Talk:Harmonic series (mathematics)

message: can someone explain better why the harmonic series converges? I couldn't follow through in this explanation.
: Umm, but the harmoic series does not converge. As the article says, it diverges. Gandalf61 09:44, Feb 6, 2005 (UTC)
I learned the harmonic series is equal to the integral:
: H_n = \int_0^1 \frac{1-x^n}{1-x}\,dx 
as well as a proof based on geometric series (proven by Euler).
Do I need to put the proof in the article?
: I do not see the need for the proof, as the one in there now is quite easy (as proofs go) to understand.  I am not aware of the proof you refer to, off the top of my head, so if indeed it is a really neat one, perhaps you could add it.  But my default reaction is no.
: Are you asking whether to include the integral identity?  If so, I would say go ahead.  But note by saying this I am not vouching for its accuracy. 
: Baccyak4H 17:29, 17 August 2006 (UTC)
replies: text: Actually, I would suggest not to include the identity, since the page on harmonic numbers has it.  It is better there.  Baccyak4H 14:11, 24 August 2006 (UTC)
thread_title: Divergence proofs

message: Can someone restrict what a and b are supposed to be? I'd assume they can be any real number, but it isn't clear. M00npirate (talk) 18:54, 7 February 2009 (UTC)
replies: text: . You are right, so I added "real number" to the article. Should the article mention it's also true for complex numbers? --68.0.124.33 (talk) 15:42, 16 May 2009 (UTC)
thread_title:  General Harmonic series 

message: Most of the proofs in this article seem to assume that you can do things like group terms (maybe this works when all terms are positive? Counterexamples are things like 1 − 1 + 1 − 1 + · · ·) and reorder terms (I really don't know what assumption is needed to justify this).  In a historical sense, it took some centuries to sort this through, so it probably makes sense to speak of a "proof" which would not be accepted today, or would only be accepted today with elaboration.  But we need some kind of disclaimer or clarification (such as "both of these statements can be made precise and formally proven, but only using well-defined mathematical concepts that arose in the 19th century" and surrounding treatment at 1 − 1 + 1 − 1 + · · ·). Kingdon (talk) 01:24, 19 August 2009 (UTC)
replies: text: I don't think the article needs 5 proofs of the divergence in any case; it has the appearance of creep. So I removed the latter two, which were probably the ones you were concerned about anyway. — Carl (CBM · talk) 01:51, 19 August 2009 (UTC)
replies: text: Can we at least have a proof by integral test? (The French version has this.) This proof is not only short but, arguably. more conceptual since it shows the connection to log. -- Taku (talk) 12:10, 19 August 2009 (UTC)
replies: text: I didn't remove the sentence on that, although it is very brief. — Carl (CBM · talk) 21:48, 19 August 2009 (UTC)

text: The approach in such "rearrangement" proofs is to first of all assume that the series is convergent. If this is true then the terms of the series can be grouped or rearranged as we like without affecting the limit. However, if a rearrangement leads to a contradiction, then, by reductio ad absurdum, our original assumption must be incorrect. Therefore the series is not convergent. Not saying the rearrangement proof should be restored - just saying that it is rigorous. Gandalf61 (talk) 16:42, 19 August 2009 (UTC)
replies: text: Proofs like that would be more clear if they pointed out that the series is positive, so absolutely convergent if convergent, and thus allowing rearrangement if convergent. But I don't think that the focus of this article should be a long list of different proofs of nonconvergence of this series; it mentions 2 right now. — Carl (CBM · talk) 21:48, 19 August 2009 (UTC)
replies: text: Hmm, I missed that consequence of the reductio ad absurdum.  But I'm fine with deleting the extra proofs, as they did break up the flow of the article. Kingdon (talk) 11:32, 24 August 2009 (UTC)
thread_title:  playing it fast and loose in "proofs" 

