Harmonic Number | Brilliant Math & Science Wiki (2024)

The harmonic numbers are the partial sums of the harmonic series. The \(n^\text{th}\) harmonic number is the sum of the reciprocals of each positive integer up to \(n\). The first few harmonic numbers are as follows:

\[\begin{align}H_1 &= 1 \\ \\H_2 &= H_1+\frac{1}{2} = \frac{3}{2} \\ \\H_3 &= H_2+\frac{1}{3} = \frac{11}{6} \\ \\H_4 &= H_3+\frac{1}{4} = \frac{25}{12} \\ \\H_5 &= H_4+\frac{1}{5} = \frac{137}{60} \\ \\&\vdots\end{align}\]

Harmonic numbers appear in many expressions involving special functions in analytic number theory, including the Riemann zeta function. They are also related closely to the natural logarithm.

The \(n^\text{th}\) harmonic number \( H_n\) is the sum of the reciprocals of the first \(n\) positive integers:

\[H_n=\sum_{k=1}^{n} \dfrac{1}{k} = \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dots + \dfrac{1}{n}. \]

It can also be expressed with the recurrence relation

\[H_n=H_{n-1}+\frac{1}{n}.\]

Note that \( H_n\) is a (left) Riemann sum for the integral \( \int_1^{n+1} \frac1{x} \, dx =\ln(n+1).\) So \( H_n\) is roughly equal to \( \ln(n).\) In fact, the difference between \( H_n\) and \( \ln(n)\) tends to a constant called the Euler-Mascheroni constant.

\( H_n\) is a rational number. However,

There does not exist an integer \(n>1\) such that \(H_n\) is an integer.

Write \( H_n = \frac{a_n}{n!}.\) Then \( a_n\) is a sum of all the products of \( n-1\) distinct integers between \( 1\) and \(n\) inclusive. By Bertrand's postulate, there is a prime \( p\) in the interval \( \big(\frac{n}2,n\big]\) for \( n \ge 2.\) Then \( p\) divides exactly one of the integers between \( 1\) and \(n\) inclusive, namely \(p\) itself \((\)because \( 2p >n).\) So \( a_n\) is a sum of all the products containing \( p\) plus one that does not, which is not divisible by \( p.\) So \( p \not\mid a_n,\) but \( p|n!,\) so \( H_n\) cannot be an integer because its denominator, when it is reduced to lowest terms, will contain a factor of \( p.\) \(_\square\)

The proof predicts that \( H_6\) will contain a factor of \( 5 \) in its denominator; indeed,

\[H_6 = \frac{720+360+240+180+144+120}{720}\]

and it is clear that the denominator is divisible by \( 5\) but the numerator is not.

It is somewhat surprising that harmonic numbers have no upper bound. That is, for any real number \(x\), there exists a harmonic number that is greater than \(x.\)

The harmonic series diverges.

Suppose that the harmonic series converges, and consider the following series:

\[{\large\sum\limits_{k=1}^\infty{2^{-\left\lceil\log_2{k}\right\rceil}}}=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\cdots,\]

where the \(\lceil \ \rceil\) grouping symbols denote the ceiling function.

Each term in this sequence is positive and less than or equal to the corresponding term in the harmonic series:

See Also

Intro to logarithm properties (article) | Khan Academy

\[\text{For any positive integer }k, \ 0<2^{-\left\lceil\log_2{k}\right\rceil} \le \frac{1}{k}.\]

Then if the harmonic series converges, this series converges as well.

However, this series does not converge. Grouping the like terms gives a repeated sum of \(\frac{1}{2}:\)

\[{\large\sum\limits_{k=1}^\infty{2^{-\left\lceil\log_2{k}\right\rceil}}}=1+\frac{1}{2}+\underbrace{\frac{1}{4}+\frac{1}{4}}_{\frac{1}{2}}+\underbrace{\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}}_{\frac{1}{2}}+\cdots.\]

The fact that this series diverges is a contradiction. Therefore, the harmonic series diverges. \(_\square\)

The harmonic numbers appear in expressions for integer values of the digamma function:

\[ \psi(n) = H_{n-1} - \gamma.\]

The harmonic numbers are used in the definition of the Euler-Mascheroni constant \(\gamma:\)

\[ \gamma = \lim_{n \to \infty}{\big(H_n - \ln(n)\big)}.\]

The harmonic numbers are used in an elementary reformulation (due to Lagarias) of the Riemann hypothesis.

The Riemann hypothesis is equivalent to the statement

\[ \sigma(n) \le H_n + \ln(H_n)e^{H_n}\]

for all integers \(n \ge 1,\) with strict inequality if \( n>1,\) where \( \sigma(n)\) denotes the sum of the positive divisors of \(n.\)

An integral representation for \( H_n\) given by Euler is

\[ H_n = \int_0^1 \frac{1 - x^n}{1 - x}\,dx.\]

\(\big(\)This is clear since \( \frac{1-x^n}{1-x} = 1+x+x^2+\cdots+x^{n-1}.\big)\)

This can be used to find an alternating series representation of \( H_n\) using the substitution \(x = 1−u:\)

\[ \begin{align}H_n &= \int_0^1 \frac{1 - x^n}{1 - x}\,dx \\&=-\int_1^0\frac{1-(1-u)^n}{u}\,du \\&= \int_0^1\frac{1-(1-u)^n}{u}\,du \\&= \int_0^1\left[\sum_{k=1}^n(-1)^{k-1}\binom nk u^{k-1}\right]\,du \\&= \sum_{k=1}^n (-1)^{k-1}\binom nk \int_0^1u^{k-1}\,du \\&= \sum_{k=1}^n(-1)^{k-1}\frac{1}{k}\binom nk .\end{align}\]

The generating function for the harmonic numbers has a relatively simple closed form

\[\sum_{n=1}^\infty H_nx^n = \frac{-\ln(1-x)}{1-x} \]

for \(-1\leq x<1.\)

We have

\[\begin{align}\sum_{n=1}^\infty H_nx^n &= \sum_{n=1}^\infty \sum_{k=1}^n \frac1{k} x^n \\&= \sum_{k=1}^\infty \frac1{k} \big(x^k+x^{k+1}+\cdots\big) \\&= \sum_{k=1}^\infty \frac1{k} \frac{x^k}{1-x} \\&= \frac1{1-x} \sum_{k=1}^\infty \frac{x^k}{k} \\&= \frac{-\ln(1-x)}{1-x},\end{align}\]

using the Maclaurin series for \(\ln(1-x).\) \(_\square\)

• Harmonic Progression