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## Summary

In mathematics, the n-th harmonic number is the sum of the reciprocals of the first n natural numbers: The harmonic number $H_{n}$ with $n=\lfloor x\rfloor$ (red line) with its asymptotic limit $\gamma +\ln(x)$ (blue line) where $\gamma$ is the Euler–Mascheroni constant.
$H_{n}=1+{\frac {1}{2}}+{\frac {1}{3}}+\cdots +{\frac {1}{n}}=\sum _{k=1}^{n}{\frac {1}{k}}.$ Starting from n = 1, the sequence of harmonic numbers begins:

$1,{\frac {3}{2}},{\frac {11}{6}},{\frac {25}{12}},{\frac {137}{60}},\dots$ Harmonic numbers are related to the harmonic mean in that the n-th harmonic number is also n times the reciprocal of the harmonic mean of the first n positive integers.

Harmonic numbers have been studied since 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 harmonic numbers roughly approximate the natural logarithm function: 143  and thus the associated harmonic series grows without limit, albeit slowly. In 1737, Leonhard Euler used the divergence of the harmonic 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 proportional to the n-th harmonic number. This leads to a variety of surprising conclusions regarding the long tail and the theory of network value.

Bertrand's postulate implies that, except for the case n = 1, the harmonic numbers are never integers.

## Identities involving harmonic numbers

By definition, the harmonic numbers satisfy the recurrence relation

$H_{n+1}=H_{n}+{\frac {1}{n+1}}.$

The harmonic numbers are connected to the Stirling numbers of the first kind by the relation

$H_{n}={\frac {1}{n!}}\left[{n+1 \atop 2}\right].$

The functions

$f_{n}(x)={\frac {x^{n}}{n!}}(\log x-H_{n})$

satisfy the property
$f_{n}'(x)=f_{n-1}(x).$

In particular
$f_{1}(x)=x(\log x-1)$

is an integral of the logarithmic function.

The harmonic numbers satisfy the series identities

$\sum _{k=1}^{n}H_{k}=(n+1)H_{n}-n$

and
$\sum _{k=1}^{n}H_{k}^{2}=(n+1)H_{n}^{2}-(2n+1)H_{n}+2n.$

These two results are closely analogous to the corresponding integral results
$\int _{0}^{x}\log y\ dy=x\log x-x$

and
$\int _{0}^{x}(\log y)^{2}\ dy=x(\log x)^{2}-2x\log x+2x.$

### Identities involving π

There are several infinite summations involving harmonic numbers and powers of π:[better source needed]

$\sum _{n=1}^{\infty }{\frac {H_{n}}{n\cdot 2^{n}}}={\frac {1}{12}}\pi ^{2}$

$\sum _{n=1}^{\infty }{\frac {H_{n}^{2}}{(n+1)^{2}}}={\frac {11}{360}}\pi ^{4}$

$\sum _{n=1}^{\infty }{\frac {H_{n}^{2}}{n^{2}}}={\frac {17}{360}}\pi ^{4}$

$\sum _{n=1}^{\infty }{\frac {H_{n}}{n^{3}}}={\frac {1}{72}}\pi ^{4}$

## Calculation

An integral representation given by Euler is

$H_{n}=\int _{0}^{1}{\frac {1-x^{n}}{1-x}}\,dx.$

The equality above is straightforward by the simple algebraic identity

${\frac {1-x^{n}}{1-x}}=1+x+\cdots +x^{n-1}.$

Using the substitution x = 1 − u, another expression for Hn is

{\begin{aligned}H_{n}&=\int _{0}^{1}{\frac {1-x^{n}}{1-x}}\,dx=\int _{0}^{1}{\frac {1-(1-u)^{n}}{u}}\,du\\[6pt]&=\int _{0}^{1}\left[-\sum _{k=1}^{n}(-1)^{k}{\binom {n}{k}}u^{k-1}\right]\,du=-\sum _{k=1}^{n}(-1)^{k}{\binom {n}{k}}\int _{0}^{1}u^{k-1}\,du\\[6pt]&=-\sum _{k=1}^{n}(-1)^{k}{\frac {1}{k}}{\binom {n}{k}}.\end{aligned}}

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: $\int _{1}^{n+1}{\frac {dx}{x}}=\ln(n+1).$

The nth harmonic number is about as large as the natural logarithm of n. The reason is that the sum is approximated by the integral

$\int _{1}^{n}{\frac {1}{x}}\,dx,$

whose value is ln n.