message: If, as the article says, harmonic series are "divergent infinite series" then why include the p-series? Since when  we have a convergent infinite sequence: we have the famous formula
: \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \ .  
In fact, if my memory serves me correctly, all p-series with p ∈ R and p > 1 converge. (Actually ζ(x) → 1 as x → +∞) ~~ Dr Dec (Talk)   ~~ 13:18, 24 September 2009 (UTC)
replies: text: The inclusion seems appropriate to me since the p-series is a simple generalization that's close enough to the subject. A separate article would be unnecessary since anything additional to say would be in the Riemann zeta function article.--RDBury (talk) 12:57, 25 September 2009 (UTC)
replies: text: But this "inclusion" includes a set of objects which contradict the definition. Taking , then an open and dense set of p-series are not divergent, and so are not harmonic series, and so the inclusion is not valid. Either the definition, that harmonic series are "divergent infinite series", needs to be changed (which is clearly not a good idea) or the example needs to be re-worded. ~~ Dr Dec (Talk) ~~  23:04, 25 October 2009 (UTC)
replies: text: The harmonic series is one specific divergent series - the sum of the reciprocals of the positive integers. I have re-worded the p-series section to make it clear that the p-series are a generalisation of the harmonic series - the harmonic series is the special case of a p-series when p=1. Gandalf61 (talk) 09:46, 26 October 2009 (UTC)
thread_title:  ''p''-series 

message: A minor suggestion and it might just be based on how my brain works, but I find this...
\sum_{n=1}^{2^k} \,\frac{1}{n} \;>\; 1 + k(\frac{1}{2})
...easier than this...
\sum_{n=1}^{2^k} \,\frac{1}{n} \;>\; 1 + \frac{k}{2}
...when making the mental jump from the expanded series above to the summation notation.  —Preceding unsigned comment added by 207.164.58.11 (talk) 17:48, 29 June 2010 (UTC)
thread_title:  Comparison Test Formula 

message: If we apply the CC test then the convergence is equivalent to asking if 1+1+1+1..... converges, and obviously it doesn't.  Its "neat" so I thought maybe someone should add a section this.  — Preceding unsigned comment added by 99.149.190.128 (talk) 04:08, 13 September 2012 (UTC)
replies: text: Cauchy's convergence test?? We can even show partial sum is not Cauchy sequence. Will add it. 91.76.22.132 (talk) 19:30, 15 May 2020 (UTC)
thread_title:  Cauchy Criterian test 

message: A period after the equation ended the sentence early, and try as I might, I can't seem to change it to a [space]comma even though there is a space before the comma in the next equation, the one for the P-series.
HELLLP!
Also, is there a way to get math html to render in edit preview mode?  I wouldn;t have had to make so many damn wrong edits if I could have previewed before saving.
Dave Bowman - Discovery Won (talk) 12:35, 8 April 2013 (UTC)
thread_title:  General harmonic series: I give up! 

message: This page demonstrates that, viewed in the traditional light, the harmonic series is divergent. However, many other divergent series have what one might term 'creative' summations using alternative techniques. (See 1 − 1 + 1 − 1 + · · ·, 1 + 2 + 4 + 8 + · · ·, 1 + 1 + 1 + 1 + · · ·, 1 + 2 + 3 + 4 + · · ·, and so forth; and Cesàro summation, Ramanujan summation, and so forth.)
So what about the harmonic series? What alternative summation techniques have mathematicians applied to it? Doops |  talk 05:14, 15 January 2014 (UTC)
replies: text: Somewhere among these blog articles https://cornellmath.wordpress.com/2007/07/28/sum-divergent-series-i/ (in comments, IIRC) I've seen it being summed to the Euler-Mascheroni constant.--Alexmagnus (talk) 00:21, 31 October 2015 (UTC)
thread_title:  alternative summation? 

message: The section says "However, the total area under the curve y = 1 / x from 1 to infinity is given by an improper integral" which is stated to equal infinity . No justification is given for that(ie. proof of divergence). JDiala (talk) 03:53, 16 February 2014 (UTC)
thread_title:  Integral Test 

message: Assuming e^x\ge x+1 for all x, and without using calculus (no limits, differentiation, integration, or epsilon-delta — that is, only using stuff a precalc student could theoretically do), it is possible to prove the following:
\ln2-\frac1n<1-\frac12+\frac13-\dotsb\frac1n<\ln2+\frac1n
Then, the Squeeze Theorem immediately tells us that the alternating harmonic series converges to \ln2.
(I don't have time to provide a proof, but try substituting things like x=\ln\frac{100}{99} and x=\ln\frac{99}{100} into the above inequality and see what happens. Remember that 1-\frac12+\frac13-\dotsb\frac1{n}=\frac1{\lfloor n/2\rfloor+1}+\frac1{\lfloor n/2\rfloor+2}+\dotsb+\frac1{n}.)  — Preceding unsigned comment added by 66.108.7.224 (talk) 10:31, 24 March 2015 (UTC)
thread_title:  Alternating Harmonic Series, no calc 

