The harmonic number
H
n
,
1
{\displaystyle H_{n,1}}
with
n
=
⌊
x
⌋
{\displaystyle n=\lfloor {x}\rfloor }
(red line) with its asymptotic limit
γ
+
ln
(
x
)
{\displaystyle \gamma +\ln(x)}
(blue line).
In mathematics , the n -th harmonic number is the sum of the reciprocals of the first n natural numbers :
H
n
=
1
+
1
2
+
1
3
+
⋯
+
1
n
=
∑
k
=
1
n
1
k
.
{\displaystyle H_{n}=1+{\frac {1}{2}}+{\frac {1}{3}}+\cdots +{\frac {1}{n}}=\sum _{k=1}^{n}{\frac {1}{k}}.}
This also equals n times the inverse of the harmonic mean of these natural numbers.
The numbers n such that the numerator of the fully reduced expression for
H
n
{\displaystyle H_{n}}
is prime are
2, 3, 5, 8, 9, 21, 26, 41, 56, 62, 69, 79, 89, 91, 122, 127, 143, 167, 201, 230, 247, 252, 290, 349, 376, 459, 489, 492, 516, 662, 687, 714, 771, 932, 944, 1061, 1281, 1352, 1489, 1730, 1969, ... (sequence A056903 in the OEIS )
Harmonic numbers were studied in antiquity and are important in various branches of number theory . They are sometimes loosely termed harmonic series , are closely related to the Riemann zeta function , and appear in the expressions of various special functions .
The associated harmonic series grows without limit, albeit very slowly, roughly approaching the natural logarithm function .[ 1] : 143 In 1737, Leonhard Euler used the divergence of this series to provide a new proof of the infinity of prime numbers . His work was extended into the complex plane by Bernhard Riemann in 1859, leading directly to the celebrated Riemann hypothesis about the distribution of prime numbers .
When the value of a large quantity of items has a Zipf's law distribution, the total value of the n most-valuable items is the n -th harmonic number. This leads to a variety of surprising conclusions in the Long Tail and the theory of network value .
Bertrand's postulate entails that, except for the case n=1 , the harmonic numbers are never integers.[ 2]
Identities involving harmonic numbers
By definition, the harmonic numbers satisfy the recurrence relation
H
n
=
H
n
−
1
+
1
n
.
{\displaystyle H_{n}=H_{n-1}+{\frac {1}{n}}.}
They also satisfy the series identity
∑
k
=
1
n
H
k
=
(
n
+
1
)
H
n
+
1
−
(
n
+
1
)
.
{\displaystyle \sum _{k=1}^{n}H_{k}=(n+1)H_{n+1}-(n+1).}
The harmonic numbers are connected to the Stirling numbers of the first kind :
H
n
=
1
n
!
[
n
+
1
2
]
.
{\displaystyle H_{n}={\frac {1}{n!}}\left[{n+1 \atop 2}\right].}
The functions
f
n
(
x
)
=
x
n
n
!
(
log
x
−
H
n
)
{\displaystyle f_{n}(x)={\frac {x^{n}}{n!}}(\log x-H_{n})}
satisfy the property
f
n
′
(
x
)
=
f
n
−
1
(
x
)
.
{\displaystyle f_{n}'(x)=f_{n-1}(x).}
In particular
f
1
(
x
)
=
x
(
log
x
−
1
)
{\displaystyle f_{1}(x)=x(\log x-1)}
is a primitive of the logarithmic function.
Identities involving π
There are several infinite summations involving harmonic numbers and powers of π :[ 3]
∑
n
=
1
∞
H
n
n
⋅
2
n
=
1
12
π
2
;
{\displaystyle \sum _{n=1}^{\infty }{\frac {H_{n}}{n\cdot 2^{n}}}={\frac {1}{12}}\pi ^{2};}
∑
n
=
1
∞
H
n
2
(
n
+
1
)
2
=
11
360
π
4
;
{\displaystyle \sum _{n=1}^{\infty }{\frac {H_{n}^{2}}{(n+1)^{2}}}={\frac {11}{360}}\pi ^{4};}
∑
n
=
1
∞
H
n
2
n
2
=
17
360
π
4
;
{\displaystyle \sum _{n=1}^{\infty }{\frac {H_{n}^{2}}{n^{2}}}={\frac {17}{360}}\pi ^{4};}
∑
n
=
1
∞
H
n
n
3
=
1
72
π
4
;
{\displaystyle \sum _{n=1}^{\infty }{\frac {H_{n}}{n^{3}}}={\frac {1}{72}}\pi ^{4};}
Calculation
An integral representation given by Euler [ 4] is
H
n
=
∫
0
1
1
−
x
n
1
−
x
d
x
.
