13
$\begingroup$

Define harmonic numbers for a complex argument $z$ as $H_z=\frac{\Gamma'(z+1)}{\Gamma(z+1)}-\Gamma'(1)$.

For $n\in\mathbb{N}$, $H_n$ are usual harmonic numbers $\sum^n_{k=1} k^{-1}$ . They are obviously rational and are known (Taeisinger 1915) to be non-integers for $n>1$.

Question: Is there a non-integer rational $q$ such that $H_q\in\mathbb{Q}$?

Improve this question
$\endgroup$
1
  • $\begingroup$ Historical details: (1) first published proof of this is by Theisinger (Monatshefte für Mathematik und Physik 26, 1915, S. 132–134), not 'Taeisinger' (2) this is misspelling is very widespread (e.g., as of 2020-05-16 it is still misspelled by the English Wikipedia in en.wikipedia.org/wiki/Harmonic_number#cite_note-5, while the German version has it right; (3) I'm of course not the first to note (2); (4) Theisinger used Bertrand's postulate, not the more elementary 2-adic reasoning. $\endgroup$
    – Peter Heinig
    Commented May 16, 2020 at 20:03

1 Answer 1

Reset to default
11
$\begingroup$

The answer is "no". Your function $H_z$ which is the same as $\psi(z+1)+\gamma$, where $\psi$ is the digamma function, and $\gamma$ is the Euler-Mascheroni constant, takes transcendental values at non-integer rationals. This is a theorem of M. Ram Murty and N. Saradha, "Transcendental values of the digamma function".

Notice that at rational values the digamma function has an explicit evaluation given by Gauss's formula.

Improve this answer
$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .