Harmonic number

==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 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 - (2 n +1) H_n + 2 n. These two results are closely analogous to the corresponding integral results \int_0^x \log y \ d y = x \log x - x and \int_0^x (\log y)^2\ d y = x (\log x)^2 - 2 x \log x + 2 x. Identities involving There are several infinite summations involving harmonic numbers and powers of Pi|: \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} ==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 , another expression for is \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} . The harmonic number can be interpreted as a Riemann sum of the integral: \int_1^{n+1} \frac{dx}{x} = \ln(n+1). The th harmonic number is about as large as the natural logarithm of . The reason is that the sum is approximated by the integral \int_1^n \frac{1}{x}\, dx, whose value is . The values of the sequence decrease monotonically towards the limit \lim_{n \to \infty} \left(H_n - \ln n\right) = \gamma, where is the Euler–Mascheroni constant. The corresponding asymptotic expansion is \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 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-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 H_n is an integer if and only if n=1, a result often attributed to Taeisinger. Indeed, using 2-adic valuation, it is not difficult to prove that for n \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=\frac{1}{2^{\lfloor\log_2(n)\rfloor}}\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 \ge 5 the numerator of H_{p-1} is divisible by p^2. Furthermore, Eisenstein proved that for all odd prime number p it holds H_{(p-1)/2} \equiv -2q_p(2) \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 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 \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 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}+\frac1{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 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) \le H_n + (\log H_n)e^{H_n}, is true for every integer with strict inequality if ; here denotes the sum of the divisors of . The eigenvalues of the nonlocal problem on L^2([-1,1]) \lambda \varphi(x) = \int_{-1}^{1} \frac{\varphi(x)-\varphi(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 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 = \sum_{k=1}^n \frac{1}{k}. The limit of H_{n, m} as is finite if , 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, ... 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 = a A - \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 . The generating function given above for 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\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}=\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 \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}=\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 \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 \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. named after Steven Roman, were introduced by Daniel Loeb and Gian-Carlo Rota in the context of a generalization of umbral calculus with logarithms. There are many possible definitions, but one of them, for n,k \geq 0, is c_n^{(0)} = 1, and c_n^{(k+1)} = \sum_{i=1}^n\frac{c_i^{(k)}}{i}. Of course, c_n^{(1)} = H_n. If n \neq 0, they satisfy c_n^{(k+1)} - \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)} = \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 n. This generalization was useful in their study to define Harmonic logarithms. ==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-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 = \psi(x+1)+\gamma, where is the digamma function, and is the Euler–Mascheroni constant. The integration process may be repeated to obtain 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 H_x=\sum_{k=2}^\infty (-1)^{k}\zeta (k)\;x^{k-1}\quad\text{ for } |x| which comes from the Taylor series for the digamma function (\zeta is the Riemann zeta function). Alternative, asymptotic formulation There is an asymptotic formulation that gives the same result as the analytic continuation of the integral just described. When seeking to approximate  for a complex number , it is effective to first compute  for some large integer . Use that as an approximation for the value of . Then use the recursion relation backwards  times, to unwind it to an approximation for . Furthermore, this approximation is exact in the limit as  goes to infinity. Specifically, for a fixed integer , it is the case that \lim_{m \rightarrow \infty} \left[H_{m+n} - H_m\right] = 0. If  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  is replaced by an arbitrary complex number , \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  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  except the negative integers, which fail because trying to use the recursion relation backwards through the value  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) , (2) for all complex numbers  except the non-positive integers, and (3) for all complex values . This last formula can be used to show that \int_0^1 H_x \, dx = \gamma, where  is the Euler–Mascheroni constant or, more generally, for every  we have: \int_0^nH_{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_{-\alpha}-H_{\alpha-1} = \pi\cot{(\pi\alpha)}. For example: \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}}{2} \ln\left(\frac{1+\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 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{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 \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} ==See also==