message: A proof by contradiction, that uses the dominated convergence theorem in the  sequence space ℓ1 goes as follows. Consider the following sequence of unit-norm elements un of ℓ1:
:u_1:=\big(1,0,0,\dots\big)
:u_2:=\Big(\frac{1}{2},\frac{1}{2},0,0,\dots\Big)
:u_3:=\Big(\frac{1}{3},\frac{1}{3},\frac{1}{3},0,0,\dots\Big),
and so on. Clearly the sequence un converges point-wise to 0; moreover it is dominated by 
:h:=\Big(1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\dots\Big)
If the harmonic series were convergent, that is, if h were an element of ℓ1, by dominated convergence we would conclude 
:1=\sum_{k=1}^\infty u_{n,k} \to0 as n\to+\infty, a contradiction.
Note: I would prefer not to quote the sequence space ℓ1 at all, but just a dominated convergence theorem for series", again a quite elementary fact.
Let me know your opinions. pma 17:31, 25 July 2019 (UTC)
thread_title: Via dominated convergence (by contradiction)

message: Lets take  H_n=\sum_{k=1}^n \frac{1}{k}=1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots +\frac{1}{n}. Now we will show that this sequence is not Cauchy sequence, i.e., that \exists \varepsilon>0: \forall k\in \mathbb{N}\ \exists n>k,\exists p \in \mathbb{N}: \left \vert H_{n+p}-H_n \right \vert\geq \varepsilon. Lets subtract \left \vert H_{n+p}-H_n \right \vert=\frac{1}{n+1}+\cdots+\frac{1}{n+p}\geq\frac{1}{n+p}+\cdots+\frac{1}{n+p}=\frac{p}{n+p}. Let p\doteq n. Then \forall n \in \mathbb{N}: \left \vert H_{2n}-H_n \right \vert\geq \frac{1}{2}. Thus, this sequence is not Cauchy sequence and is divergent. By definition the series also diverges.
This is a very important result as it uses very simple math (the first or second theme in calculus) and the definition of series convergence. The integral test is dubious and requires rather good knowledge of practical applications and anyway they come after series usually or with them; Oresme test is not correct anyway unless you will define summation rules properly in your method (inserting zeros and rearranging). This comes much later. 2A00:1370:812C:DB93:4C5A:7358:27B4:89F (talk) 20:25, 15 May 2020 (UTC)
replies: text: A few points.  What you've added is written very poorly: bad English and bad math style (full of dense symbols that aren't necessary here).  In terms of what a reader is likely to have seen, basic series tests like direct comparison and the integral test are usually introduced earlier and to a much wider audience than the Cauchy criterion for convergence of sequences.  In spirit, this doesn't really offer any insight into what's going on beyond the direct comparison prooffinding runs which add up to at least 1/2 arbitrarily far into the series.  So it doesn't really add anything new by including it.  Coupled with the fact that the source isn't in English (which isn't automatically disallowed, but is discouraged when English sources are readily available), there are too many reasons not to include this. –Deacon Vorbis (carbon • videos) 21:39, 15 May 2020 (UTC)
replies: text: I added another proof that is used everywhere, telescoping proof. Cauchy criterion is introduced the same time here. LOL. It is very simple. You must show at least two proofs of Harmonic Sequencies on exams usually... Just saying. I will not find English source for you, I did WP:ONUS technically. Why are you reverting? 2A00:1370:812C:DB93:80CE:FB82:3BB2:B836 (talk) 21:59, 15 May 2020 (UTC)
replies: text: Also, "full of dense symbols" is just formal definition of non-Cauchy sequence, one can skip it. 2A00:1370:812C:DB93:80CE:FB82:3BB2:B836 (talk) 22:05, 15 May 2020 (UTC)
replies: text:  I agree with Deacon Vorbis. --JBL (talk) 22:23, 15 May 2020 (UTC)
replies: text: look into second proof then. Again this is just wrong, both "proofs" are garbage. 2A00:1FA0:46E1:941E:147F:5EC2:B1C:7931 (talk) 22:29, 15 May 2020 (UTC)
replies: text:  Please stop.  There are zillions of possible proofs of the divergence of the harmonic series.  Your second proof is cute; that does not mean it needs to be in the article, particularly uncited.  The Cauchy criterion proof is a pointlessly technical belaboring of the beautiful idea of the Oresme proof and does not add anything to an encyclopedic understanding of the harmonic series. --JBL (talk) 22:41, 15 May 2020 (UTC)
replies: text: "Cute" Really? The proof that is used in all textbooks here (at least in 6 of them) is just cute to you? Wow. Okay, if you want to have pathetic article for school so be it. 2A00:1370:812C:DB93:80CE:FB82:3BB2:B836 (talk) 23:15, 15 May 2020 (UTC)
thread_title:  Proof using the sequence of partial sums 

