harmonic number

By definition, the harmonic numbers satisfy the recurrence relation

$$H\_\{n\; +\; 1\}\; =\; H\_\{n\}\; +\; \backslash frac\{1\}\{n\; +\; 1\}.$$

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

$$H\_n\; =\; \backslash frac\{1\}\{n!\}\backslash left[\{n+1\; \backslash atop\; 2\}\backslash right].$$

The harmonic numbers satisfy the series identities

$$\backslash sum\_\{k=1\}^n\; H\_k\; =\; (n+1)\; H\_\{n\}\; -\; n$$

and

$$\backslash sum\_\{k=1\}^n\; H\_k^2\; =\; (n+1)H\_\{n\}^2\; -\; (2\; n\; +1)\; H\_n\; +\; 2\; n.$$

These two results are closely analogous to the corresponding integral results

$$\backslash int\_0^x\; \backslash log\; y\; \backslash \; d\; y\; =\; x\; \backslash log\; x\; -\; x$$

and

$$\backslash int\_0^x\; (\backslash log\; y)^2\backslash \; d\; y\; =\; x\; (\backslash log\; x)^2\; -\; 2\; x\; \backslash log\; x\; +\; 2\; x.$$

There are several infinite summations involving harmonic numbers and powers of {{pi}}:{{Cite web |last=Weisstein |first=Eric W. |title=Harmonic Number |url=https://mathworld.wolfram.com/HarmonicNumber.html |access-date=2024-09-30 |website=mathworld.wolfram.com |language=en}}{{better source|date=February 2022}}

$$\backslash begin\{align\}$$

\sum_{n=1}^\infty \frac{H_n}{n\cdot 2^n} &= \frac{\pi^2}{12} \\

\sum_{n=1}^\infty \frac{H_n^2}{n^2} &= \frac{17}{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}{n^3} &= \frac{\pi^4}{72}

\end{align}

An integral representation given by Euler{{citation|title=How Euler Did It|series=MAA Spectrum|first=C. Edward|last=Sandifer|publisher=Mathematical Association of America|year=2007|isbn=9780883855638|page=206|url=https://books.google.com/books?id=sohHs7ExOsYC&pg=PA206}}. is

$$H\_n\; =\; \backslash int\_0^1\; \backslash frac\{1\; -\; x^n\}\{1\; -\; x\}\backslash ,dx.$$

The equality above is straightforward by the simple algebraic identity

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

Using the substitution {{math|1=x = 1 − u}}, another expression for {{math|H_n}} is

$$\backslash begin\{align\}$$

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 \binom nk (-u)^{k-1}\right]\,du

= \sum_{k=1}^n \binom nk \int_0^1 (-u)^{k-1}\,du \\[6pt]

&= \sum_{k=1}^n \binom nk \frac{(-1)^{k-1}}{k}.

\end{align}

File:Integral Test.svg. The harmonic number {{math|H_n}} can be interpreted as a Riemann sum of the integral: $\backslash int\_1^\{n+1\}\; \backslash frac\{dx\}\{x\}\; =\; \backslash ln(n+1).$]]

The {{mvar|n}}th harmonic number is about as large as the natural logarithm of {{mvar|n}}. The reason is that the sum is approximated by the integral

$$\backslash int\_1^n\; \backslash frac\{1\}\{x\}\backslash ,\; dx,$$

whose value is {{math|ln n}}.

The values of the sequence {{math|H_n − ln n}} decrease monotonically towards the limit

$$\backslash lim\_\{n\; \backslash to\; \backslash infty\}\; \backslash left(H\_n\; -\; \backslash ln\; n\backslash right)\; =\; \backslash gamma,$$

where {{math|γ ≈ 0.5772156649}} is the Euler–Mascheroni constant. The corresponding asymptotic expansion is

$$\backslash begin\{align\}$$

H_n &\sim \ln{n}+\gamma+\frac{1}{2n}-\sum_{k=1}^\infty \frac{B_{2k}}{2k n^{2k}}\\

&=\ln{n}+\gamma+\frac{1}{2n}-\frac{1}{12n^2}+\frac{1}{120n^4}-\cdots,

\end{align}

where {{math|B_k}} are the Bernoulli numbers.

