Harmonic number

For other uses, see Harmonic number (disambiguation).

The harmonic number $H_{n,1}$$H_{{n,1}}$ with $n=\lfloor {x}\rfloor$$n=\lfloor {x}\rfloor$ (red line) with its asymptotic limit $\gamma +\ln(x)$$\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+{\frac {1}{2}}+{\frac {1}{3}}+\cdots +{\frac {1}{n}}=\sum _{k=1}^{n}{\frac {1}{k}}.$$H_{n}=1+{\frac {1}{2}}+{\frac {1}{3}}+\cdots +{\frac {1}{n}}=\sum _{k=1}^{n}{\frac {1}{k}}.$

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 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 harmonic numbers roughly approximate the natural logarithm function 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 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.

Contents

Identities involving harmonic numbersEdit

By definition, the harmonic numbers satisfy the recurrence relation

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

The harmonic numbers are connected to the Stirling numbers of the first kind:

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

The functions

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

satisfy the property

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

In particular

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

is an integral of the logarithmic function.

The harmonic numbers satisfy the series identity

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

Identities involving πEdit

There are several infinite summations involving harmonic numbers and powers of π:

$\sum _{n=1}^{\infty }{\frac {H_{n}}{n\cdot 2^{n}}}={\frac {1}{12}}\pi ^{2};$$\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+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}^{2}}{n^{2}}}={\frac {17}{360}}\pi ^{4};$

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

CalculationEdit

An integral representation given by Euler is

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

The equality above is obvious by the simple algebraic identity

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

${\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}}$${\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,\ldots ,x_{n}=n$$x_{1}=1,\ldots ,x_{n}=n$ and using the fact that $\Pi _{k}(1,\ldots ,n)=(-1)^{n-k}(k-1)!(n-k)!$$\Pi _{k}(1,\ldots ,n)=(-1)^{n-k}(k-1)!(n-k)!$

$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}}.$$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}=\sum _{k=2}^{\infty }(-1)^{k}\zeta (k)\;x^{k-1}\quad {\text{ for }}|x|<1$$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 H_n can be interpreted as a Riemann sum of the integral: $\int _{1}^{n+1}{\frac {1}{x}}\mathrm {d} x=\ln(n+1)$$\int _{1}^{n+1}{\frac {1}{x}}\mathrm {d} 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}{1 \over x}\,dx$$\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\to +\infty }\left(H_{n}-\ln n\right)=\gamma ,$$\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}+\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 ,$$H_{n}=\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}$$B_{k}$ are the Bernoulli numbers.

Special values for fractional argumentsEdit

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\,.$$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 }}\,,$$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 }}\,.$$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}{2}}=2-2\ln {2}$

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

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

Asymptotic formulationEdit

For every positive integer x, we have that

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

Adding H_x to both sides gives

${\begin{aligned}H_{x}&=\lim _{n\rightarrow +\infty }\left[H_{n}-(H_{n+x}-H_{x})\right]\\&=\lim _{n\rightarrow +\infty }\left[\left(\sum _{k=1}^{n}{\frac {1}{k}}\right)-\left(\sum _{k=1}^{n}{\frac {1}{x+k}}\right)\right]\\&=\sum _{k=1}^{+\infty }\left[{\frac {1}{k}}-{\frac {1}{x+k}}\right]=x\sum _{k=1}^{\infty }{\frac {1}{k(x+k)}}\,.\end{aligned}}$${\begin{aligned}H_{x}&=\lim _{n\rightarrow +\infty }\left[H_{n}-(H_{n+x}-H_{x})\right]\\&=\lim _{n\rightarrow +\infty }\left[\left(\sum _{k=1}^{n}{\frac {1}{k}}\right)-\left(\sum _{k=1}^{n}{\frac {1}{x+k}}\right)\right]\\&=\sum _{k=1}^{+\infty }\left[{\frac {1}{k}}-{\frac {1}{x+k}}\right]=x\sum _{k=1}^{\infty }{\frac {1}{k(x+k)}}\,.\end{aligned}}$

Despite being derived for positive integers x, this last expression for H_x is well defined for any complex number x except the negative integers. The function H_x is the unique function of x for which (1) H₀ = 0, (2) H_x = H_x−1 + 1/x for all complex values x except the non-positive integers, and (3) lim_n→+∞ (H_n+x − H_n) = 0 for all complex values x.

Based on this last formula, it can be shown that:

$\int _{0}^{1}H_{x}\,dx=\gamma \,,$$\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!)}\,.$$\int _{0}^{n}H_{x}\,dx=n\gamma +\ln {(n!)}\,.$

Generating functionsEdit

A generating function for the harmonic numbers is

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

${\mbox{Ein}}(z)={\mbox{E}}_{1}(z)+\gamma +\ln z=\Gamma (0,z)+\gamma +\ln z\,$${\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.

