Talk:Harmonic number

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I think we should put the Riemann Hypothesis section back... Look at it this way: someone studying the harmonic numbers using our encyclopedia would miss out on the connection to the Riemann Hypothesis. Scythe33 20:10, 7 August 2005 (UTC)

What's the connection? There are a variety of statements equivalent to RH, which involve all sorts of constants, sums, limits and formulas. I didn't see any direct connection at all; the statement was a bound on the divisor function. I'd rather see all of the various RH-eqivalent statements written up in RH article directly; preferably at greater length even. (I mean, everything is connected to RH if you dig deep enough; I don't think its right to footnote everything to say "this is connected" ... almost every article I've edited on WP is a topic that is one-off from RH...) linas 20:49, 8 August 2005 (UTC)
I think the RH is much deeper and more complex than the harmonic numbers, and that the section on the RH equivalent is more relevent on the RH page itself. EulerGamma 23:16, 4 September 2006 (UTC)

Anyway, I edited out this:

"Note that n may be equal to \infty, provided m > 1.
And if m \le 1, while n=\infty, the harmonic series does not :converge and hence the harmonic number does not exist."

because infinity is not a real number, and thus is not a natural number. One must be very careful when saying "</math>n = \infty</math>" in any formal mathematical context. EulerGamma 23:16, 4 September 2006 (UTC)

Also, I see that it talks about a recurrence relation with Euler's Identity. This is called induction, and all the article shows is that if it is true for n, it is true for n+1. It fails to mention that it is true for n = 1 (even though this is quite obvious). I have an idea for how to show the identity (does anybody think it is okay for inclusion?):

H_n=\sum_{k=1}^n \frac{1}{k}
=\sum_{k=1}^n \int_0^1 x^{k-1} dx
=\int_0^1 \sum_{k=1}^n x^{k-1} dx
=\int_0^1 \frac{1-x^n}{1-x}dx

Another thing that I am thinking might be good for inclusion, is that (by the above method):

\sum_{k=1}^n \frac{a^k}{k} = \int_0^a \frac{1-x^n}{1-x}dx
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