{{Reflist|group=note}}

A generating function for the harmonic numbers is

$$\backslash sum\_\{n=1\}^\backslash infty\; z^n\; H\_n\; =\; \backslash frac\; \{-\backslash ln(1-z)\}\{1-z\},$$

where ln(z) is the natural logarithm. An exponential generating function is

$$\backslash sum\_\{n=1\}^\backslash infty\; \backslash frac\; \{z^n\}\{n!\}\; H\_n\; =\; e^z\; \backslash sum\_\{k=1\}^\backslash infty\; \backslash frac\{(-1)^\{k-1\}\}\{k\}\; \backslash frac\; \{z^k\}\{k!\}\; =\; e^z\; \backslash operatorname\{Ein\}(z)$$

where Ein(z) is the entire exponential integral. The exponential integral may also be expressed as

$$\backslash operatorname\{Ein\}(z)\; =\; \backslash mathrm\{E\}\_1(z)\; +\; \backslash gamma\; +\; \backslash ln\; z\; =\; \backslash Gamma\; (0,z)\; +\; \backslash gamma\; +\; \backslash ln\; z$$

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

The harmonic numbers have several interesting arithmetic properties. It is well-known that $H\_n$ is an integer if and only if $n=1$, a result often attributed to Taeisinger.{{Cite book|title=CRC Concise Encyclopedia of Mathematics|last=Weisstein|first=Eric W.|publisher=Chapman & Hall/CRC|year=2003|isbn=978-1-58488-347-0|location=Boca Raton, FL|pages=3115}} Indeed, using 2-adic valuation, it is not difficult to prove that for $n\; \backslash ge\; 2$ the numerator of $H\_n$ is an odd number while the denominator of $H\_n$ is an even number. More precisely,

$$H\_n=\backslash frac\{1\}\{2^\{\backslash lfloor\backslash log\_2(n)\backslash rfloor\}\}\backslash frac\{a\_n\}\{b\_n\}$$

with some odd integers $a\_n$ and $b\_n$.

As a consequence of Wolstenholme's theorem, for any prime number $p\; \backslash ge\; 5$ the numerator of $H\_\{p-1\}$ is divisible by $p^2$. Furthermore, Eisenstein{{Cite journal|last=Eisenstein|first=Ferdinand Gotthold Max|year=1850|title=Eine neue Gattung zahlentheoretischer Funktionen, welche von zwei Elementen ahhängen und durch gewisse lineare Funktional-Gleichungen definirt werden|journal=Berichte Königl. Preuβ. Akad. Wiss. Berlin|volume=15|pages=36–42}} proved that for all odd prime number $p$ it holds

$$H\_\{(p-1)/2\}\; \backslash equiv\; -2q\_p(2)\; \backslash pmod\; p$$

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

In 1991, Eswarathasan and Levine{{Cite journal|last1=Eswarathasan|first1=Arulappah|last2=Levine|first2=Eugene|year=1991|title=p-integral harmonic sums|journal=Discrete Mathematics|volume=91|issue=3|pages=249–257|doi=10.1016/0012-365X(90)90234-9|doi-access=free}} 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

$$\backslash \{p-1,p^2-p,p^2-1\backslash \}\backslash subseteq\; J\_p$$

for all prime numbers $p\; \backslash ge\; 5,$ and they defined harmonic primes to be the primes $p$ 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{{Cite journal|last=Boyd|first=David W.|year=1994|title=A p-adic study of the partial sums of the harmonic series|url=http://projecteuclid.org/euclid.em/1048515811|journal=Experimental Mathematics|volume=3|issue=4|pages=287–302|doi=10.1080/10586458.1994.10504298|citeseerx=10.1.1.56.7026}} 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{{Cite journal|last=Sanna|first=Carlo|year=2016|title=On the p-adic valuation of harmonic numbers|journal=Journal of Number Theory|volume=166|pages=41–46|doi=10.1016/j.jnt.2016.02.020|hdl=2318/1622121|url=https://iris.unito.it/bitstream/2318/1622121/1/padicharm.pdf|doi-access=free}} showed that $J\_p$ has zero asymptotic density, while Bing-Ling Wu and Yong-Gao Chen{{Cite journal|last1=Chen|first1=Yong-Gao|last2=Wu|first2=Bing-Ling|year=2017|title=On certain properties of harmonic numbers|journal=Journal of Number Theory|volume=175|pages=66–86|doi=10.1016/j.jnt.2016.11.027|doi-access=}} proved that the number of elements of $J\_p$ not exceeding $x$ is at most $3x^\{\backslash frac\{2\}\{3\}+\backslash frac1\{25\; \backslash log\; p\}\}$, for all $x\; \backslash geq\; 1$.