{\displaystyle H_{n}=\int _{0}^{1}{\frac {1-x^{n}}{1-x}}\,dx.}
The equality above is obvious by the simple algebraic identity
1
−
x
n
1
−
x
=
1
+
x
+
⋯
+
x
n
−
1
.
{\displaystyle {\frac {1-x^{n}}{1-x}}=1+x+\cdots +x^{n-1}.}
Using the simple integral transform x = 1−u , an elegant combinatorial expression for Hn is
H
n
=
∫
0
1
1
−
x
n
1
−
x
d
x
=
−
∫
1
0
1
−
(
1
−
u
)
n
u
d
u
=
∫
0
1
1
−
(
1
−
u
)
n
u
d
u
=
∫
0
1
[
∑
k
=
1
n
(
−
1
)
k
−
1
(
n
k
)
u
k
−
1
]
d
u
=
∑
k
=
1
n
(
−
1
)
k
−
1
(
n
k
)
∫
0
1
u
k
−
1
d
u
=
∑
k
=
1
n
(
−
1
)
k
−
1
1
k
(
n
k
)
.
{\displaystyle {\begin{aligned}H_{n}&=\int _{0}^{1}{\frac {1-x^{n}}{1-x}}\,dx\\&=-\int _{1}^{0}{\frac {1-(1-u)^{n}}{u}}\,du\\&=\int _{0}^{1}{\frac {1-(1-u)^{n}}{u}}\,du\\&=\int _{0}^{1}\left[\sum _{k=1}^{n}(-1)^{k-1}{\binom {n}{k}}u^{k-1}\right]\,du\\&=\sum _{k=1}^{n}(-1)^{k-1}{\binom {n}{k}}\int _{0}^{1}u^{k-1}\,du\\&=\sum _{k=1}^{n}(-1)^{k-1}{\frac {1}{k}}{\binom {n}{k}}.\end{aligned}}}
The same representation can be produced by using the third Retkes identity by setting
x
1
=
1
,
…
,
x
n
=
n
{\displaystyle x_{1}=1,\ldots ,x_{n}=n}
and using the fact that
Π
k
(
1
,
…
,
n
)
=
(
−
1
)
n
−
k
(
k
−
1
)
!
(
n
−
k
)
!
{\displaystyle \Pi _{k}(1,\ldots ,n)=(-1)^{n-k}(k-1)!(n-k)!}
H
n
=
H
n
,
1
=
∑
k
=
1
n
1
k
=
(
−
1
)
n
−
1
n
!
∑
k
=
1
n
1
k
2
Π
k
(
1
,
…
,
n
)
=
∑
k
=
1
n
(
−
1
)
k
−
1
1
k
(
n
k
)
.
{\displaystyle H_{n}=H_{n,1}=\sum _{k=1}^{n}{\frac {1}{k}}=(-1)^{n-1}n!\sum _{k=1}^{n}{\frac {1}{k^{2}\Pi _{k}(1,\ldots ,n)}}=\sum _{k=1}^{n}(-1)^{k-1}{\frac {1}{k}}{\binom {n}{k}}.}
The Taylor series for the harmonic numbers is
H
x
=
∑
k
=
2
∞
(
−
1
)
k
ζ
(
k
)
x
k
−
1
for
|
x
|
<
1
{\displaystyle H_{x}=\sum _{k=2}^{\infty }(-1)^{k}\zeta (k)\;x^{k-1}\quad {\text{ for }}|x|<1}
which comes from the Taylor series for the Digamma function.
