Ψ(y))2 for certain values of x and y, J. Comp. Appl. Math. 37, 125141, 1991.
A. Erdélyi, W. Magnus, F. Oberhettinger and F. G. Tricomi, Higher
Transcendental Functions, Vol. 1., §1.11 p. 27-31, Krieger, New York,
1981.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics,
Addison Wesley, 1993.
M. Abramovitz and I. A. Stegun (eds.), Handbook of Mathematical
Functions, Dover Publications, New York, 1972.
Department of Algebra and Number Theory, Institute of Mathematics, University of Debrecen, Hungary
E-mail address: imezo@math.klte.hu