# Talk:Harmonic series (mathematics)

can someone explain better why the harmonic series converges? I couldn't follow through in this explanation. — Preceding unsigned comment added by 141.44.163.16 (talk) 23:19, 4 February 2005 (UTC)

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
          
        
        =
        
          ∫
          
            0
          
          
            1
          
        
        
          
            
              1
              −
              
                x
                
                  n
                
              
            
            
              1
              −
              x
            
          
        
        
        d
        x
      
    
    {\displaystyle 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? — Preceding unsigned comment added by Scythe33 (talk • contribs) 21:35, 4 June 2005 (UTC)

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)
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)
Is there a reason that Jakob Bernoulli's 1689 proof isn't cited? It's rather clever and generally regarded as the standard early proof; I hadn't heard of the medieval guy's proof before, though. — Preceding unsigned comment added by 69.114.7.161 (talk) 05:51, 11 June 2007 (UTC)

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)

 Done. 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)

1   1   3     3   1   11     11   1   50     50   1   275     275   1   1770          ∞
- + - = -     - + - = --     -- + - = --     -- + - = ---     --- + - = ----        = -
1   2   2     2   3   6      6    4   24     24   5   120     120   6   720 ...       ∞

Robo37 (talk) 19:20, 16 June 2009 (UTC)

Maybe, and maybe not.  See limit.  Baccyak4H (Yak!) 19:22, 16 June 2009 (UTC)
More precisely: in some cases, when the numerator and denominator both diverge to infinity, the fraction itself diverges to infinity.  In some such cases, the fraction converges to a finite number.  In some such cases, the fraction oscillates or otherwise fails to converge without diverging to infinity.

Robo37, why do you write 50/24 instead of 25/12, and 275/120 instead of 55/24, and 1770/720 instead of 59/24?  If you write the fractions in lowest terms, it's not so clear that the numerator and denominator diverge to infinity.  If the two denominators are 6 and 4, then the least common denominator is 12, not 24.  If they are 120 and 6, then it is 120, not 720.

Baccyak4H, when you use the word "maybe", it makes it sound as if you don't know the answer. Michael Hardy (talk) 19:40, 16 June 2009 (UTC)

Hmm, it does.  Of course, I did know the answer ("maybe" referred to the section heading alone; another such series might not make the header "true").  Baccyak4H (Yak!) 04:28, 17 June 2009 (UTC)
I put 50/24 instead of 25/12 275/120 instead of 55/24 and 1770/720 instead of 59/24 simply because multiplying the denominators together was the easiest thing to do. Changing the fractions make little difference, since you are still left with:
3    11    25    55    59    437    ∞
-    --    --    --    --    ---    -
2... 6 ... 12... 24... 24... 168... ∞

Thinking about it, the same method can be used to prove that ∞/∞ is any positive number.
                     9     9     9       9       9         9     9   99     9    999      9
1 = 0.999999999... = -- + --- + ---- + ----- + ------    = -- + ---  --- + ----  ---- + -----
                     10   100   1000   10000   100000...   10   100, 100   1000, 1000   10000,

9999      9     99999       ∞
----- + ------  ------    = -
10000   100000, 100000...   ∞

                     1   9     9     9       9       9         1   9   19    9   199    9
2 = 1.999999999... = - + -- + --- + ---- + ----- + ------    = - + --  -- + ---  --- + ----
                     1   10   100   1000   10000   100000...   1   10, 10   100, 100   1000,

1999     9    19999      ∞
---- + -----  -----    = -
1000   10000, 10000...   ∞

1                          4     9     9       9       9         4     9   49     9    499
- = 0.5 = 0.499999999... = -- + --- + ---- + ----- + ------    = -- + ---  --- + ----  ---- +
2                          10   100   1000   10000   100000...   10   100, 100   1000, 1000

  9    4999      9     49999       ∞
-----  ----- + ------  ------    = -
10000, 10000   100000, 100000...   ∞

1                           2     4     9       9       9         2     4   24     9    249
- = 0.25 = 0.249999999... = -- + --- + ---- + ----- + ------    = -- + ---  --- + ----  ---- +
4                           10   100   1000   10000   100000...   10   100, 100   1000, 1000

  9    2499      9     24999       ∞
-----  ----- + ------  ------    = -
10000, 10000   100000, 100000...   ∞

1                 3     3     3       3       3         3     3   33     3    333
- = 0.333333333 = -- + --- + ---- + ----- + ------    = -- + ---  --- + ----  ---- +
3                 10   100   1000   10000   100000...   10   100, 100   1000, 1000

