Talk:Harmonic number

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Mathematics rating:
 C Class
 Mid Importance
Field:  Analysis

Riemann Hypothesis

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 ${\displaystyle n}$ may be equal to ${\displaystyle \infty }$, provided ${\displaystyle m>1}$.
And if ${\displaystyle m\leq 1}$, while ${\displaystyle n=\infty }$, 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 "[/itex]n = \infty[/itex]" 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?):

${\displaystyle H_{n}=\sum _{k=1}^{n}{\frac {1}{k}}}$
${\displaystyle =\sum _{k=1}^{n}\int _{0}^{1}x^{k-1}dx}$
${\displaystyle =\int _{0}^{1}\sum _{k=1}^{n}x^{k-1}dx}$
${\displaystyle =\int _{0}^{1}{\frac {1-x^{n}}{1-x}}dx}$

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

${\displaystyle \sum _{k=1}^{n}{\frac {a^{k}}{k}}=\int _{0}^{a}{\frac {1-x^{n}}{1-x}}dx}$

LaTeX issue

I can't get the first math line of the section "Special values for fractional arguments" to be rendered properly like the other lines. I don't know why. Chymicus (talk) 23:20, 10 April 2008 (UTC)

How to compute specials values of fractional argument

If ${\displaystyle n,m}$ are natural numbers, then:

${\displaystyle H_{\frac {m}{n}}={\frac {n}{m}}-\ln \left(n\right)+H_{m-1}+\sum \limits _{p=1}^{n-1}{e^{\frac {i2\pi pm}{n}}\left(\ln \left(1-e^{-{\frac {i2\pi p}{n}}}\right)+\sum \limits _{k=1}^{m-1}{\frac {e^{-k{\frac {i2\pi p}{n}}}}{k}}\right)}}$ .

It can be drived from the identity ${\displaystyle H_{x}=\sum \limits _{k=1}^{\infty }{{\frac {1}{k}}-{\frac {1}{k+x}}}}$

--77.125.151.16 (talk) 14:13, 3 January 2010 (UTC)

Error

The following statement does not make sense:

for every x > 0, integer or not. We have:
${\displaystyle \int _{0}^{1}H_{x}\,dx=\gamma ,}$
where γ is the Euler–Mascheroni constant

Since the harmonic numbers are monotonically growing this cannot be true. I don't know what the correct one is, but this is pretty certainly wrong. — Preceding unsigned comment added by 129.78.233.211 (talkcontribs)

I think you may have misunderstood what is meant here. The phrase "for every x > 0, integer or not" belongs to the infinite series definition of the continuous harmonic function on the previous line. The next line then says that the integral of this function between 0 and 1 is γ. I have modified the article to make this clearer. Gandalf61 (talk) 08:08, 20 May 2013 (UTC)

Um

Does anyone notice that ${\displaystyle H_{n}={\frac {1}{\frac {n\cdot (n+1)}{2}}}}$, Maybe that can be added. 190.60.93.218 (talk) 13:34, 1 August 2013 (UTC)

Not sure why you think that, but your formula fails for n = 2, when ${\displaystyle H_{2}=1+{\frac {1}{2}}={\frac {3}{2}}}$. Actually, you formula obviously fails because it is less than 1 for n > 1, whereas ${\displaystyle H_{n}}$ is greater than 1 for n > 1. Gandalf61 (talk) 13:41, 1 August 2013 (UTC)
I have made a terrible mistake, however I'm glad you've corrected me, after all I'm just learning. 190.60.93.218 (talk) 15:02, 1 August 2013 (UTC)

Sum of Harmonic series as an infinite series - original research?

This section concerning Aakash Praliya's derivation appears to be original research due to its wording and since the image of the derivation (see below) was uploaded by the user AakashPraliya2. Note that almost the same wording also appears on teh article Harmonic_series_(mathematics)

— Preceding unsigned comment added by Falcor84 (talkcontribs) 19:28, 12 October 2013 (UTC)