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

There are several infinite summations involving harmonic numbers and powers of π:^{[4][better source needed]} $${\displaystyle {\begin{aligned}\sum _{n=1}^{\infty }{\frac {H_{n}}{n\cdot 2^{n}}}&={\frac {\pi ^{2}}{12}}\\\sum _{n=1}^{\infty }{\frac {H_{n}^{2}}{n^{2}}}&={\frac {17}{360}}\pi ^{4}\\\sum _{n=1}^{\infty }{\frac {H_{n}^{2}}{(n+1)^{2}}}&={\frac {11}{360}}\pi ^{4}\\\sum _{n=1}^{\infty }{\frac {H_{n}}{n^{3}}}&={\frac {\pi ^{4}}{72}}\end{aligned}}}$$ 

An integral representation given by Euler^[5] 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 $${\displaystyle {\begin{aligned}H_{n}&=\int _{0}^{1}{\frac {1-x^{n}}{1-x}}\,dx=\int _{0}^{1}{\frac {1-(1-u)^{n}}{u}}\,du\\[6pt]&=\int _{0}^{1}\left[\sum _{k=1}^{n}{\binom {n}{k}}(-u)^{k-1}\right]\,du=\sum _{k=1}^{n}{\binom {n}{k}}\int _{0}^{1}(-u)^{k-1}\,du\\[6pt]&=\sum _{k=1}^{n}{\binom {n}{k}}{\frac {(-1)^{k-1}}{k}}.\end{aligned}}}$$ 

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 $${\displaystyle \sum _{n=1}^{\infty }{\frac {z^{n}}{n!}}H_{n}=e^{z}\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{k}}{\frac {z^{k}}{k!}}=e^{z}\operatorname {Ein} (z)}$$  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.^[6] 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^[7] 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^[8] 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^[9] 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^[10] showed that ${\displaystyle J_{p}}$  has zero asymptotic density, while Bing-Ling Wu and Yong-Gao Chen^[11] 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^[12] 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 on ${\displaystyle L^{2}([-1,1])}$  $${\displaystyle \lambda \varphi (x)=\int _{-1}^{1}{\frac {\varphi (x)-\varphi (y)}{|x-y|}}\,dy}$$  are given by ${\displaystyle \lambda =2H_{n}}$ , where by convention ${\displaystyle H_{0}=0}$ , and the corresponding eigenfunctions are given by the Legendre polynomials ${\displaystyle \varphi (x)=P_{n}(x)}$ .^[13]

The nth generalized harmonic number of order m is given by $${\displaystyle H_{n,m}=\sum _{k=1}^{n}{\frac {1}{k^{m}}}.}$$ 

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

The special case m = 0 gives ${\displaystyle H_{n,0}=n.}$  The special case m = 1 reduces to the usual harmonic number: $${\displaystyle H_{n,1}=H_{n}=\sum _{k=1}^{n}{\frac {1}{k}}.}$$ 

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

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

Some integrals of generalized harmonic numbers are $${\displaystyle \int _{0}^{a}H_{x,2}\,dx=a{\frac {\pi ^{2}}{6}}-H_{a}}$$  and $${\displaystyle \int _{0}^{a}H_{x,3}\,dx=aA-{\frac {1}{2}}H_{a,2},}$$  where A is Apéry's constant ζ(3), and $${\displaystyle \sum _{k=1}^{n}H_{k,m}=(n+1)H_{n,m}-H_{n,m-1}{\text{ for }}m\geq 0.}$$ 

Every generalized harmonic number of order m can be written as a function of harmonic numbers of order ${\displaystyle m-1}$  using $${\displaystyle H_{n,m}=\sum _{k=1}^{n-1}{\frac {H_{k,m-1}}{k(k+1)}}+{\frac {H_{n,m-1}}{n}}}$$    for example: ${\displaystyle H_{4,3}={\frac {H_{1,2}}{1\cdot 2}}+{\frac {H_{2,2}}{2\cdot 3}}+{\frac {H_{3,2}}{3\cdot 4}}+{\frac {H_{4,2}}{4}}}$ 