Graph demonstrating a connection between harmonic numbers and the natural logarithm . The harmonic number Hn can be interpreted as a Riemann sum of the integral:
∫
1
n
+
1
1
x
d
x
=
ln
(
n
+
1
)
{\displaystyle \int _{1}^{n+1}{\frac {1}{x}}\mathrm {d} x=\ln(n+1)}
The n th harmonic number is about as large as the natural logarithm of n . The reason is that the sum is approximated by the integral
∫
1
n
1
x
d
x
{\displaystyle \int _{1}^{n}{1 \over x}\,dx}
whose value is ln(n ).
The values of the sequence H n - ln(n ) decrease monotonically towards the limit
lim
n
→
∞
(
H
n
−
ln
n
)
=
γ
,
{\displaystyle \lim _{n\to \infty }\left(H_{n}-\ln n\right)=\gamma ,}
where γ ≈ 0.5772156649 is the Euler–Mascheroni constant . The corresponding asymptotic expansion as n → ∞ is
H
n
∼
ln
n
+
γ
+
1
2
n
−
∑
k
=
1
∞
B
2
k
2
k
n
2
k
=
ln
n
+
γ
+
1
2
n
−
1
12
n
2
+
1
120
n
4
−
⋯
,
{\displaystyle H_{n}\sim \ln {n}+\gamma +{\frac {1}{2n}}-\sum _{k=1}^{\infty }{\frac {B_{2k}}{2kn^{2k}}}=\ln {n}+\gamma +{\frac {1}{2n}}-{\frac {1}{12n^{2}}}+{\frac {1}{120n^{4}}}-\cdots ,}
where
B
k
{\displaystyle B_{k}}
are the Bernoulli numbers .
Special values for fractional arguments
There are the following special analytic values for fractional arguments between 0 and 1, given by the integral
H
α
=
∫
0
1
1
−
x
α
1
−
x
d
x
.
{\displaystyle H_{\alpha }=\int _{0}^{1}{\frac {1-x^{\alpha }}{1-x}}\,dx\,.}
More values may be generated from the recurrence relation
H
α
=
H
α
−
1
+
1
α
,
{\displaystyle H_{\alpha }=H_{\alpha -1}+{\frac {1}{\alpha }}\,,}
or from the reflection relation
H
1
−
α
−
H
α
=
π
cot
(
π
α
)
−
1
α
+
1
1
−
α
.
{\displaystyle H_{1-\alpha }-H_{\alpha }=\pi \cot {(\pi \alpha )}-{\frac {1}{\alpha }}+{\frac {1}{1-\alpha }}\,.}
For example:
H
3
4
=
4
3
−
3
ln
2
+
π
2
{\displaystyle H_{\frac {3}{4}}={\tfrac {4}{3}}-3\ln {2}+{\tfrac {\pi }{2}}}
H
2
3
=
3
2
(
1
−
ln
3
)
+
3
π
6
{\displaystyle H_{\frac {2}{3}}={\tfrac {3}{2}}(1-\ln {3})+{\sqrt {3}}{\tfrac {\pi }{6}}}
H
1
2
=
2
−
2
ln
2
{\displaystyle H_{\frac {1}{2}}=2-2\ln {2}}
H
1
3
=
3
−
π
2
3
−
3
2
ln
3
{\displaystyle H_{\frac {1}{3}}=3-{\tfrac {\pi }{2{\sqrt {3}}}}-{\tfrac {3}{2}}\ln {3}}
H
1
4
=
4
−
π
2
−
3
ln
2
{\displaystyle H_{\frac {1}{4}}=4-{\tfrac {\pi }{2}}-3\ln {2}}
H
1
6
=
6
−
π
2
3
−
2
ln
2
−
3
2
ln
3
{\displaystyle H_{\frac {1}{6}}=6-{\tfrac {\pi }{2}}{\sqrt {3}}-2\ln {2}-{\tfrac {3}{2}}\ln {3}}
H
1
8
=
8
−
π
2
−
4
ln
2
−
1
2
{
π
+
ln
(
2
+
2
)
−
ln
(
2
−
2
)
}
{\displaystyle H_{\frac {1}{8}}=8-{\tfrac {\pi }{2}}-4\ln {2}-{\tfrac {1}{\sqrt {2}}}\left\{\pi +\ln \left(2+{\sqrt {2}}\right)-\ln \left(2-{\sqrt {2}}\right)\right\}}