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

$$\backslash psi(n)\; =\; H\_\{n-1\}\; -\; \backslash gamma.$$

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 {{mvar|γ}} using the limit introduced earlier:

$$\backslash gamma\; =\; \backslash lim\_\{n\; \backslash rightarrow\; \backslash infty\}\{\backslash left(H\_n\; -\; \backslash ln(n)\backslash right)\},$$

although

$$\backslash gamma\; =\; \backslash lim\_\{n\; \backslash to\; \backslash infty\}\{\backslash left(H\_n\; -\; \backslash ln\backslash left(n+\backslash frac\{1\}\{2\}\backslash right)\backslash right)\}$$

converges more quickly.

In 2002, Jeffrey Lagarias proved{{cite journal |author=Jeffrey Lagarias |title=An Elementary Problem Equivalent to the Riemann Hypothesis |journal=Amer. Math. Monthly |volume=109 |issue=6 |year=2002 |pages=534–543 |arxiv=math.NT/0008177 |doi=10.2307/2695443|jstor=2695443 }} that the Riemann hypothesis is equivalent to the statement that

$$\backslash sigma(n)\; \backslash le\; H\_n\; +\; (\backslash log\; H\_n)e^\{H\_n\},$$

is true for every integer {{math|n ≥ 1}} with strict inequality if {{math|n > 1}}; here {{math|σ(n)}} denotes the sum of the divisors of {{mvar|n}}.

The eigenvalues of the nonlocal problem on $L^2([-1,1])$

$$\backslash lambda\; \backslash varphi(x)\; =\; \backslash int\_\{-1\}^\{1\}\; \backslash frac\{\backslash varphi(x)-\backslash varphi(y)\}$$

are given by $\backslash lambda\; =\; 2H\_n$, where by convention $H\_0\; =\; 0$, and the corresponding eigenfunctions are given by the Legendre polynomials $\backslash varphi(x)\; =\; P\_n(x)$.{{cite journal |author=E.O. Tuck |title=Some methods for flows past blunt slender bodies |journal=J. Fluid Mech. |volume=18 |year=1964 |issue=4 |pages=619–635 |doi=10.1017/S0022112064000453|bibcode=1964JFM....18..619T |s2cid=123120978 }}

The nth generalized harmonic number of order m is given by

$$H\_\{n,m\}=\backslash sum\_\{k=1\}^n\; \backslash frac\{1\}\{k^m\}.$$

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

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

$$H\_\{n,\; 1\}\; =\; H\_n\; =\; \backslash sum\_\{k=1\}^n\; \backslash frac\{1\}\{k\}.$$

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

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

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, ... {{OEIS|id=A128670}}

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

$$\backslash int\_0^a\; H\_\{x,2\}\; \backslash ,\; dx\; =\; a\; \backslash frac\; \{\backslash pi^2\}\{6\}-H\_\{a\}$$

and

$$\backslash int\_0^a\; H\_\{x,3\}\; \backslash ,\; dx\; =\; a\; A\; -\; \backslash frac\; \{1\}\{2\}\; H\_\{a,2\},$$ where A is Apéry's constant ζ(3),

and

$$\backslash sum\_\{k=1\}^n\; H\_\{k,m\}=(n+1)H\_\{n,m\}-\; H\_\{n,m-1\}\; \backslash text\{\; for\; \}\; m\; \backslash geq\; 0\; .$$