The values of the sequence Hn − ln n decrease monotonically towards the limit

$\lim _{n\to \infty }\left(H_{n}-\ln n\right)=\gamma ,$

where γ ≈ 0.5772156649 is the Euler–Mascheroni constant. The corresponding asymptotic expansion is
{\begin{aligned}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 ,\end{aligned}}

where Bk are the Bernoulli numbers.

## Generating functions

A generating function for the harmonic numbers is

$\sum _{n=1}^{\infty }z^{n}H_{n}={\frac {-\ln(1-z)}{1-z}},$

where ln(z) is the natural logarithm. An exponential generating function is
$\sum _{n=1}^{\infty }{\frac {z^{n}}{n!}}H_{n}=-e^{z}\sum _{k=1}^{\infty }{\frac {1}{k}}{\frac {(-z)^{k}}{k!}}=e^{z}\operatorname {Ein} (z)$

where Ein(z) is the entire exponential integral. The exponential integral may also be expressed as
$\operatorname {Ein} (z)=\mathrm {E} _{1}(z)+\gamma +\ln z=\Gamma (0,z)+\gamma +\ln z$

where Γ(0, z) is the incomplete gamma function.

## Arithmetic properties

The harmonic numbers have several interesting arithmetic properties. It is well-known that ${\textstyle H_{n}}$  is an integer if and only if ${\textstyle n=1}$ , a result often attributed to Taeisinger. Indeed, using 2-adic valuation, it is not difficult to prove that for ${\textstyle n\geq 2}$  the numerator of ${\textstyle H_{n}}$  is an odd number while the denominator of ${\textstyle H_{n}}$  is an even number. More precisely,

$H_{n}={\frac {1}{2^{\lfloor \log _{2}(n)\rfloor }}}{\frac {a_{n}}{b_{n}}}$

with some odd integers ${\textstyle a_{n}}$  and ${\textstyle b_{n}}$ .

As a consequence of Wolstenholme's theorem, for any prime number $p\geq 5$  the numerator of $H_{p-1}$ is divisible by ${\textstyle p^{2}}$ . Furthermore, Eisenstein proved that for all odd prime number ${\textstyle p}$  it holds

$H_{(p-1)/2}\equiv -2q_{p}(2){\pmod {p}}$

where ${\textstyle q_{p}(2)=(2^{p-1}-1)/p}$  is a Fermat quotient, with the consequence that ${\textstyle p}$  divides the numerator of $H_{(p-1)/2}$  if and only if ${\textstyle p}$  is a Wieferich prime.

In 1991, Eswarathasan and Levine defined $J_{p}$  as the set of all positive integers $n$  such that the numerator of $H_{n}$  is divisible by a prime number $p.$  They proved that

$\{p-1,p^{2}-p,p^{2}-1\}\subseteq J_{p}$

for all prime numbers $p\geq 5,$  and they defined harmonic primes to be the primes ${\textstyle p}$  such that $J_{p}$  has exactly 3 elements.

Eswarathasan and Levine also conjectured that $J_{p}$  is a finite set for all primes $p,$  and that there are infinitely many harmonic primes. Boyd verified that $J_{p}$  is finite for all prime numbers up to $p=547$  except 83, 127, and 397; and he gave a heuristic suggesting that the density of the harmonic primes in the set of all primes should be $1/e$ . Sanna showed that $J_{p}$  has zero asymptotic density, while Bing-Ling Wu and Yong-Gao Chen proved that the number of elements of $J_{p}$  not exceeding $x$  is at most $3x^{{\frac {2}{3}}+{\frac {1}{25\log p}}}$ , for all $x\geq 1$ .

## Applications

The harmonic numbers appear in several calculation formulas, such as the digamma function

$\psi (n)=H_{n-1}-\gamma .$

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:
$\gamma =\lim _{n\rightarrow \infty }{\left(H_{n}-\ln(n)\right)},$

although
$\gamma =\lim _{n\to \infty }{\left(H_{n}-\ln \left(n+{\frac {1}{2}}\right)\right)}$

converges more quickly.

In 2002, Jeffrey Lagarias proved that the Riemann hypothesis is equivalent to the statement that

$\sigma (n)\leq H_{n}+(\log H_{n})e^{H_{n}},$

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

$\lambda \varphi (x)=\int _{-1}^{1}{\frac {\varphi (x)-\varphi (y)}{|x-y|}}\,dy$

are given by $\lambda =2H_{n}$ , where by convention $H_{0}=0$ , and the corresponding eigenfunctions are given by the Legendre polynomials $\varphi (x)=P_{n}(x)$ .

