Talk:Harmonic number

message: 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)
thread_title:  LaTeX issue 

message: If n,m are natural numbers, then:
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 
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)
thread_title:  How to compute specials values of fractional argument 

message: The following statement does not make sense:
:for every x > 0, integer or not. We have:  
:: \int_0^1H_{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 (talk • contribs) 
: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)
thread_title:  Error 

message: Does anyone notice that 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)
replies: text: Not sure why you think that, but your formula fails for n = 2, when H_2 = 1 + \frac{1}{2} = \frac{3}{2}. Actually, you formula obviously fails because it is less than 1 for n > 1, whereas H_n is greater than 1 for n > 1. Gandalf61 (talk) 13:41, 1 August 2013 (UTC)
replies: text: 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)
thread_title:  Um 

message: 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 (talk • contribs) 19:28, 12 October 2013 (UTC)
thread_title:  Sum of Harmonic series as an infinite series - original research? 

message: The article claims that the harmonic numbers are related to the Stirling numbers as:
:  H_n = \frac{1}{n!}\left[{n+1 \atop 2}\right]. 
The first few terms of \left[{n+1 \atop 2}\right] are
: 1, 3, 6, 10, 15, 21, 28, 36, 45
and the first few terms of n! are
: 1, 2, 6, 24, 120, 720, 5040, 40320, 362880
The product of the two tends quickly toward zero, which is not what H_n does.  Maybe the square brackets are being used to indicate something other than "combinations?" --LesPaul75talk 22:50, 9 March 2019 (UTC)
replies: text: In the table at Stirling numbers of the first kind, the first few values of \textstyle\left[{n + 1 \atop 2}\right] are 1, 3, 11, 50, 274, ..., which agrees with the statement in the article.  –Deacon Vorbis (carbon • videos) 23:07, 9 March 2019 (UTC)
replies: text: Yes; see for instance here: \textstyle \left[{n \atop k}\right] is the number of permutations of n elements with exactly k cycles, called the (unsigned) Stirling number of the first kind. Boris Tsirelson (talk) 23:11, 9 March 2019 (UTC)
thread_title:  Stirling numbers, combinations 

message: The statement that x(\log(x)-1) is an integral of \log(x) is a routine calculation for a first semester calculus student and should not need a citation.Johnwaylandbales (talk) 16:07, 16 September 2019 (UTC)
thread_title:  The statement that <math>x(\log(x)-1)</math> is an integral of <math>\log(x)</math> does not need a citation.