ApplicationsEdit

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

$\psi (n)=H_{n-1}-\gamma .$$\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)},$$\gamma =\lim _{n\rightarrow +\infty }{\left(H_{n}-\ln(n)\right)},$

although

$\gamma =\lim _{n\to +\infty }{\left(H_{n}-\ln \left(n+{1 \over 2}\right)\right)}$$\gamma =\lim _{n\to +\infty }{\left(H_{n}-\ln \left(n+{1 \over 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}+\ln(H_{n})e^{H_{n}},$$\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

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

are given by $\lambda =2H_{n}$$\lambda =2H_{n}$, where by convention, $H_{0}=0.$$H_{0}=0.$

GeneralizationEdit

Generalized harmonic numbersEdit

The generalized harmonic number of order n of m is given by

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

The special case of m = 0 gives $H_{n,0}=n$$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}=\sum _{k=1}^{n}{\frac {1}{k}}.$$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\rightarrow +\infty }H_{n,m}=\zeta (m).$$\lim _{n\rightarrow +\infty }H_{n,m}=\zeta (m).$

The related sum $\sum _{k=1}^{n}k^{m}$$\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

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

$\sum _{k=1}^{n}H_{k,m}=(n+1)H_{n,m}-H_{n,m-1}$$\sum _{k=1}^{n}H_{k,m}=(n+1)H_{n,m}-H_{n,m-1}$   for $m\geq 0$$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}=\sum _{k=1}^{n-1}{\frac {H_{k,m-1}}{k(k+1)}}+{\frac {H_{n,m-1}}{n}}$$H_{n,m}=\sum _{k=1}^{n-1}{\frac {H_{k,m-1}}{k(k+1)}}+{\frac {H_{n,m-1}}{n}}$   for example: $H_{4,3}={\frac {H_{1,2}}{1\cdot 2}}+{\frac {H_{2,2}}{2\cdot 3}}+{\frac {H_{3,2}}{3\cdot 4}}+{\frac {H_{4,2}}{4}}$$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 {\mathrm {Li} _{m}(z)}{1-z}},$$\sum _{n=1}^{\infty }z^{n}H_{n,m}={\frac {\mathrm {Li} _{m}(z)}{1-z}},$

where $\mathrm {Li} _{m}(z)$$\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$$p,q>0$ integer, and $m>1$$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}}}$$H_{q/p,m}=\zeta (m)-p^{m}\sum _{k=1}^{\infty }{\frac {1}{(q+pk)^{m}}}$

where $\zeta (m)$$\zeta (m)$ is the Riemann zeta function. The relevant recurrence relation is:

$H_{a,m}=H_{a-1,m}+{\frac {1}{a^{m}}}$$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}$$H_{{\frac {1}{4}},2}=16-8G-{\tfrac {5}{6}}\pi ^{2}$ where G is the Catalan's constant

$H_{{\frac {1}{2}},2}=4-{\tfrac {\pi ^{2}}{3}}$$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 {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}{4}},3}=64-27\zeta (3)-\pi ^{3}$

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

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

Multiplication formulasEdit

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_{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},$$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}.$$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_{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),$$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)$$\zeta (n)$ is the Riemann zeta function.

Generalization to the complex planeEdit

When | x−1 | < 1, we can write the integrand (1−x^s)/(1−x) as an infinite series. We start by writing

$x^{s}=(1+(x-1))^{s}=\sum _{k=0}^{+\infty }{s \choose k}(x-1)^{k}\,,$$x^{s}=(1+(x-1))^{s}=\sum _{k=0}^{+\infty }{s \choose k}(x-1)^{k}\,,$

which is the binomial expansion for the suitably extended binomial coefficients. Then

${\frac {1-x^{s}}{1-x}}={\frac {1-\sum _{k=0}^{+\infty }{s \choose k}(x-1)^{k}}{1-x}}=\sum _{k=1}^{+\infty }{s \choose k}(x-1)^{k-1}\,.$${\frac {1-x^{s}}{1-x}}={\frac {1-\sum _{k=0}^{+\infty }{s \choose k}(x-1)^{k}}{1-x}}=\sum _{k=1}^{+\infty }{s \choose k}(x-1)^{k-1}\,.$

