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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.

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  • $\begingroup$ For powers $k\le -1$ these generalised harmonic numbers are certainly integers. $\endgroup$ – Dietrich Burde Nov 16 '15 at 20:45
  • $\begingroup$ Sure, I wasn't clear. I meant the generalized harmonic number when $k\geq 1$. For $k=1$ I have many proofs, for $k\geq 2$ ? $\endgroup$ – Tolaso Nov 16 '15 at 20:48
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    $\begingroup$ If $k\ge2$, the sum is between $1$ and $2$, so not an integer. $\endgroup$ – Matthew Conroy Nov 16 '15 at 21:14
  • $\begingroup$ I get that $\mathcal{H}_n^{(k)}\geq 1$ for $k\geq 2$. How do you prove the other inequality? $\endgroup$ – Tolaso Nov 17 '15 at 0:28
  • $\begingroup$ $\sum_{n\geq 1}\frac1{n^2}=\frac{\pi^2}{6}$. $\endgroup$ – PITTALUGA Nov 17 '15 at 9:10

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