H
1
12
=
12
−
3
(
ln
2
+
ln
3
2
)
−
π
(
1
+
3
2
)
+
2
3
ln
(
2
−
3
)
{\displaystyle H_{\frac {1}{12}}=12-3\left(\ln {2}+{\tfrac {\ln {3}}{2}}\right)-\pi \left(1+{\tfrac {\sqrt {3}}{2}}\right)+2{\sqrt {3}}\ln \left({\sqrt {2-{\sqrt {3}}}}\right)}
For positive integers p and q with p < q , we have:
H
p
q
=
q
p
+
2
∑
k
=
1
⌊
q
−
1
2
⌋
cos
(
2
π
p
k
q
)
ln
(
sin
(
π
k
q
)
)
−
π
2
cot
(
π
p
q
)
−
ln
(
2
q
)
{\displaystyle H_{\frac {p}{q}}={\frac {q}{p}}+2\sum _{k=1}^{\lfloor {\frac {q-1}{2}}\rfloor }\cos({\frac {2\pi pk}{q}})\ln({\sin({\frac {\pi k}{q}})})-{\frac {\pi }{2}}\cot({\frac {\pi p}{q}})-\ln(2q)}
For every x > 0, integer or not, we have:
H
x
=
x
∑
k
=
1
∞
1
k
(
x
+
k
)
.
{\displaystyle H_{x}=x\sum _{k=1}^{\infty }{\frac {1}{k(x+k)}}\,.}
Based on this, it can be shown that:
∫
0
1
H
x
d
x
=
γ
,
{\displaystyle \int _{0}^{1}H_{x}\,dx=\gamma \,,}
where γ is the Euler–Mascheroni constant or, more generally, for every n we have:
∫
0
n
H
x
d
x
=
n
γ
+
ln
(
n
!
)
.
{\displaystyle \int _{0}^{n}H_{x}\,dx=n\gamma +\ln {(n!)}\,.}
Generating functions
A generating function for the harmonic numbers is
∑
n
=
1
∞
z
n
H
n
=
−
ln
(
1
−
z
)
1
−
z
,
{\displaystyle \sum _{n=1}^{\infty }z^{n}H_{n}={\frac {-\ln(1-z)}{1-z}},}
where ln(z ) is the natural logarithm . An exponential generating function is
∑
n
=
1
∞
z
n
n
!
H
n
=
−
e
z
∑
k
=
1
∞
1
k
(
−
z
)
k
k
!
=
e
z
Ein
(
z
)
{\displaystyle \sum _{n=1}^{\infty }{\frac {z^{n}}{n!}}H_{n}=-e^{z}\sum _{k=1}^{\infty }{\frac {1}{k}}{\frac {(-z)^{k}}{k!}}=e^{z}{\mbox{Ein}}(z)}
where Ein(z ) is the entire exponential integral . Note that
Ein
(
z
)
=
E
1
(
z
)
+
γ
+
ln
z
=
Γ
(
0
,
z
)
+
γ
+
ln
z
{\displaystyle {\mbox{Ein}}(z)={\mbox{E}}_{1}(z)+\gamma +\ln z=\Gamma (0,z)+\gamma +\ln z\,}
where Γ(0, z ) is the incomplete gamma function .
Applications
The harmonic numbers appear in several calculation formulas, such as the digamma function
ψ
(
n
)
=
H
n
−
1
−
γ
.
{\displaystyle \psi (n)=H_{n-1}-\gamma .}
This relation is also frequently used to define the extension of the harmonic numbers to non-integer n . The harmonic numbers are also frequently used to define γ using the limit introduced earlier:
γ
=
lim
n
→
∞
(
H
n
−
ln
(
n
)
)
,
{\displaystyle \gamma =\lim _{n\rightarrow \infty }{\left(H_{n}-\ln(n)\right)},}
although
γ
=
lim
n
→
∞
(
H
n
−
ln
(
n
+
1
2
)
)
{\displaystyle \gamma =\lim _{n\to \infty }{\left(H_{n}-\ln \left(n+{1 \over 2}\right)\right)}}
converges more quickly.