  3    3333      3     33333       ∞
-----  ----- + ------  ------    = -
10000, 10000   100000, 100000...   ∞

1                 1     6     6       6       6         1     6   16     6    166
- = 0.166666666 = -- + --- + ---- + ----- + ------    = -- + ---  --- + ----  ---- +
6                 10   100   1000   10000   100000...   10   100, 100   1000, 1000

  6    1666      6     16666       ∞
-----  ----- + ------  ------    = -
10000, 10000   100000, 100000...   ∞

1                 1     4     2       8       5         1     4   14     2    142
- = 0.142857142 = -- + --- + ---- + ----- + ------    = -- + ---  --- + ----  ---- +
7                 10   100   1000   10000   100000...   10   100, 100   1000, 1000

  8    1428      5     14285       ∞
-----  ----- + ------  ------    = -
10000, 10000   100000, 100000...   ∞

                  3    1     4     1       5     3   1   31    4   314
π = 3.141592654 = - +  -- + --- + ---- + ----- = - + --  -- + ---  --- +
                  1    10   100   1000   10000   1   10, 10   100, 100

  1    3141      5     31415       ∞
-----  ----- + ------  ------    = -
10000, 10000   100000, 100000...   ∞

Robo37 (talk) 20:03, 17 June 2009 (UTC)

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)

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)
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)
I didn't remove the sentence on that, although it is very brief. — Carl (CBM · talk) 21:48, 19 August 2009 (UTC)
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)
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)
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)

If, as the article says, harmonic series are "divergent infinite series" then why include the p-series? Since when p = 2 we have a convergent infinite sequence: we have the famous formula

  
    
      
        
          ∑
          
            n
            =
            1
          
          
            ∞
          
        
        
          
            1
            
              n
              
                2
              
            
          
        
        =
        
          
            
              π
              
                2
              
            
            6
          
        
         
        .
      
    
    {\displaystyle \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)

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)
But this "inclusion" includes a set of objects which contradict the definition. Taking p ∈ R, 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)
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)

I have thought about this statement and calculated that it is way off. If we imagine the sum of the sequence 1 + (1/2) + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8)... we can see: each group of terms (in brackets) is equal to 1/2. In every next group, the number of terms is twice as big (because each term is halved). If we have n groups, then to find the number of terms, we have to add 1 + 2 + 4 + 8... + 2n - 1, which is a geometric progression with a common ratio of 2. We will start the addition from the other end, to make it a progression just like 1 + 1/2 + 1/4 + 1/8... = 2. So, for very large n, the number of terms will be about twice the number of groups. So, if we have 1043 terms, then we will have 5 × 1042 groups, each of which equals to 1/2. Thus this sum will equal to about 2.5 × 1042, which is unimaginably far larger than 100! And knowing that the harmonic series grows faster than the series here (which was used to prove the divergence of the harmonic series), this means that the sum of that many terms of the harmonic series will be ever bigger.Majopius (talk) 02:20, 6 February 2010 (UTC)

Sorry—you've mangled your math.  If you have n groups, you've got the sum > n×(1/2).  But how many terms does it take to make n groups?  If you have 1043 terms, then how many groups do you have, each exceeding 1/2?

If you have n groups, then you have 2n − 1 terms.  If you have 200 groups (so that the sum exceeds 100), then you have 2200 − 1 terms, or about 1.6 × 1060.

Now let's try something less crude:

  
    
      
        
          
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    {\displaystyle {1 \over 2}+{1 \over 3}+{1 \over 4}+\cdots +{1 \over n}<\int _{1}^{n}{dx \over x}<1+{1 \over 2}+{1 \over 3}+{1 \over 4}+\cdots +{1 \over n-1}}
  

and

  
    
      
        
          ∫
          
            1
          
          
            n
          
        
        
          
            
              d
              x
            
            x
          
        
        =
        
          log
          
            e
          
        
        ⁡
        n
        =
        
          
            
              
                log
                
                  10
                
              
              ⁡
              n
            
            
              
                log
                
                  10
                
              
              ⁡
              e
            
          
        
        .
      
    
    {\displaystyle \int _{1}^{n}{dx \over x}=\log _{e}n={\log _{10}n \over \log _{10}e}.}
  

If n = 1043 then this is 43/log10 e = about 99.