message: To be precise, the second sentence:  
::Proofs were given in the 17th century by Pietro Mengoli, Johann Bernoulli, and Jacob Bernoulli 
is slightly inaccurate, according to the given links: Mengoli's proof dates 17th century, but Johann Bernoulli's came in the middle of the 18th century, [no, his 1742 book was just published posthumous] and Jacob Bernoulli did not authored another proof, but just published his brother's. It could also be observed that the Bernoulli brothers were not aware of the preceding proofs, given that Jacob writes "My brother first discovered this.." pma 21:14, 12 October 2020 (UTC)
replies: text: Bernoulli's argument is also nice; since it is not easy to find it in the linked Latin text, I'll report it in the footnote, like Mengoli's, and see as it looks like.pma 11:43, 13 October 2020 (UTC)
thread_title: History

message: Reference 2, Kullman (2001), is cited to support the claim that Johann Bernoulli proved the series' divergence, but it only mentions Jacob Bernoulli (under the name Jacques Bernoulli).
Reference 14, Havil (2003), appears not to actually use the term harmonic number — it just uses H_n, in all the instances I can find. "No property is more unexpected than H_n's divergence, and it is this that Oresme proved", etc. I'm actually a bit puzzled by this, since the terminology is so well established.
In the "Crossing a desert" example, leuca is undefined, though in context it's clearly a measure of distance. The source says that one leuca is 1500 double paces, or about 1.5 miles. Perhaps this should be worked into the text. XOR'easter (talk) 01:07, 9 March 2022 (UTC)
replies: text: The statement looks right, but the source provided does not include a specific mention of Mertens' theorems. Maybe it should be supplemented? XOR'easter (talk) 18:35, 12 March 2022 (UTC)
thread_title:  Comments 

message: The link here: P series links to this page, but this page doesn't mention p-series anymore. DeathOfBalance (talk) 23:24, 31 July 2022 (UTC)
replies: text: Convergence_tests#p-series_test also points here. 66.44.49.56 (talk) 02:18, 1 August 2022 (UTC)

text: So it happened in this edit by , part of the process that led to this article reaching good article status, which reframed it as a section about the Riemann zeta function.  As far as I can tell, the only remaining discussion of the "p-series test" (a standard, if not terribly important, part of single-variable calculus courses) in Wikipedia is in the article Convergence tests.  It seems to me like that is the natural home for content related to the convergence test.  I'll re-target the link at P series. JBL (talk) 18:54, 1 August 2022 (UTC)
thread_title:  p-series links here but is gone. 

message: Hi,
Sorry if this is not how to do it - it's my first time to suggest something on Wikipedia, and I don't want to mess up an article, especially one with a "good article" designation, by "just editing"!  But here's a concern: the article, at the very beginning, says:
"The first n terms of the series sum to approximately ln n + gamma, where ln is the natural logarithm and gamma is the Euler–Mascheroni constant."
However, I think this is wrong (easy to see with Excel). Instead, ln(n) + 1/(2n) + gamma, see:
https://mathschallenge.net/full/harmonic_sum_approximation
Also, I think it's confusing to use "n" here because "n" is the variable in the 1/n sum term above, I think it would be better to use, e.g., "k" in this statement. 129.240.66.34 (talk) 10:21, 17 March 2024 (UTC)
replies: text: The lead says . This is correct, since the difference between the sum of the  first terms and \ln n + \gamma is less than 1/n and tends to zero when n tends to infinity. However, the term  is important in some contexts, since it reduces the approximation error from less than  to less than , which is much smaller. See . D.Lazard (talk) 11:05, 17 March 2024 (UTC)
thread_title:  Approximation