In 2002, Jeffrey Lagarias proved[ 5] that the Riemann hypothesis is equivalent to the statement that
σ
(
n
)
≤
H
n
+
ln
(
H
n
)
e
H
n
,
{\displaystyle \sigma (n)\leq H_{n}+\ln(H_{n})e^{H_{n}},}
is true for every integer n ≥ 1 with strict inequality if n > 1; here σ(n ) denotes the sum of the divisors of n .
The eigenvalues of the nonlocal problem
λ
ϕ
(
x
)
=
∫
−
1
1
ϕ
(
x
)
−
ϕ
(
y
)
|
x
−
y
|
d
y
{\displaystyle \lambda \phi (x)=\int _{-1}^{1}{\frac {\phi (x)-\phi (y)}{|x-y|}}dy}
are given by
λ
=
2
H
n
{\displaystyle \lambda =2H_{n}}
, where by convention,
H
0
=
0.
{\displaystyle H_{0}=0.}
Generalization
Generalized harmonic numbers
The generalized harmonic number of order n of m is given by
H
n
,
m
=
∑
k
=
1
n
1
k
m
.
{\displaystyle H_{n,m}=\sum _{k=1}^{n}{\frac {1}{k^{m}}}.}
The limit as n tends to infinity exists if m > 1.
Other notations occasionally used include
H
n
,
m
=
H
n
(
m
)
=
H
m
(
n
)
.
{\displaystyle H_{n,m}=H_{n}^{(m)}=H_{m}(n).}
The special case of m = 0 gives
H
n
,
0
=
n
{\displaystyle H_{n,0}=n}
The special case of m = 1 is simply called a harmonic number and is frequently written without the superscript, as
H
n
=
∑
k
=
1
n
1
k
.
{\displaystyle H_{n}=\sum _{k=1}^{n}{\frac {1}{k}}.}
Smallest natural number k such that kn does not divide the denominator of generalized harmonic number H (k , n ) nor the denominator of alternating generalized harmonic number H' (k , n ) are
77, 20, 94556602, 42, 444, 20, 104, 42, 76, 20, 77, 110, 3504, 20, 903, 42, 1107, 20, 104, 42, 77, 20, 2948, 110, 136, 20, 76, 42, 903, 20, 77, 42, 268, 20, 7004, 110, 1752, 20, 19203, 42, 77, 20, 104, 42, 76, 20, 370, 110, 1107, 20, ... (sequence A128670 in the OEIS )
In the limit of n → ∞, the generalized harmonic number converges to the Riemann zeta function
lim
n
→
∞
H
n
,
m
=
ζ
(
m
)
.
{\displaystyle \lim _{n\rightarrow \infty }H_{n,m}=\zeta (m).}
The related sum
∑
k
=
1
n
k
m
{\displaystyle \sum _{k=1}^{n}k^{m}}
occurs in the study of Bernoulli numbers ; the harmonic numbers also appear in the study of Stirling numbers .
Some integrals of generalized harmonic are
∫
0
a
H
x
,
2
d
x
=
a
π
2
6
−
H
a
{\displaystyle \int _{0}^{a}H_{x,2}\,dx=a{\frac {\pi ^{2}}{6}}-H_{a}}
and
∫
0
a
H
x
,
3
d
x
=
a
A
−
1
2
H
a
,
2
,
{\displaystyle \int _{0}^{a}H_{x,3}\,dx=aA-{\frac {1}{2}}H_{a,2},}
where A is the Apéry's constant , i.e. ζ(3).
