# Why is there no general form for the harmonic numbers?

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
• You might be interested in this: oeis.org/A001008 – Daniel R Sep 28 '15 at 18:49
• 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

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

• 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
• 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

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 $f(n)=H_n$.

$$H_n=\frac1{n!}\left(\sum_{k=1}^n\prod_{1\le j\le n, j\neq k}j\right)$$

This is a closed form: there are no trascendental operations involved, and it can be computed in finte time. Ugly, but closed.

• "without an integral, a sum, product or a limit as part of the definition" – leonbloy Sep 28 '15 at 18:56
• I'm curious why that should be considered better (or "more closed form") than $H_n=\sum_{k=1}^n 1/k$ ? – leonbloy Sep 28 '15 at 18:57