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

$$H\_\{n,m\}\; =\; \backslash sum\_\{k=1\}^\{n-1\}\; \backslash frac\; \{H\_\{k,m-1\}\}\{k(k+1)\}\; +\; \backslash frac\; \{H\_\{n,m-1\}\}\{n\}$$   for example: $H\_\{4,3\}\; =\; \backslash frac\; \{H\_\{1,2\}\}\{1\; \backslash cdot\; 2\}\; +\; \backslash frac\; \{H\_\{2,2\}\}\{2\; \backslash cdot\; 3\}\; +\; \backslash frac\; \{H\_\{3,2\}\}\{3\; \backslash cdot\; 4\}\; +\; \backslash frac\; \{H\_\{4,2\}\}\{4\}$

A generating function for the generalized harmonic numbers is

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

where $\backslash operatorname\{Li\}\_m(z)$ is the polylogarithm, and {{math|{{mabs|z}} < 1}}. The generating function given above for {{math|1=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:

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

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

$$H\_\{a,m\}=H\_\{a-1,m\}+\backslash frac\{1\}\{a^m\}.$$

Some special values are$$\backslash begin\{align\}$$

H_{\frac{1}{4},2} &= 16-\tfrac{5}{6}\pi^2 -8G\\

H_{\frac{1}{2},2} &= 4-\frac{\pi^2}{3} \\

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

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

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

H_{\frac{3}{4},3} &= \left(\frac{4}{3}\right)^3+\pi^3 -27\zeta(3)

\end{align}where G is Catalan's constant. In the special case that $p\; =\; 1$, we get

$$H\_\{n,m\}=\backslash zeta(m,\; 1)\; -\; \backslash zeta(m,\; n+1),$$

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

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

$$\backslash begin\{align\}$$

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,

\end{align}

or, more generally,

$$H\_\{nx\}=\backslash frac\{1\}\{n\}\backslash left(H\_x+H\_\{x-\backslash frac\{1\}\{n\}\}+H\_\{x-\backslash frac\{2\}\{n\}\}+\backslash cdots\; +H\_\{x-\backslash frac\{n-1\}\{n\}\}\; \backslash right)\; +\; \backslash ln\; n.$$

For generalized harmonic numbers, we have

$$\backslash begin\{align\}$$

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),

\end{align}

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

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

$$H\_n^\{(0)\}\; =\; \backslash frac1n.$$

Then the nth hyperharmonic number of order r (r>0) is defined recursively as

$$H\_n^\{(r)\}\; =\; \backslash sum\_\{k=1\}^n\; H\_k^\{(r-1)\}.$$

In particular, $H\_n^\{(1)\}$ is the ordinary harmonic number $H\_n$.

The Roman Harmonic numbers,{{Cite journal |last=Sesma |first=J. |date=2017 |title=The Roman harmonic numbers revisited |url=http://dx.doi.org/10.1016/j.jnt.2017.05.009 |journal=Journal of Number Theory |volume=180 |pages=544–565 |doi=10.1016/j.jnt.2017.05.009 |issn=0022-314X|arxiv=1702.03718 }} named after Steven Roman, were introduced by Daniel Loeb and Gian-Carlo Rota in the context of a generalization of umbral calculus with logarithms.{{Cite journal |last1=Loeb |first1=Daniel E |last2=Rota |first2=Gian-Carlo |date=1989 |title=Formal power series of logarithmic type |journal=Advances in Mathematics |volume=75 |issue=1 |pages=1–118 |doi=10.1016/0001-8708(89)90079-0 |issn=0001-8708|doi-access=free }} There are many possible definitions, but one of them, for $n,k\; \backslash geq\; 0$, is$$c\_n^\{(0)\}\; =\; 1,$$and$$c\_n^\{(k+1)\}\; =\; \backslash sum\_\{i=1\}^n\backslash frac\{c\_i^\{(k)\}\}\{i\}.$$Of course,$$c\_n^\{(1)\}\; =\; H\_n.$$

If $n\; \backslash neq\; 0$, they satisfy$$c\_n^\{(k+1)\}\; -\; \backslash frac\{c\_n^\{(k)\}\}\{n\}\; =\; c\_\{n-1\}^\{(k+1)\}.$$Closed form formulas are$$c\_n^\{(k)\}\; =\; n!\; (-1)^k\; s(-n,k),$$where $s(-n,k)$ is Stirling numbers of the first kind generalized to negative first argument, and$$c\_n^\{(k)\}\; =\; \backslash sum\_\{j=1\}^n\; \backslash binom\{n\}\{j\}\; \backslash frac\{(-1)^\{j-1\}\}\{j^k\},$$which was found by Donald Knuth.

