The harmonic number of any order is not an integer

Asked Nov 16 '15 at 20:17

Active Nov 16 '15 at 20:17

Viewed 814 times

In this site it has been proved that the harmonic number defined as:

$$\mathcal{H}_n=\sum_{k=1}^{n} \frac{1}{k}$$

is not an integer. One proof I have seen uses Bertrand's postulate while others use elementary number thoery. See here, here and here and I guess elsewhere. Of course Bertrand's postulate is a killer for this question. Now, how would one prove that the harmonic number of any order , namely,

$$\mathcal{H}_n^{(k)}= \sum_{i=1}^{n} \frac{1}{i^k}$$

is also not an integer ? I cannot think of something.

edited Apr 13 '17 at 12:20

Community

asked Nov 16 '15 at 20:17

Tolaso

6,656

For powers $k\le -1$ these generalised harmonic numbers are certainly integers. – Dietrich Burde Nov 16 '15 at 20:45
Sure, I wasn't clear. I meant the generalized harmonic number when $k\geq 1$. For $k=1$ I have many proofs, for $k\geq 2$ ? – Tolaso Nov 16 '15 at 20:48
6

If $k\ge2$, the sum is between $1$ and $2$, so not an integer. – Matthew Conroy Nov 16 '15 at 21:14
I get that $\mathcal{H}_n^{(k)}\geq 1$ for $k\geq 2$. How do you prove the other inequality? – Tolaso Nov 17 '15 at 00:28
$\sum_{n\geq 1}\frac1{n^2}=\frac{\pi^2}{6}$. – PITTALUGA Nov 17 '15 at 09:10
You can bound these sums by comparison with an integral. That will suffice. For instance, $\sum 1/n^2 < 1 + \int_{1}^{\infty} 1/x^2 , dx = 2$. – Matthew Conroy Nov 17 '15 at 20:01

The harmonic number of any order is not an integer

0 Answers0