## Generalizations

### Generalized harmonic numbers

The nth generalized harmonic number of order m is given by

$H_{n,m}=\sum _{k=1}^{n}{\frac {1}{k^{m}}}.$

(In some sources, this may also be denoted by ${\textstyle H_{n}^{(m)}}$  or ${\textstyle H_{m}(n).}$ )

The special case m = 0 gives $H_{n,0}=n.$  The special case m = 1 reduces to the usual harmonic number:

$H_{n,1}=H_{n}=\sum _{k=1}^{n}{\frac {1}{k}}.$

The limit of ${\textstyle H_{n,m}}$  as n → ∞ is finite if m > 1, with the generalized harmonic number bounded by and converging to the Riemann zeta function

$\lim _{n\rightarrow \infty }H_{n,m}=\zeta (m).$

The 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) is, for n=1, 2, ... :

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)

The related sum $\sum _{k=1}^{n}k^{m}$  occurs in the study of Bernoulli numbers; the harmonic numbers also appear in the study of Stirling numbers.

Some integrals of generalized harmonic numbers are

$\int _{0}^{a}H_{x,2}\,dx=a{\frac {\pi ^{2}}{6}}-H_{a}$

and
$\int _{0}^{a}H_{x,3}\,dx=aA-{\frac {1}{2}}H_{a,2},$

where A is Apéry's constant ζ(3), and
$\sum _{k=1}^{n}H_{k,m}=(n+1)H_{n,m}-H_{n,m-1}{\text{ for }}m\geq 0.$

Every generalized harmonic number of order m can be written as a function of harmonic numbers of order $m-1$  using

$H_{n,m}=\sum _{k=1}^{n-1}{\frac {H_{k,m-1}}{k(k+1)}}+{\frac {H_{n,m-1}}{n}}$

for example: $H_{4,3}={\frac {H_{1,2}}{1\cdot 2}}+{\frac {H_{2,2}}{2\cdot 3}}+{\frac {H_{3,2}}{3\cdot 4}}+{\frac {H_{4,2}}{4}}$

A generating function for the generalized harmonic numbers is

$\sum _{n=1}^{\infty }z^{n}H_{n,m}={\frac {\operatorname {Li} _{m}(z)}{1-z}},$

where $\operatorname {Li} _{m}(z)$  is the polylogarithm, and |z| < 1. The generating function given above for m = 1 is a special case of this formula.

A fractional argument for generalized harmonic numbers can be introduced as follows:

For every $p,q>0$  integer, and $m>1$  integer or not, we have from polygamma functions:

$H_{q/p,m}=\zeta (m)-p^{m}\sum _{k=1}^{\infty }{\frac {1}{(q+pk)^{m}}}$

where $\zeta (m)$  is the Riemann zeta function. The relevant recurrence relation is
$H_{a,m}=H_{a-1,m}+{\frac {1}{a^{m}}}.$

Some special values are
$H_{{\frac {1}{4}},2}=16-8G-{\tfrac {5}{6}}\pi ^{2}$

where G is Catalan's constant,
$H_{{\frac {1}{2}},2}=4-{\tfrac {\pi ^{2}}{3}}$

$H_{{\frac {3}{4}},2}=8G+{\tfrac {16}{9}}-{\tfrac {5}{6}}\pi ^{2}$

$H_{{\frac {1}{4}},3}=64-27\zeta (3)-\pi ^{3}$

$H_{{\frac {1}{2}},3}=8-6\zeta (3)$

$H_{{\frac {3}{4}},3}={({\tfrac {4}{3}})}^{3}-27\zeta (3)+\pi ^{3}$

In the special case that $p=1$ , we get

$H_{n,m}=\zeta (m,1)-\zeta (m,n+1),$

where $\zeta (m,n)$  is the Hurwitz zeta function. This relationship is used to calculate harmonic numbers numerically.