So: Yes, really. The article's claim is correct. Michael Hardy (talk) 05:02, 6 February 2010 (UTC)
I did not mangle anything! I was attentive when thinking about it, and got this result. Oh well, its a paradox.Majopius (talk) 18:33, 6 February 2010 (UTC)
You said "...the number of terms will be about twice the number of groups". Actually, the number of terms is about two to the power of the number of groups. Gandalf61 (talk) 14:36, 9 February 2011 (UTC)
For primes  Sum(1/p)  already diverges: is this worth mentioning?195.96.229.83 (talk) 11:33, 9 February 2011 (UTC)

A minor suggestion and it might just be based on how my brain works, but I find this...

  
    
      
        
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            n
            =
            1
          
          
            
              2
              
                k
              
            
          
        
        
        
          
            1
            n
          
        
        
        >
        
        1
        +
        k
        (
        
          
            1
            2
          
        
        )
      
    
    {\displaystyle \sum _{n=1}^{2^{k}}\,{\frac {1}{n}}\;>\;1+k({\frac {1}{2}})}
  

...easier than this...

  
    
      
        
          ∑
          
            n
            =
            1
          
          
            
              2
              
                k
              
            
          
        
        
        
          
            1
            n
          
        
        
        >
        
        1
        +
        
          
            k
            2
          
        
      
    
    {\displaystyle \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) 

Shouldn't the inequality be a weak inequality? After all, the two sides are equal at k = 1.  — Preceding unsigned comment added by 134.10.12.152 (talk) 22:13, 8 June 2011 (UTC) 

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) 

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)

The article currently defines the general harmonic series as:

  
    
      
        
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            n
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            0
          
          
            ∞
          
        
        
          
            1
            
              a
              n
              +
              b
            
          
        
        .
        
      
    
    {\displaystyle \sum _{n=0}^{\infty }{\frac {1}{an+b}}.\!}
  

Couldn't the 0 be replaced with an additional variable, such as 
  
    
      
        c
      
    
    {\displaystyle c}
  
? After all, I'm pretty sure you can always remove a finite number of terms from the start of a divergent series and it has to remain divergent. ± Lenoxus (" *** ") 15:04, 20 March 2013 (UTC)

Yes, but

  
    
      
        
          ∑
          
            n
            =
            c
          
          
            ∞
          
        
        
          
            1
            
              a
              n
              +
              b
            
          
        
        =
        
          ∑
          
            n
            =
            0
          
          
            ∞
          
        
        
          
            1
            
              a
              (
              n
              +
              c
              )
              +
              b
            
          
        
        =
        
          ∑
          
            n
            =
            0
          
          
            ∞
          
        
        
          
            1
            
              a
              n
              +
              d
            
          
        
        .
        
      
    
    {\displaystyle \sum _{n=c}^{\infty }{\frac {1}{an+b}}=\sum _{n=0}^{\infty }{\frac {1}{a(n+c)+b}}=\sum _{n=0}^{\infty }{\frac {1}{an+d}}.\!}
  

where d = ac + b. So if you start the summation at c it is still part of the same family. And, as you say, still a divergent series. Gandalf61 (talk) 15:13, 20 March 2013 (UTC)

Ah, very nice. We can change the 0 but may as well change the b. Thanks. ± Lenoxus (" *** ") 16:55, 20 March 2013 (UTC)

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)

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)

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)

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)

Assuming 
  
    
      
        
          e
          
            x
          
        
        ≥
        x
        +
        1
      
    
    {\displaystyle e^{x}\geq 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:

  
    
      
        ln
        ⁡
        2
        −
        
          
            1
            n
          
        
        <
        1
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            1
            2
          
        
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            n
          
        
        <
        ln
        ⁡
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            1
            n
          
        
      
    
    {\displaystyle \ln 2-{\frac {1}{n}}<1-{\frac {1}{2}}+{\frac {1}{3}}-\dotsb {\frac {1}{n}}<\ln 2+{\frac {1}{n}}}
  

Then, the Squeeze Theorem immediately tells us that the alternating harmonic series converges to 
  
    
      
        ln
        ⁡
        2
      
    
    {\displaystyle \ln 2}
  
.
(I don't have time to provide a proof, but try substituting things like 
  
    
      
        x
        =
        ln
        ⁡
        
          
            100
            99
          
        
      
    
    {\displaystyle x=\ln {\frac {100}{99}}}
  
 and 
  
    
      
        x
        =
        ln
        ⁡
        
          
            99
            100
          
        
      