and
∑
k
=
1
n
H
k
,
m
=
(
n
+
1
)
H
n
,
m
−
H
n
,
m
−
1
{\displaystyle \sum _{k=1}^{n}H_{k,m}=(n+1)H_{n,m}-H_{n,m-1}}
for
m
≥
0
{\displaystyle m\geq 0}
Every generalized harmonic number of order m can be written as a function of harmonic of order m-1 using:
H
n
,
m
=
∑
k
=
1
n
−
1
H
k
,
m
−
1
k
(
k
+
1
)
+
H
n
,
m
−
1
n
{\displaystyle H_{n,m}=\sum _{k=1}^{n-1}{\frac {H_{k,m-1}}{k(k+1)}}+{\frac {H_{n,m-1}}{n}}}
for example:
H
4
,
3
=
H
1
,
2
1
⋅
2
+
H
2
,
2
2
⋅
3
+
H
3
,
2
3
⋅
4
+
H
4
,
2
4
{\displaystyle H_{4,3}={\frac {H_{1,2}}{1\cdot 2}}+{\frac {H_{2,2}}{2\cdot 3}}+{\frac {H_{3,2}}{3\cdot 4}}+{\frac {H_{4,2}}{4}}}
A generating function for the generalized harmonic numbers is
∑
n
=
1
∞
z
n
H
n
,
m
=
L
i
m
(
z
)
1
−
z
,
{\displaystyle \sum _{n=1}^{\infty }z^{n}H_{n,m}={\frac {\mathrm {Li} _{m}(z)}{1-z}},}
where
L
i
m
(
z
)
{\displaystyle \mathrm {Li} _{m}(z)}
is the polylogarithm , and |z | < 1. The generating function given above for m = 1 is a special case of this formula.
Fractional argument for generalized harmonic numbers can be introduced as follows:
For every
p
,
q
>
0
{\displaystyle p,q>0}
integer, and
m
>
1
{\displaystyle m>1}
integer or not, we have from polygamma functions:
H
q
/
p
,
m
=
ζ
(
m
)
−
p
m
∑
k
=
1
∞
1
(
q
+
p
k
)
m
{\displaystyle H_{q/p,m}=\zeta (m)-p^{m}\sum _{k=1}^{\infty }{\frac {1}{(q+pk)^{m}}}}
where
ζ
(
m
)
{\displaystyle \zeta (m)}
is the Riemann zeta function . The relevant recurrence relation is:
H
a
,
m
=
H
a
−
1
,
m
+
1
a
m
{\displaystyle H_{a,m}=H_{a-1,m}+{\frac {1}{a^{m}}}}
Some special values are:
H
1
4
,
2
=
16
−
8
G
−
5
6
π
2
{\displaystyle H_{{\frac {1}{4}},2}=16-8G-{\tfrac {5}{6}}\pi ^{2}}
where G is the Catalan's constant
H
1
2
,
2
=
4
−
π
2
3
{\displaystyle H_{{\frac {1}{2}},2}=4-{\tfrac {\pi ^{2}}{3}}}
H
3
4
,
2
=
8
G
+
16
9
−
5
6
π
2
{\displaystyle H_{{\frac {3}{4}},2}=8G+{\tfrac {16}{9}}-{\tfrac {5}{6}}\pi ^{2}}
H
1
4
,
3
=
64
−
27
ζ
(
3
)
−
π
3
{\displaystyle H_{{\frac {1}{4}},3}=64-27\zeta (3)-\pi ^{3}}
H
1
2
,
3
=
8
−
6
ζ
(
3
)
{\displaystyle H_{{\frac {1}{2}},3}=8-6\zeta (3)}
H
3
4
,
3
=
(
4
3
)
3
−
27
ζ
(
3
)
+
π
3
{\displaystyle H_{{\frac {3}{4}},3}={({\tfrac {4}{3}})}^{3}-27\zeta (3)+\pi ^{3}}
The multiplication theorem applies to harmonic numbers. Using polygamma functions, we obtain
H
2
x
=
1
2
(
H
x
+
H
x
−
1
2
)
+
ln
2
{\displaystyle H_{2x}={\frac {1}{2}}\left(H_{x}+H_{x-{\frac {1}{2}}}\right)+\ln {2}}
H
3
x
=
1
3
(
H
x
+
H
x
−
1
3
+
H
x
−
2
3
)
+
ln
3
,
{\displaystyle H_{3x}={\frac {1}{3}}\left(H_{x}+H_{x-{\frac {1}{3}}}+H_{x-{\frac {2}{3}}}\right)+\ln {3},}
or, more generally,
H
n
x
=
1
n
(
H
x
+
H
x
−
1
n
+
H
x
−
2
n
+
⋯
+
H
x
−
n
−
1
n
)
+
ln
n
.