### Multiplication formulas

The multiplication theorem applies to harmonic numbers. Using polygamma functions, we obtain

$H_{2x}={\frac {1}{2}}\left(H_{x}+H_{x-{\frac {1}{2}}}\right)+\ln 2,$

$H_{3x}={\frac {1}{3}}\left(H_{x}+H_{x-{\frac {1}{3}}}+H_{x-{\frac {2}{3}}}\right)+\ln 3,$

or, more generally,
$H_{nx}={\frac {1}{n}}\left(H_{x}+H_{x-{\frac {1}{n}}}+H_{x-{\frac {2}{n}}}+\cdots +H_{x-{\frac {n-1}{n}}}\right)+\ln n.$

For generalized harmonic numbers, we have

$H_{2x,2}={\frac {1}{2}}\left(\zeta (2)+{\frac {1}{2}}\left(H_{x,2}+H_{x-{\frac {1}{2}},2}\right)\right)$

$H_{3x,2}={\frac {1}{9}}\left(6\zeta (2)+H_{x,2}+H_{x-{\frac {1}{3}},2}+H_{x-{\frac {2}{3}},2}\right),$

where $\zeta (n)$  is the Riemann zeta function.

### Hyperharmonic numbers

The next generalization was discussed by J. H. Conway and R. K. Guy in their 1995 book The Book of Numbers.: 258  Let

$H_{n}^{(0)}={\frac {1}{n}}.$

Then the nth hyperharmonic number of order r (r>0) is defined recursively as
$H_{n}^{(r)}=\sum _{k=1}^{n}H_{k}^{(r-1)}.$

In particular, $H_{n}^{(1)}$  is the ordinary harmonic number $H_{n}$ .

## Harmonic numbers for real and complex values

The formulae given above,

$H_{x}=\int _{0}^{1}{\frac {1-t^{x}}{1-t}}\,dt=-\sum _{k=1}^{\infty }{x \choose k}{\frac {(-1)^{k}}{k}}$

are an integral and a series representation for a function that interpolates the harmonic numbers and, via analytic continuation, extends the definition to the complex plane other than the negative integers x. The interpolating function is in fact closely related to the digamma function
$H_{x}=\psi (x+1)+\gamma ,$

where ψ(x) is the digamma function, and γ is the Euler–Mascheroni constant. The integration process may be repeated to obtain
$H_{x,2}=-\sum _{k=1}^{\infty }{\frac {(-1)^{k}}{k}}{x \choose k}H_{k}.$

The Taylor series for the harmonic numbers is

$H_{x}=\sum _{k=2}^{\infty }(-1)^{k}\zeta (k)\;x^{k-1}\quad {\text{ for }}|x|<1$

which comes from the Taylor series for the digamma function ($\zeta$  is the Riemann zeta function).

### Approximation using the Taylor series expansion

The harmonic number can be approximated using the first few terms of the Taylor series expansion:

$H_{n}=\gamma +\ln n+{\frac {1}{2n}}+O\left({\frac {1}{n^{2}}}\right)\simeq \gamma +\ln n+{\frac {1}{2n}}$

where $\gamma =0.57721...$  is the Euler–Mascheroni constant.

### Alternative, asymptotic formulation

When seeking to approximate Hx for a complex number x, it is effective to first compute Hm for some large integer m. Use that as an approximation for the value of Hm+x. Then use the recursion relation Hn = Hn−1 + 1/n backwards m times, to unwind it to an approximation for Hx. Furthermore, this approximation is exact in the limit as m goes to infinity.

Specifically, for a fixed integer n, it is the case that

$\lim _{m\rightarrow \infty }\left[H_{m+n}-H_{m}\right]=0.$

If n is not an integer then it is not possible to say whether this equation is true because we have not yet (in this section) defined harmonic numbers for non-integers. However, we do get a unique extension of the harmonic numbers to the non-integers by insisting that this equation continue to hold when the arbitrary integer n is replaced by an arbitrary complex number x.

$\lim _{m\rightarrow \infty }\left[H_{m+x}-H_{m}\right]=0,$

Swapping the order of the two sides of this equation and then subtracting them from Hx gives
{\begin{aligned}H_{x}&=\lim _{m\rightarrow \infty }\left[H_{m}-(H_{m+x}-H_{x})\right]\\[6pt]&=\lim _{m\rightarrow \infty }\left[\left(\sum _{k=1}^{m}{\frac {1}{k}}\right)-\left(\sum _{k=1}^{m}{\frac {1}{x+k}}\right)\right]\\[6pt]&=\lim _{m\rightarrow \infty }\sum _{k=1}^{m}\left({\frac {1}{k}}-{\frac {1}{x+k}}\right)=x\sum _{k=1}^{\infty }{\frac {1}{k(x+k)}}\,.\end{aligned}}

This infinite series converges for all complex numbers x except the negative integers, which fail because trying to use the recursion relation Hn = Hn−1 + 1/n backwards through the value n = 0 involves a division by zero. By this construction, the function that defines the harmonic number for complex values is the unique function that simultaneously satisfies (1) H0 = 0, (2) Hx = Hx−1 + 1/x for all complex numbers x except the non-positive integers, and (3) limm→+∞ (Hm+xHm) = 0 for all complex values x.