In fact, these numbers were defined in a more general manner using Roman numbers and Roman factorials, that include negative values for $n$. This generalization was useful in their study to define Harmonic logarithms.

{{unreferenced section|date=May 2019}}

The formulae given above,

$$H\_x\; =\; \backslash int\_0^1\; \backslash frac\{1-t^x\}\{1-t\}\; \backslash ,\; dt=\; \backslash sum\_\{k=1\}^\backslash infty\; \{x\; \backslash choose\; k\}\; \backslash frac\{(-1)^\{k-1\}\}\{k\}$$

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

$$H\_x\; =\; \backslash psi(x+1)+\backslash gamma,$$

where {{math|ψ(x)}} is the digamma function, and {{math|γ}} is the Euler–Mascheroni constant. The integration process may be repeated to obtain

$$H\_\{x,2\}=\; \backslash sum\_\{k=1\}^\backslash infty\; \backslash frac\; \{(-1)^\{k-1\}\}\{k\}\; \{x\; \backslash choose\; k\}\; H\_k.$$

The Taylor series for the harmonic numbers is

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

There is an asymptotic formulation that gives the same result as the analytic continuation of the integral just described. When seeking to approximate {{math|H{{sub|x}}}} for a complex number {{math|x}}, it is effective to first compute {{math|H{{sub|m}}}} for some large integer {{math|m}}. Use that as an approximation for the value of {{math|H{{sub|m+x}}}}. Then use the recursion relation {{math|1=H{{sub|n}} = H{{sub|n−1}} + 1/n}} backwards {{math|m}} times, to unwind it to an approximation for {{math|H{{sub|x}}}}. Furthermore, this approximation is exact in the limit as {{math|m}} goes to infinity.

Specifically, for a fixed integer {{math|n}}, it is the case that

$$\backslash lim\_\{m\; \backslash rightarrow\; \backslash infty\}\; \backslash left[H\_\{m+n\}\; -\; H\_m\backslash right]\; =\; 0.$$

If {{math|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 {{math|n}} is replaced by an arbitrary complex number {{math|x}},

$$\backslash lim\_\{m\; \backslash rightarrow\; \backslash infty\}\; \backslash left[H\_\{m+x\}\; -\; H\_m\backslash right]\; =\; 0\backslash ,.$$

Swapping the order of the two sides of this equation and then subtracting them from {{math|H_x}} gives

\begin{align}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{align}

This infinite series converges for all complex numbers {{math|x}} except the negative integers, which fail because trying to use the recursion relation {{math|1=H{{sub|n}} = H{{sub|n−1}} + 1/n}} backwards through the value {{math|1=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) {{math|1=H{{sub|0}} = 0}}, (2) {{math|1=H{{sub|x}} = H{{sub|x−1}} + 1/x}} for all complex numbers {{math|x}} except the non-positive integers, and (3) {{math|1=lim_m→+∞ (H_m+x − H_m) = 0}} for all complex values {{math|x}}.

This last formula can be used to show that

$$\backslash int\_0^1\; H\_x\; \backslash ,\; dx\; =\; \backslash gamma,$$

where {{math|γ}} is the Euler–Mascheroni constant or, more generally, for every {{math|n}} we have:

$$\backslash int\_0^nH\_\{x\}\backslash ,dx\; =\; n\backslash gamma\; +\; \backslash ln(n!)\; .$$

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

$$H\_\backslash alpha\; =\; \backslash int\_0^1\backslash frac\{1-x^\backslash alpha\}\{1-x\}\backslash ,dx\backslash ,\; .$$

More values may be generated from the recurrence relation

$$H\_\backslash alpha\; =\; H\_\{\backslash alpha-1\}+\backslash frac\{1\}\{\backslash alpha\}\backslash ,,$$

or from the reflection relation

$$H\_\{-\backslash alpha\}-H\_\{\backslash alpha-1\}\; =\; \backslash pi\backslash cot\{(\backslash pi\backslash alpha)\}.$$