A generating function for the generalized harmonic numbers is $${\displaystyle \sum _{n=1}^{\infty }z^{n}H_{n,m}={\frac {\operatorname {Li} _{m}(z)}{1-z}},}$$  where ${\displaystyle \operatorname {Li} _{m}(z)}$  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 ${\displaystyle p,q>0}$  integer, and ${\displaystyle m>1}$  integer or not, we have from polygamma functions: $${\displaystyle H_{q/p,m}=\zeta (m)-p^{m}\sum _{k=1}^{\infty }{\frac {1}{(q+pk)^{m}}}}$$  where ${\displaystyle \zeta (m)}$  is the Riemann zeta function. The relevant recurrence relation is $${\displaystyle H_{a,m}=H_{a-1,m}+{\frac {1}{a^{m}}}.}$$  Some special values are$${\displaystyle {\begin{aligned}H_{{\frac {1}{4}},2}&=16-{\tfrac {5}{6}}\pi ^{2}-8G\\H_{{\frac {1}{2}},2}&=4-{\frac {\pi ^{2}}{3}}\\H_{{\frac {3}{4}},2}&={\frac {16}{9}}-{\frac {5}{6}}\pi ^{2}+8G\\H_{{\frac {1}{4}},3}&=64-\pi ^{3}-27\zeta (3)\\H_{{\frac {1}{2}},3}&=8-6\zeta (3)\\H_{{\frac {3}{4}},3}&=\left({\frac {4}{3}}\right)^{3}+\pi ^{3}-27\zeta (3)\end{aligned}}}$$ where G is Catalan's constant. In the special case that ${\displaystyle p=1}$ , we get $${\displaystyle H_{n,m}=\zeta (m,1)-\zeta (m,n+1),}$$ 

where ${\displaystyle \zeta (m,n)}$  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 $${\displaystyle {\begin{aligned}H_{2x}&={\frac {1}{2}}\left(H_{x}+H_{x-{\frac {1}{2}}}\right)+\ln 2\\H_{3x}&={\frac {1}{3}}\left(H_{x}+H_{x-{\frac {1}{3}}}+H_{x-{\frac {2}{3}}}\right)+\ln 3,\end{aligned}}}$$  or, more generally, $${\displaystyle H_{nx}={\frac {1}{n}}\left(H_{x}+H_{x-{\frac {1}{n}}}+H_{x-{\frac {2}{n}}}+\cdots +H_{x-{\frac {n-1}{n}}}\right)+\ln n.}$$ 

For generalized harmonic numbers, we have $${\displaystyle {\begin{aligned}H_{2x,2}&={\frac {1}{2}}\left(\zeta (2)+{\frac {1}{2}}\left(H_{x,2}+H_{x-{\frac {1}{2}},2}\right)\right)\\H_{3x,2}&={\frac {1}{9}}\left(6\zeta (2)+H_{x,2}+H_{x-{\frac {1}{3}},2}+H_{x-{\frac {2}{3}},2}\right),\end{aligned}}}$$  where ${\displaystyle \zeta (n)}$  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.^[2]: 258 Let $${\displaystyle H_{n}^{(0)}={\frac {1}{n}}.}$$  Then the nth hyperharmonic number of order r (r>0) is defined recursively as $${\displaystyle H_{n}^{(r)}=\sum _{k=1}^{n}H_{k}^{(r-1)}.}$$  In particular, ${\displaystyle H_{n}^{(1)}}$  is the ordinary harmonic number ${\displaystyle H_{n}}$ .

The Roman Harmonic numbers,^[14] named after Steven Roman, were introduced by Daniel Loeb and Gian-Carlo Rota in the context of a generalization of umbral calculus with logarithms.^[15] There are many possible definitions, but one of them, for ${\displaystyle n,k\geq 0}$ , is$${\displaystyle c_{n}^{(0)}=1,}$$ and$${\displaystyle c_{n}^{(k+1)}=\sum _{i=1}^{n}{\frac {c_{i}^{(k)}}{i}}.}$$ Of course,$${\displaystyle c_{n}^{(1)}=H_{n}.}$$ 

If ${\displaystyle n\neq 0}$ , they satisfy$${\displaystyle c_{n}^{(k+1)}-{\frac {c_{n}^{(k)}}{n}}=c_{n-1}^{(k+1)}.}$$ Closed form formulas are$${\displaystyle c_{n}^{(k)}=n!(-1)^{k}s(-n,k),}$$ where ${\displaystyle s(-n,k)}$  is Stirling numbers of the first kind generalized to negative first argument, and$${\displaystyle c_{n}^{(k)}=\sum _{j=1}^{n}{\binom {n}{j}}{\frac {(-1)^{j-1}}{j^{k}}},}$$ 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 ${\displaystyle n}$ . This generalization was useful in their study to define Harmonic logarithms.

