Do harmonic numbers have a “closed-form” expression?

Question

One of the joys of high-school mathematics is summing a complicated series to get a “closed-form” expression. And of course many of us have tried summing the harmonic series $H_n =\sum \limits_{k \leq n} \frac{1}{k}$, and failed. But should we necessarily fail?

More precisely, is it known that $H_n$ cannot be written in terms of the elementary functions, say, the rational functions, $\exp(x)$ and $\ln x$? If so, how is such a theorem proved?

Note. When I started writing the question, I was going to ask if it is known that the harmonic function cannot be represented simply as a rational function? But this is easy to see, since $H_n$ grows like $\ln n+O(1)$, whereas no rational function grows logarithmically.

Added note: This earlier question asks a similar question for “elementary integration”. I guess I am asking if there is an analogous theory of “elementary summation”.

Answer 1

There is a theory of elementary summation; the phrase generally used is "summation in finite terms." An important reference is Michael Karr, Summation in finite terms, Journal of the Association for Computing Machinery 28 (1981) 305-350, DOI: 10.1145/322248.322255. Quoting,

This paper describes techniques which greatly broaden the scope of what is meant by 'finite terms'...these methods will show that the following sums have no formula as a rational function of $n$: $$\sum_{i=1}^n{1\over i},\quad \sum_{i=1}^n{1\over i^2},\quad \sum_{i=1}^n{2^i\over i},\quad \sum_{i=1}^ni!$$

Undoubtedly the particular problem of $H_n$ goes back well before 1981. The references in Karr's paper may be of some help here.

Answer 2

Harmonic numbers can be represented in terms of integrals of elementary functions: $$H_n=\frac{\int_0^{\infty} x^n e^{-x} \log x \; dx}{\int_0^{\infty} x^n e^{-x} dx}-\int_0^{\infty} e^{-x} \log x \; dx.$$ This formula could also be used to generalize harmonic numbers to fractional or even complex arguments. These generalized harmonic numbers retain some of their useful properties, for example, $$H_z=H_{z-1}+\frac{1}{z}.$$

Answer 3

This is probably not what you were looking for (since it isn't in terms of rational or elementary functions), but for the harmonic numbers we have

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

where $\left[{n \atop k}\right]$ are the (unsigned) Stirling numbers of the first kind (page 261 from the book Concrete Mathematics by Graham, Knuth and Patashnik - second edition).

For the generalized harmonic numbers I like this formula - even though it does involve an integral and Riemann zeta...

Maybe you prefer this

Answer 4

$$H_n = \frac{\binom{(n+1)!+n}{n}-1}{(n+1)!}-(n+1)\Biggl\lfloor \frac{\binom{(n+1)!+n}{n}-1}{(n+1)(n+1)!}\Biggr\rfloor $$

Answer 5

The following series shows the relationship between the harmonic numbers and the logarithm of odd integers.

$$ log(2n+1)=H_n+\sum_{k=1}^{\infty}\left(\sum_{i=-n}^{n}\frac{1}{(2n+1)k+i}-\frac{1}{k}\right) $$ https://math.stackexchange.com/a/1602945/134791

Equivalently, $$ H_n=log(2n+1)-\sum_{k=1}^{\infty}\left(\sum_{i=-n}^{n}\frac{1}{(2n+1)k+i}-\frac{1}{k}\right) $$

An integral form is given by

$$H_n=\log(2n+1)-\int_{0}^{1} \frac{x^n(1-x)}{\sum_{k=0}^{2n}x^k} \left( \frac{n(n+1)}{2}x^{n-1}+\sum_{k=0}^{n-2}\frac{(k+1)(k+2)}{2}\left(x^k+x^{2(n-1)-k}\right)\right)dx$$

An integral to prove that $\log(2n+1) \ge H_n$

Answer 6

Using this extension of real numbers one can write

$$H_n=\gamma + \operatorname{reg} \ln(\omega_++n)$$

where $\operatorname{reg}$ is taking regular (finite) part of an extended number and $\omega_+$ is a special infinite element $\omega_+=\sum_0^\infty 1=\pi\delta(0)+1/2$.

Moreover, one can write it as

$$H_n=\ln(\omega_++n)+\gamma-\int_1^\infty \frac{dx}x=\ln\left(\frac{\omega_++n}{\omega_+}\right)$$