Krdytkyu

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”.

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 $n$:
$$sum_{i=1}^n{1over i},quad sum_{i=1}^n{1over i^2},quad sum_{i=1}^n{2^iover i},quad sum_{i=1}^ni!$$

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

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,
$$H_z=H_{z-1}+frac{1}{z}.$$

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

$$H_n = frac{binom{(n+1)!+n}{n}-1}{(n+1)!}-(n+1)Biggllfloor frac{binom{(n+1)!+n}{n}-1}{(n+1)(n+1)!}Biggrrfloor $$

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 $log(2n+1) ge H_n$