The formulae given above, $${\displaystyle H_{x}=\int _{0}^{1}{\frac {1-t^{x}}{1-t}}\,dt=\sum _{k=1}^{\infty }{x \choose k}{\frac {(-1)^{k-1}}{k}}}$$  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 $${\displaystyle H_{x}=\psi (x+1)+\gamma ,}$$  where ψ(x) is the digamma function, and γ is the Euler–Mascheroni constant. The integration process may be repeated to obtain $${\displaystyle H_{x,2}=\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{k}}{x \choose k}H_{k}.}$$ 

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).

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.

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

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: $${\displaystyle {\begin{aligned}H_{\frac {1}{2}}&=2-2\ln 2\\H_{\frac {1}{3}}&=3-{\frac {\pi }{2{\sqrt {3}}}}-{\frac {3}{2}}\ln 3\\H_{\frac {2}{3}}&={\frac {3}{2}}+{\frac {\pi }{2{\sqrt {3}}}}-{\frac {3}{2}}\ln 3\\H_{\frac {1}{4}}&=4-{\frac {\pi }{2}}-3\ln 2\\H_{\frac {3}{4}}&={\frac {4}{3}}+{\frac {\pi }{2}}-3\ln 2\\H_{\frac {1}{6}}&=6-{\frac {\sqrt {3}}{2}}\pi -2\ln 2-{\frac {3}{2}}\ln 3\\H_{\frac {1}{8}}&=8-{\frac {1+{\sqrt {2}}}{2}}\pi -4\ln {2}-{\frac {1}{\sqrt {2}}}\left(\ln \left(2+{\sqrt {2}}\right)-\ln \left(2-{\sqrt {2}}\right)\right)\\H_{\frac {1}{12}}&=12-\left(1+{\frac {\sqrt {3}}{2}}\right)\pi -3\ln {2}-{\frac {3}{2}}\ln {3}+{\sqrt {3}}\ln \left(2-{\sqrt {3}}\right)\end{aligned}}}$$ 

Which are computed via Gauss's digamma theorem, which essentially states that for positive integers p and q with p < q $${\displaystyle H_{\frac {p}{q}}={\frac {q}{p}}+2\sum _{k=1}^{\lfloor {\frac {q-1}{2}}\rfloor }\cos \left({\frac {2\pi pk}{q}}\right)\ln \left({\sin \left({\frac {\pi k}{q}}\right)}\right)-{\frac {\pi }{2}}\cot \left({\frac {\pi p}{q}}\right)-\ln \left(2q\right)}$$ 

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

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• Arthur T. Benjamin; Gregory O. Preston; Jennifer J. Quinn (2002). "A Stirling Encounter with Harmonic Numbers" (PDF). Mathematics Magazine. 75 (2): 95–103. CiteSeerX 10.1.1.383.722. doi:10.2307/3219141. JSTOR 3219141. Archived from the original (PDF) on 2009-06-17. Retrieved 2005-08-08.
• Donald Knuth (1997). "Section 1.2.7: Harmonic Numbers". The Art of Computer Programming. Vol. 1: Fundamental Algorithms (Third ed.). Addison-Wesley. pp. 75–79. ISBN 978-0-201-89683-1.
• Ed Sandifer, How Euler Did It — Estimating the Basel problem Archived 2005-05-13 at the Wayback Machine (2003)
• Paule, Peter; Schneider, Carsten (2003). "Computer Proofs of a New Family of Harmonic Number Identities" (PDF). Adv. 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. doi:10.37236/1856.
• Ayhan Dil; István Mező (2008). "A Symmetric Algorithm for Hyperharmonic and Fibonacci Numbers". Applied Mathematics and Computation. 206 (2): 942–951. arXiv:0803.4388. doi:10.1016/j.amc.2008.10.013. S2CID 12130670.

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

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