    
    {\displaystyle x=\ln {\frac {99}{100}}}
  
 into the above inequality and see what happens. Remember that 
  
    
      
        1
        −
        
          
            1
            2
          
        
        +
        
          
            1
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        −
        ⋯
        
          
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        =
        
          
            1
            
              ⌊
              n
              
                /
              
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              ⌋
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    {\displaystyle 1-{\frac {1}{2}}+{\frac {1}{3}}-\dotsb {\frac {1}{n}}={\frac {1}{\lfloor n/2\rfloor +1}}+{\frac {1}{\lfloor n/2\rfloor +2}}+\dotsb +{\frac {1}{n}}}
  
.)  — Preceding unsigned comment added by 66.108.7.224 (talk) 10:31, 24 March 2015 (UTC) 

I thought it would not be bad to have another few proofs of the divergence, maybe not in full details as the two in the current version. The current article is nice, but (IMHO) it is still too close to the standard tradition of calculus textbooks.
I added two more proofs, that were deleted -- for various objections mentioned on the history page; these may have some good reasons, but none of them seems really strong). In particular, this material is definitely not "original reserch" (recall the "Paris is the capital of France" example). Instead it is some old arguments maybe not universally known: but the point we should discuss is not if they are well-known enough to enter an encyclopedia, but if they are worth. 
My proposal is to leave were it is the "comparison proof", and move in a last section named "other proofs" the "integral test" together with the other two I provided. I copy them here below.
The scope of this last section should not be to add strength to the proof (one proof is enough), but instead to show connections with other tools and ideas in mathematics.  


### Comparison II
Whether the  harmonic series is finite or not,  we can write it as a sum of two series:  

  
    
      
        
          
            
              
                
                  
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    {\displaystyle {\begin{aligned}{\bigg (}1&+{\frac {1}{3}}+{\frac {1}{5}}+{\frac {1}{7}}+{\frac {1}{9}}+\cdots {\bigg )}+\\{\bigg (}{\frac {1}{2}}&+{\frac {1}{4}}+{\frac {1}{6}}+{\frac {1}{8}}+{\frac {1}{10}}+\cdots {\bigg )}=\\1&+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+\dots \end{aligned}}}
  

The terms of the second series are strictly less than the corresponding terms of the first one, so the sum of the second series is strictly less than the sum of the first one, provided the sum is finite. But in fact these two sums are equal: indeed, the second series, thus the first one too, is exactly one half of the harmonic series. Since the sum can't be finite we conclude it is infinite.
Note: one main objection was "It relies on results about rearranging series that are much more sophisticated than what is being proven": this is quite wrong (there is no rearrangement, which by the way for positive series is almost as elementary as the notion itself of sum of a positive series, but less more that term-wise sum of series)
In a simpler version:


### Comparison II -bis
Majorize each odd term oh the harmonic series with the following even term:

  
    
      
        
          
            
              
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                    2
                  
                
                
                  
                    )
                  
                
                +
                
                  
                    (
                  
                
                
                  
                    1
                    4
                  
                
                +
                
                  
                    1
                    4
                  
                
                
                  
                    )
                  
                
                +
                
                  
                    (
                  
                
                
                  
                    1
                    6
                  
                
                +
                
                  
                    1
                    6
                  
                
                
                  
                    )
                  
                
                ⋯
                =
              
            
            
              
                1
              
              
                
                +
                
                  
                    1
                    2
                  
                
                +
                
                  
                    1
                    3
                  
                
                +
                
                  
                    1
                    4
                  
                
                +
                
                  
                    1
                    5
                  
                
                +
                
                  
                    1
                    6
                  
                
                …
              
            
          
        
      
    
    {\displaystyle {\begin{aligned}1&+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+{\frac {1}{6}}\cdots \geq \\{\frac {1}{2}}&+{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{4}}+{\frac {1}{6}}+{\frac {1}{6}}\cdots =\\{\bigg (}{\frac {1}{2}}&+{\frac {1}{2}}{\bigg )}+{\bigg (}{\frac {1}{4}}+{\frac {1}{4}}{\bigg )}+{\bigg (}{\frac {1}{6}}+{\frac {1}{6}}{\bigg )}\cdots =\\1&+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+{\frac {1}{6}}\dots \end{aligned}}}
  

However, if the harmonic sum where finite, the above inequality would be strict, which is impossible, therefore the sum is infinite.
Note: grouping the pairs of consecutive equal terms in the second sum just corresponds to taking a subsequence of the sequence of partial sums, which should remove any fear about rearrangements that may confuse the freshman


