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?

Answer 1

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

Answer 2

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 $f(n)=H_n$.