The integral from some value a ∈ (0, 1) is then

$\int _{a}^{1}{\frac {1-x^{s}}{1-x}}=-\sum _{k=1}^{+\infty }{s \choose k}{\frac {(a-1)^{k}}{k}}\,.$$\int _{a}^{1}{\frac {1-x^{s}}{1-x}}=-\sum _{k=1}^{+\infty }{s \choose k}{\frac {(a-1)^{k}}{k}}\,.$

By choosing a = 0, this formula gives both 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 s. The interpolating function is in fact the digamma function

$H_{s}=\psi (s+1)+\gamma =\int _{0}^{1}{\frac {1-x^{s}}{1-x}}\,dx,$$H_{s}=\psi (s+1)+\gamma =\int _{0}^{1}{\frac {1-x^{s}}{1-x}}\,dx,$

where $\psi (x)$$\psi (x)$ is the digamma, and γ is the Euler-Mascheroni constant. The integration process may be repeated to obtain

$H_{s,2}=-\sum _{k=1}^{\infty }{\frac {(-1)^{k}}{k}}{s \choose k}H_{k}.$$H_{s,2}=-\sum _{k=1}^{\infty }{\frac {(-1)^{k}}{k}}{s \choose k}H_{k}.$

Relation to the Riemann zeta functionEdit

Some derivatives of fractional harmonic numbers are given by:

${\frac {d^{n}H_{x}}{dx^{n}}}=(-1)^{n+1}n!\left[\zeta (n+1)-H_{x,n+1}\right]$${\frac {d^{n}H_{x}}{dx^{n}}}=(-1)^{n+1}n!\left[\zeta (n+1)-H_{x,n+1}\right]$

${\frac {d^{n}H_{x,2}}{dx^{n}}}=(-1)^{n+1}(n+1)!\left[\zeta (n+2)-H_{x,n+2}\right]$${\frac {d^{n}H_{x,2}}{dx^{n}}}=(-1)^{n+1}(n+1)!\left[\zeta (n+2)-H_{x,n+2}\right]$

${\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].$${\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}=\sum _{n=1}^{\infty }(-1)^{n+1}x^{n}\zeta (n+1)$$H_{x}=\sum _{n=1}^{\infty }(-1)^{n+1}x^{n}\zeta (n+1)$

$H_{x,2}=\sum _{n=1}^{\infty }(-1)^{n+1}(n+1)x^{n}\zeta (n+2)$$H_{x,2}=\sum _{n=1}^{\infty }(-1)^{n+1}(n+1)x^{n}\zeta (n+2)$

$H_{x,3}={\frac {1}{2}}\sum _{n=1}^{\infty }(-1)^{n+1}(n+1)(n+2)x^{n}\zeta (n+3).$$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_{\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_{\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_{{\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_{{\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_{{\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).$$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 numbersEdit

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

$H_{n}^{(0)}={\frac {1}{n}}.$$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)}.$$H_{n}^{(r)}=\sum _{k=1}^{n}H_{k}^{(r-1)}.$

In special, $H_{n}=H_{n}^{(1)}$$H_{n}=H_{n}^{(1)}$.

See alsoEdit

• Watterson estimator
• Tajima's D
• Coupon collector's problem
• Jeep problem
• Riemann zeta function

ReferencesEdit

• Arthur T. Benjamin; Gregory O. Preston; Jennifer J. Quinn (2002). "A Stirling Encounter with Harmonic Numbers" (PDF). Mathematics Magazine. 75 (2): 95–103. doi:10.2307/3219141. 
• Donald Knuth (1997). "Section 1.2.7: Harmonic Numbers". The Art of Computer Programming. Volume 1: Fundamental Algorithms (Third ed.). Addison-Wesley. pp. 75–79. ISBN 0-201-89683-4. 
• Ed Sandifer, How Euler Did It — Estimating the Basel problem (2003)
• Paule, Peter; Schneider, Carsten (2003). "Computer Proofs of a New Family of Harmonic Number Identities" (PDF). Adv. in Appl. Math. 31 (2): 359–378. doi:10.1016/s0196-8858(03)00016-2. 
• Wenchang Chu (2004). "A Binomial Coefficient Identity Associated with Beukers' Conjecture on Apery Numbers" (PDF). The Electronic Journal of Combinatorics. 11: N15. 
• Ayhan Dil; István Mező (2008). "A Symmetric Algorithm for Hyperharmonic and Fibonacci Numbers". Applied Mathematics and Computation. 206 (2): 942–951. doi:10.1016/j.amc.2008.10.013. 
• Zoltán Retkes (2008). "An extension of the Hermite–Hadamard Inequality". Acta Sci. Math. (Szeged). 74: 95–106. 

External linksEdit

• Weisstein, Eric W. "Harmonic Number". MathWorld. 

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