Note that this last formula can be used to show that

$\int _{0}^{1}H_{x}\,dx=\gamma ,$

where γ is the Euler–Mascheroni constant or, more generally, for every n we have:
$\int _{0}^{n}H_{x}\,dx=n\gamma +\ln(n!).$

### Special values for fractional arguments

There are the following special analytic values for fractional arguments between 0 and 1, given by the integral

$H_{\alpha }=\int _{0}^{1}{\frac {1-x^{\alpha }}{1-x}}\,dx\,.$

More values may be generated from the recurrence relation

$H_{\alpha }=H_{\alpha -1}+{\frac {1}{\alpha }}\,,$

or from the reflection relation
$H_{1-\alpha }-H_{\alpha }=\pi \cot {(\pi \alpha )}-{\frac {1}{\alpha }}+{\frac {1}{1-\alpha }}\,.$

For example:

$H_{\frac {1}{2}}=2-2\ln {2}$

$H_{\frac {1}{3}}=3-{\tfrac {\pi }{2{\sqrt {3}}}}-{\tfrac {3}{2}}\ln {3}$

$H_{\frac {2}{3}}={\tfrac {3}{2}}(1-\ln {3})+{\sqrt {3}}{\tfrac {\pi }{6}}$

$H_{\frac {1}{4}}=4-{\tfrac {\pi }{2}}-3\ln {2}$

$H_{\frac {3}{4}}={\tfrac {4}{3}}-3\ln {2}+{\tfrac {\pi }{2}}$

$H_{\frac {1}{6}}=6-{\tfrac {\pi }{2}}{\sqrt {3}}-2\ln {2}-{\tfrac {3}{2}}\ln {3}$

$H_{\frac {1}{8}}=8-{\tfrac {\pi }{2}}-4\ln {2}-{\tfrac {1}{\sqrt {2}}}\left\{\pi +\ln \left(2+{\sqrt {2}}\right)-\ln \left(2-{\sqrt {2}}\right)\right\}$

$H_{\frac {1}{12}}=12-3\left(\ln {2}+{\tfrac {\ln {3}}{2}}\right)-\pi \left(1+{\tfrac {\sqrt {3}}{2}}\right)+2{\sqrt {3}}\ln \left({\sqrt {2-{\sqrt {3}}}}\right)$

For positive integers p and q with p < q, we have:

$H_{\frac {p}{q}}={\frac {q}{p}}+2\sum _{k=1}^{\lfloor {\frac {q-1}{2}}\rfloor }\cos \left({\frac {2\pi pk}{q}}\right)\ln \left({\sin \left({\frac {\pi k}{q}}\right)}\right)-{\frac {\pi }{2}}\cot \left({\frac {\pi p}{q}}\right)-\ln \left(2q\right)$

### Relation to the Riemann zeta function

Some derivatives of fractional harmonic numbers are given by

{\begin{aligned}{\frac {d^{n}H_{x}}{dx^{n}}}&=(-1)^{n+1}n!\left[\zeta (n+1)-H_{x,n+1}\right]\\[6pt]{\frac {d^{n}H_{x,2}}{dx^{n}}}&=(-1)^{n+1}(n+1)!\left[\zeta (n+2)-H_{x,n+2}\right]\\[6pt]{\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].\end{aligned}}

And using Maclaurin series, we have for x < 1 that

{\begin{aligned}H_{x}&=\sum _{n=1}^{\infty }(-1)^{n+1}x^{n}\zeta (n+1)\\[5pt]H_{x,2}&=\sum _{n=1}^{\infty }(-1)^{n+1}(n+1)x^{n}\zeta (n+2)\\[5pt]H_{x,3}&={\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n+1}(n+1)(n+2)x^{n}\zeta (n+3).\end{aligned}}

For fractional arguments between 0 and 1 and for a > 1,

{\begin{aligned}H_{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)\\[6pt]H_{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)\\[6pt]H_{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).\end{aligned}}