Why is there no general form for the harmonic numbers?

Question

The Harmonic numbers $H_n$ are given by the sum of the reciprocals of the natural numbers up to a given $n$, ie:

$H_1 = 1$

$H_2 = 1 + 1/2 = 3/2$

$H_3 = 1 + 1/2 + 1/3 = 11/6$

$H_n$ for noninteger $n$ can be given by the integral definition $$\int_0^1 \frac{1-x^n}{1-x}dx$$

ie: $H_{1/2} = 2-2\ln2$, or $\ln\frac{e^2}{4}$

But as far as I can tell, no general formula (ie: without an integral, a sum, product or a limit as part of the definition) for any $n$ exists. Is there a specific reason why? A proof that one does not exist? Or have we just not found one yet?

It is hard to imagine something with a nice closed form, that grows like $\log(n)$ but takes only rational values. — Jack D'Aurizio, Sep 28 '15 at 18:43
For arbitrary $n$, this can be expressed in terms of the digamma function: $H_n=\psi(n+1)+\gamma$ — Julian Rosen, Sep 28 '15 at 18:54
What constitutes a valid 'general formula' is a very subjective thing. — Winther, Sep 28 '15 at 18:56
There is a satisfactory theory of when an integral $\int_a^x f(t),dt$, where $f$ is an elementary function, can be expressed in terms of elementary functions. It is natural to ask whether there is a corresponding theory for sums. The problem must have been considered, but I know of no results. — André Nicolas, Sep 28 '15 at 19:31
@AndréNicolas Take a look at A=B: https://www.math.upenn.edu/~wilf/AeqB.html. — Yuval Filmus, Sep 28 '15 at 20:40
Quite a while ago, I read it, with pleasure. A procedure is given, that very often works. I do not know of any accompanying "Galois" theory. — André Nicolas, Sep 28 '15 at 20:49
@AndréNicolas I know very little about expressing antiderivatives as elementary functions, but I am interested in your comment, particularly the mention of a Galois theory for such things. Can you suggest a reference? — Julian Rosen, Sep 28 '15 at 22:27
Maybe a start would be the Wikipedia article on Differential Galois Theory. — André Nicolas, Sep 28 '15 at 22:38
@JackD'Aurizio: if you allow the floor function, that becomes easy. :) — , Sep 29 '15 at 09:28

lhf · Accepted Answer · 2015-09-28T18:57:54.523

15

This is a partial answer.

Since $H_n \sim \log n$, there is no formula for $H_n$ using a rational function of $n$ because $p(x)/q(x) \sim x^k$, for some integer $k$, and $\log x$ is never asymptotic to $x^k$.

Here, $f(x) \sim g(x)$ when $\displaystyle \lim_{x\to\infty} \dfrac{f(x)}{g(x)}=1$.

edited Sep 28 '15 at 18:57

answered Sep 28 '15 at 18:56

lhf

216,483

5

For the same reason, it isn't possible to express $H_n$ as an algebraic function of $n$. – Julian Rosen Sep 28 '15 at 18:57
@JulianRosen Can you explain in more detail? If $f$ is algebraic why is it necessarily the case that $f(x) \sim x^{\alpha}$? – user7530 Sep 28 '15 at 20:43
1

An algebraic function $f(x)$ can be expanded as a Puiseux series in $x^{-1}$, i.e. $f(x)=\sum_{n\geq 0} a_i x^{n_i}$, with $a_i\in\mathbb{C}$, $n_i$ a sequence of rational numbers with bounded denominator heading to $-\infty$. Then $|f(x)|\sim |x|^{\alpha}$ as $x\to\infty$, where $\alpha=\min{n_i:a_i\neq 0}\in\mathbb{Q}$. The function $f(x)$ satisfies an identity $\sum_{i=0}^n g_i(x)f(x)^i=0$, with $g_0(x),\ldots,g_n(x)$ rational functions, and the value of $\alpha$ can be computed using the Newton polygon. – Julian Rosen Sep 28 '15 at 21:35

score 6 · Answer 2 · answered Sep 28 '15 at 23:42

As lhf explained, every rational function grows like some integer power of $x$ as $x\to\infty$. Since $H_n\sim\log n$ as $n\to\infty$, there can be no rational function $f$ with $f(n)=H_n$.

This can be generalized to algebraic functions, which are functions satisfying an equation $$ \tag{$\star$} \sum_{i=0}^n a_i(x) f(x)^i =0 $$ for some polynomials $a_0,\ldots,a_n$ not all $0$. If $f$ satisfies $(\star)$, then $|f(x)|\sim C x^\alpha$ as $x\to\infty$ for some constants $C$, $\alpha$, with $\alpha$ a rational number of denominator at most $n$ (the value of $\alpha$ can be computed using Newton polygons). In particular, since there is no $\alpha$ for which $\log(n)\sim C n^\alpha$, there can be no algebraic functions $f$ with $f(n)=H_n$.

Why is there no general form for the harmonic numbers?

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