{\displaystyle H_{nx}={\frac {1}{n}}\left(H_{x}+H_{x-{\frac {1}{n}}}+H_{x-{\frac {2}{n}}}+\cdots +H_{x-{\frac {n-1}{n}}}\right)+\ln {n}.}
For generalized harmonic numbers, we have
H
2
x
,
2
=
1
2
(
ζ
(
2
)
+
1
2
(
H
x
,
2
+
H
x
−
1
2
,
2
)
)
{\displaystyle H_{2x,2}={\frac {1}{2}}\left(\zeta (2)+{\frac {1}{2}}\left(H_{x,2}+H_{x-{\frac {1}{2}},2}\right)\right)}
H
3
x
,
2
=
1
9
(
6
ζ
(
2
)
+
H
x
,
2
+
H
x
−
1
3
,
2
+
H
x
−
2
3
,
2
)
,
{\displaystyle H_{3x,2}={\frac {1}{9}}\left(6\zeta (2)+H_{x,2}+H_{x-{\frac {1}{3}},2}+H_{x-{\frac {2}{3}},2}\right),}
where
ζ
(
n
)
{\displaystyle \zeta (n)}
is the Riemann zeta function .
Generalization to the complex plane
Euler's integral formula for the harmonic numbers follows from the integral identity
∫
a
1
1
−
x
s
1
−
x
d
x
=
−
∑
k
=
1
∞
1
k
(
s
k
)
(
a
−
1
)
k
,
{\displaystyle \int _{a}^{1}{\frac {1-x^{s}}{1-x}}\,dx=-\sum _{k=1}^{\infty }{\frac {1}{k}}{s \choose k}(a-1)^{k},}
which holds for general complex-valued s , for the suitably extended binomial coefficients . By choosing a = 0, this formula gives both an integral and a series representation for a function that interpolates the harmonic numbers and extends a definition to the complex plane. This integral relation is easily derived by manipulating the Newton series
∑
k
=
0
∞
(
s
k
)
(
−
x
)
k
=
(
1
−
x
)
s
,
{\displaystyle \sum _{k=0}^{\infty }{s \choose k}(-x)^{k}=(1-x)^{s},}
which is just the Newton's generalized binomial theorem . The interpolating function is in fact the digamma function
H
s
=
ψ
(
s
+
1
)
+
γ
=
∫
0
1
1
−
x
s
1
−
x
d
x
,
{\displaystyle H_{s}=\psi (s+1)+\gamma =\int _{0}^{1}{\frac {1-x^{s}}{1-x}}\,dx,}
where
ψ
(
x
)
{\displaystyle \psi (x)}
is the digamma, and γ is the Euler-Mascheroni constant. The integration process may be repeated to obtain
H
s
,
2
=
−
∑
k
=
1
∞
(
−
1
)
k
k
(
s
k
)
H
k
.
{\displaystyle H_{s,2}=-\sum _{k=1}^{\infty }{\frac {(-1)^{k}}{k}}{s \choose k}H_{k}.}
Relation to the Riemann zeta function
Some derivatives of fractional harmonic numbers are given by:
d
n
H
x
d
x
n
=
(
−
1
)
n
+
1
n
!
[
ζ
(
n
+
1
)
−
H
x
,
n
+
1
]
{\displaystyle {\frac {d^{n}H_{x}}{dx^{n}}}=(-1)^{n+1}n!\left[\zeta (n+1)-H_{x,n+1}\right]}
d
n
H
x
,
2
d
x
n
=
(
−
1
)
n
+
1
(
n
+
1
)
!
[
ζ
(
n
+
2
)
−
H
x
,
n
+
2
]
{\displaystyle {\frac {d^{n}H_{x,2}}{dx^{n}}}=(-1)^{n+1}(n+1)!\left[\zeta (n+2)-H_{x,n+2}\right]}
d
n
H
x
,
3
d
x
n
=
(
−
1
)
n
+
1
1
2
(
n
+
2
)
!
[
ζ
(
n
+
3
)
−
H
x
,
n
+
3
]
.