For example:

$$\backslash begin\{align\}$$

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

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

H_{\frac{2}{3}} &= \frac{3}{2}+\frac{\pi}{2\sqrt{3}} - \frac{3}{2}\ln 3 \\

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

H_{\frac{1}{5}} &= 5 - \frac{\pi}{2} \sqrt{1+\frac{2}{\sqrt{5}}} - \frac{5}{4} \ln 5 - \frac{\sqrt{5}}{4} \ln\left(\frac{3+\sqrt{5}}{2}\right) \\

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

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

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

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

\end{align}

Which are computed via Gauss's digamma theorem, which essentially states that for positive integers p and q with p < q

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

Some derivatives of fractional harmonic numbers are given by

\begin{align}

\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{align}

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

\begin{align}

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{align}

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

\begin{align}

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\cdot3\zeta(4)-\frac{3\cdot4}{a}\zeta(5)+\frac{4\cdot5}{a^2}\zeta(6)-\frac{5\cdot6}{a^3}\zeta(7)+\cdots\right).

\end{align}

• Watterson estimator
• Tajima's D
• Coupon collector's problem
• Jeep problem
• 100 prisoners problem
• Riemann zeta function
• List of sums of reciprocals
• False discovery rate#Benjamini–Yekutieli procedure
• Block-stacking problem

{{Reflist|refs=

{{Cite book

| last1 = John H. | first1 = Conway

| last2 = Richard K. | first2 = Guy

| title = The book of numbers

| year = 1995

| publisher = Copernicus

}}

}}

• {{cite journal |author1=Arthur T. Benjamin |author2=Gregory O. Preston |author3=Jennifer J. Quinn |url=http://www.math.hmc.edu/~benjamin/papers/harmonic.pdf |title=A Stirling Encounter with Harmonic Numbers |year=2002 |journal=Mathematics Magazine |volume=75 |issue=2 |pages=95–103 |doi=10.2307/3219141 |jstor=3219141 |citeseerx=10.1.1.383.722 |access-date=2005-08-08 |archive-url=https://web.archive.org/web/20090617002133/http://www.math.hmc.edu/~benjamin/papers/harmonic.pdf |archive-date=2009-06-17 |url-status=dead }}
• {{cite book |author=Donald Knuth |author-link=Donald Knuth |title=The Art of Computer Programming |volume=1: Fundamental Algorithms |edition=Third |publisher=Addison-Wesley |year=1997 |isbn=978-0-201-89683-1 |chapter=Section 1.2.7: Harmonic Numbers |pages=75–79}}
• Ed Sandifer, [http://www.maa.org/editorial/euler/How%20Euler%20Did%20It%2002%20Estimating%20the%20Basel%20Problem.pdf How Euler Did It — Estimating the Basel problem] {{Webarchive|url=https://web.archive.org/web/20050513105944/http://www.maa.org/editorial/euler/How%20Euler%20Did%20It%2002%20Estimating%20the%20Basel%20Problem.pdf |date=2005-05-13 }} (2003)
• {{cite journal

| last1 = Paule | first1 = Peter | author1-link = Peter Paule

| last2 = Schneider | first2 = Carsten

| doi = 10.1016/s0196-8858(03)00016-2

| issue = 2

| journal = Adv. Appl. Math.

| pages = 359–378

| title = Computer Proofs of a New Family of Harmonic Number Identities

| url = http://www.risc.uni-linz.ac.at/publications/download/risc_200/HarmonicNumberIds.pdf

| volume = 31

| year = 2003}}

• {{cite journal |author=Wenchang Chu |url=http://www.combinatorics.org/Volume_11/PDF/v11i1n15.pdf |title=A Binomial Coefficient Identity Associated with Beukers' Conjecture on Apery Numbers |year=2004 |journal=The Electronic Journal of Combinatorics |volume=11 |pages=N15|doi=10.37236/1856 |doi-access=free }}

• {{MathWorld|urlname=HarmonicNumber |title=Harmonic Number}}

{{PlanetMath attribution|id=3421|title=Harmonic number}}

{{DEFAULTSORT:Harmonic Number}}

Category:Number theory