Harmonic number

By definition, the harmonic numbers satisfy the recurrence relation

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

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

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

The functions

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

satisfy the property

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

In particular

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

is an integral of the logarithmic function.

The harmonic numbers satisfy the series identities

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

and

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

These two results are closely analogous to the corresponding integral results

$${\displaystyle \int _{0}^{x}\log y\ dy=x\log x-x}$$

and

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

Identities involving π

There are several infinite summations involving harmonic numbers and powers of π:[3]

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

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

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

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

An integral representation given by Euler[4] is

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

The equality above is straightforward by the simple algebraic identity

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

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

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

Graph demonstrating a connection between harmonic numbers and the natural logarithm. The harmonic number H_n can be interpreted as a Riemann sum of the integral: ${\displaystyle \int _{1}^{n+1}{\frac {dx}{x}}=\ln(n+1).}$

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

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

whose value is ln n.

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

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

where γ ≈ 0.5772156649 is the Euler–Mascheroni constant. The corresponding asymptotic expansion is

$${\displaystyle {\begin{aligned}H_{n}&\sim \ln {n}+\gamma +{\frac {1}{2n}}-\sum _{k=1}^{\infty }{\frac {B_{2k}}{2kn^{2k}}}\\&=\ln {n}+\gamma +{\frac {1}{2n}}-{\frac {1}{12n^{2}}}+{\frac {1}{120n^{4}}}-\cdots ,\end{aligned}}}$$

where B_k are the Bernoulli numbers.

A generating function for the harmonic numbers is

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

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

$${\displaystyle \sum _{n=1}^{\infty }{\frac {z^{n}}{n!}}H_{n}=-e^{z}\sum _{k=1}^{\infty }{\frac {1}{k}}{\frac {(-z)^{k}}{k!}}=e^{z}\operatorname {Ein} (z)}$$

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

$${\displaystyle \operatorname {Ein} (z)=\mathrm {E} _{1}(z)+\gamma +\ln z=\Gamma (0,z)+\gamma +\ln z}$$

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

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.[5] 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,

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

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

As a consequence of Wolstenholme's theorem, for any prime number ${\displaystyle p\geq 5}$ the numerator of ${\displaystyle H_{p-1}}$is divisible by ${\textstyle p^{2}}$. Furthermore, Eisenstein[6] proved that for all odd prime number ${\textstyle p}$ it holds

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

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

In 1991, Eswarathasan and Levine[7] defined ${\displaystyle J_{p}}$ as the set of all positive integers ${\displaystyle n}$ such that the numerator of ${\displaystyle H_{n}}$ is divisible by a prime number ${\displaystyle p.}$ They proved that

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

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

Eswarathasan and Levine also conjectured that ${\displaystyle J_{p}}$ is a finite set for all primes ${\displaystyle p,}$ and that there are infinitely many harmonic primes. Boyd[8] verified that ${\displaystyle J_{p}}$ is finite for all prime numbers up to ${\displaystyle p=547}$ 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 ${\displaystyle 1/e}$. Sanna[9] showed that ${\displaystyle J_{p}}$ has zero asymptotic density, while Bing-Ling Wu and Yong-Gao Chen[10] proved that the number of elements of ${\displaystyle J_{p}}$ not exceeding ${\displaystyle x}$ is at most ${\displaystyle 3x^{{\frac {2}{3}}+{\frac {1}{25\log p}}}}$, for all ${\displaystyle x\geq 1}$.

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

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

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:

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

although

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

converges more quickly.

In 2002, Jeffrey Lagarias proved[11] that the Riemann hypothesis is equivalent to the statement that

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

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

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

are given by ${\displaystyle \lambda =2H_{n}}$, where by convention ${\displaystyle H_{0}=0}$, and the corresponding eigenfunctions are given by the Legendre polynomials ${\displaystyle \varphi (x)=P_{n}(x)}$.[12]

The formulae given above,

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

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

$${\displaystyle H_{x}=\psi (x+1)+\gamma ,}$$

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

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

The Taylor series for the harmonic numbers is

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

which comes from the Taylor series for the digamma function (${\displaystyle \zeta }$ 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:[13]

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

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

Alternative, asymptotic formulation

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

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

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

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.

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

Swapping the order of the two sides of this equation and then subtracting them from H_x gives

$${\displaystyle {\begin{aligned}H_{x}&=\lim _{m\rightarrow \infty }\left[H_{m}-(H_{m+x}-H_{x})\right]\\[6pt]&=\lim _{m\rightarrow \infty }\left[\left(\sum _{k=1}^{m}{\frac {1}{k}}\right)-\left(\sum _{k=1}^{m}{\frac {1}{x+k}}\right)\right]\\[6pt]&=\lim _{m\rightarrow \infty }\sum _{k=1}^{m}\left({\frac {1}{k}}-{\frac {1}{x+k}}\right)=x\sum _{k=1}^{\infty }{\frac {1}{k(x+k)}}\,.\end{aligned}}}$$

This infinite series converges for all complex numbers x except the negative integers, which fail because trying to use the recursion relation H_n = H_n−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) H_x = H_x−1 + 1/x for all complex numbers x except the non-positive integers, and (3) lim_m→+∞ (H_m+x − H_m) = 0 for all complex values x.

Note that this last formula can be used to show that

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

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

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

Special values for fractional arguments

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

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

More values may be generated from the recurrence relation

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

or from the reflection relation

$${\displaystyle H_{1-\alpha }-H_{\alpha }=\pi \cot {(\pi \alpha )}-{\frac {1}{\alpha }}+{\frac {1}{1-\alpha }}\,.}$$

For example:

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

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

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

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

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

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

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

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

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

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

Relation to the Riemann zeta function

Some derivatives of fractional harmonic numbers are given by

$${\displaystyle {\begin{aligned}{\frac {d^{n}H_{x}}{dx^{n}}}&=(-1)^{n+1}n!\left[\zeta (n+1)-H_{x,n+1}\right]\\[6pt]{\frac {d^{n}H_{x,2}}{dx^{n}}}&=(-1)^{n+1}(n+1)!\left[\zeta (n+2)-H_{x,n+2}\right]\\[6pt]{\frac {d^{n}H_{x,3}}{dx^{n}}}&=(-1)^{n+1}{\frac {1}{2}}(n+2)!\left[\zeta (n+3)-H_{x,n+3}\right].\end{aligned}}}$$

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

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

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

$${\displaystyle {\begin{aligned}H_{1/a}&={\frac {1}{a}}\left(\zeta (2)-{\frac {1}{a}}\zeta (3)+{\frac {1}{a^{2}}}\zeta (4)-{\frac {1}{a^{3}}}\zeta (5)+\cdots \right)\\[6pt]H_{1/a,\,2}&={\frac {1}{a}}\left(2\zeta (3)-{\frac {3}{a}}\zeta (4)+{\frac {4}{a^{2}}}\zeta (5)-{\frac {5}{a^{3}}}\zeta (6)+\cdots \right)\\[6pt]H_{1/a,\,3}&={\frac {1}{2a}}\left(2\cdot 3\zeta (4)-{\frac {3\cdot 4}{a}}\zeta (5)+{\frac {4\cdot 5}{a^{2}}}\zeta (6)-{\frac {5\cdot 6}{a^{3}}}\zeta (7)+\cdots \right).\end{aligned}}}$$

• 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

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