{\displaystyle {\frac {d^{n}H_{x,3}}{dx^{n}}}=(-1)^{n+1}{\frac {1}{2}}(n+2)!\left[\zeta (n+3)-H_{x,n+3}\right].}
And using Maclaurin series , we have for x < 1:
H
x
=
∑
n
=
1
∞
(
−
1
)
n
+
1
x
n
ζ
(
n
+
1
)
{\displaystyle H_{x}=\sum _{n=1}^{\infty }(-1)^{n+1}x^{n}\zeta (n+1)}
H
x
,
2
=
∑
n
=
1
∞
(
−
1
)
n
+
1
(
n
+
1
)
x
n
ζ
(
n
+
2
)
{\displaystyle H_{x,2}=\sum _{n=1}^{\infty }(-1)^{n+1}(n+1)x^{n}\zeta (n+2)}
H
x
,
3
=
1
2
∑
n
=
1
∞
(
−
1
)
n
+
1
(
n
+
1
)
(
n
+
2
)
x
n
ζ
(
n
+
3
)
.
{\displaystyle H_{x,3}={\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n+1}(n+1)(n+2)x^{n}\zeta (n+3).}
For fractional arguments between 0 and 1, and for a > 1:
H
1
a
=
1
a
(
ζ
(
2
)
−
1
a
ζ
(
3
)
+
1
a
2
ζ
(
4
)
−
1
a
3
ζ
(
5
)
+
⋯
)
{\displaystyle H_{\frac {1}{a}}={\frac {1}{a}}\left(\zeta (2)-{\frac {1}{a}}\zeta (3)+{\frac {1}{a^{2}}}\zeta (4)-{\frac {1}{a^{3}}}\zeta (5)+\cdots \right)}
H
1
a
,
2
=
1
a
(
2
ζ
(
3
)
−
3
a
ζ
(
4
)
+
4
a
2
ζ
(
5
)
−
5
a
3
ζ
(
6
)
+
⋯
)
{\displaystyle H_{{\frac {1}{a}},2}={\frac {1}{a}}\left(2\zeta (3)-{\frac {3}{a}}\zeta (4)+{\frac {4}{a^{2}}}\zeta (5)-{\frac {5}{a^{3}}}\zeta (6)+\cdots \right)}
H
1
a
,
3
=
1
2
a
(
2
⋅
3
ζ
(
4
)
−
3
⋅
4
a
ζ
(
5
)
+
4
⋅
5
a
2
ζ
(
6
)
−
5
⋅
6
a
3
ζ
(
7
)
+
⋯
)
.
{\displaystyle H_{{\frac {1}{a}},3}={\frac {1}{2a}}\left(2\cdot 3\zeta (4)-{\frac {3\cdot 4}{a}}\zeta (5)+{\frac {4\cdot 5}{a^{2}}}\zeta (6)-{\frac {5\cdot 6}{a^{3}}}\zeta (7)+\cdots \right).}
Hyperharmonic numbers
The next generalization was discussed by J. H. Conway and R. K. Guy in their 1995 book The Book of Numbers .[ 1] : 258 Let
H
n
(
0
)
=
1
n
.
{\displaystyle H_{n}^{(0)}={\frac {1}{n}}.}
Then the nth hyperharmonic number of order r (r>0 ) is defined recursively as
H
n
(
r
)
=
∑
k
=
1
n
H
k
(
r
−
1
)
.
{\displaystyle H_{n}^{(r)}=\sum _{k=1}^{n}H_{k}^{(r-1)}.}
In special,
H
n
=
H
n
(
1
)
{\displaystyle H_{n}=H_{n}^{(1)}}
.
See also
Notes
^ a b
John H., Conway; Richard K., Guy (1995). The book of numbers . Copernicus.
^ Ronald L., Graham; Donald E., Knuth; Oren, Patashnik (1994). Concrete Mathematics . Addison-Wesley.
^ Sondow, Jonathan and Weisstein, Eric W. "Harmonic Number." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/HarmonicNumber.html
^ Sandifer, C. Edward (2007), How Euler Did It , MAA Spectrum, Mathematical Association of America, p. 206, ISBN 9780883855638 .
^ Jeffrey Lagarias (2002). "An Elementary Problem Equivalent to the Riemann Hypothesis". Amer. Math. Monthly . 109 : 534–543. arXiv :math.NT/0008177 . doi :10.2307/2695443 .
References
External links
This article incorporates material from Harmonic number on PlanetMath , which is licensed under the Creative Commons Attribution/Share-Alike License .