### Via dominated convergence (by contradiction)
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
          
        
        :=
        
          
            (
          
        
        1
        ,
        0
        ,
        0
        ,
        …
        
          
            )
          
        
      
    
    {\displaystyle u_{1}:={\big (}1,0,0,\dots {\big )}}
  

  
    
      
        
          u
          
            2
          
        
        :=
        
          
            (
          
        
        
          
            1
            2
          
        
        ,
        
          
            1
            2
          
        
        ,
        0
        ,
        0
        ,
        …
        
          
            )
          
        
      
    
    {\displaystyle u_{2}:={\Big (}{\frac {1}{2}},{\frac {1}{2}},0,0,\dots {\Big )}}
  

  
    
      
        
          u
          
            3
          
        
        :=
        
          
            (
          
        
        
          
            1
            3
          
        
        ,
        
          
            1
            3
          
        
        ,
        
          
            1
            3
          
        
        ,
        0
        ,
        0
        ,
        …
        
          
            )
          
        
        ,
      
    
    {\displaystyle 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
        :=
        
          
            (
          
        
        1
        ,
        
          
            1
            2
          
        
        ,
        
          
            1
            3
          
        
        ,
        
          
            1
            4
          
        
        ,
        …
        
          
            )
          
        
      
    
    {\displaystyle 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
        =
        
          ∑
          
            k
            =
            1
          
          
            ∞
          
        
        
          u
          
            n
            ,
            k
          
        
        →
        0
      
    
    {\displaystyle 1=\sum _{k=1}^{\infty }u_{n,k}\to 0}
  
 as 
  
    
      
        n
        →
        +
        ∞
      
    
    {\displaystyle 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)


### Discussion
In MOS:MATH#proofs, one can read Include proofs when they expose or illuminate the concept or idea; don't include them when they serve only to establish the correctness of a result. This means that it is not useful to provide all existing proofs. The first proof that is given here is probably the simplest possible (the is the one that requires the minimal mathematical background), and deserve thus to be there. The second one is also useful, as appearing in many textbooks, and is a good introduction on the strong relation between harmonic series and logarithm.
For adding other proofs, it does not suffice that they exist. One must explain how they "illuminate" the subject of the article. Such an explanation has not been given so far. Moreover both involve concepts that a reader is not supposed to know at this part of the article. The new proposed proof (Comparison 2) is also more complicated of the first previous first proof, as needing to understand that x + y = x and y > 0 imply that x is infinite.
For these reasons, I oppose to the addition of any of these new proofs. D.Lazard (talk) 11:54, 26 July 2019 (UTC)

Thank you D.Lazard. Let me distinguish, because I partially agree. I personally warmly agree that wikipedia should not include a bunch of detailed proofs. Rather, what is important, in the very etymological sense, is to give the information that a certain proof was made (if it is worth), where or how to find the details, and the connections with other ideas. In particular, these articles should not mimic a calculus textbook (btw: IMHO the style of the current version goes a bit too much in that direction, but sure I don't want to add to the controversy). The idea I had in mind is to quote and briefly describe the relevant existing proofs in a last section, not in full details, possibly with the coordinates and links to fill them up, and possibly with a source (unfortunately, at the moment we have it only for the first one; I guess a successful search would bring us to to Euler or Bernoulli). To make an example: several proofs are listed for the Pythagorean theorem, for the fundamental theorem of algebra, for the Euclid's theorem on the infinitude of primes &c. (You'll say: but these theorems are quite more important. I'll answer: true, yet the harmonic series also has its place in the development of analysis, and deserves its list of proofs).pma 20:59, 26 July 2019 (UTC)
In particular, I would give much more emphasis to the historical proofs already mentioned, at the moment almost hidden in notes pma 22:58, 26 July 2019 (UTC)
I agree with D.Lazard.  In addition to the objections he raises, all evidence suggests you came up with these proofs yourself, which is a clear-cut violation of WP:OR. --JBL (talk) 20:59, 27 July 2019 (UTC)
Do not trust too much what all evidence suggests to you, for she often just likes to make fun of simpletons ;) pma 22:02, 12 October 2020 (UTC)
I'm also obliged to respond to the request One must explain how they "illuminate" the subject of the article. I need then to enlarge the discussion. let's talk of Applications: — Preceding unsigned comment added by PMajer (talk • contribs) 
"first proof that is given here is probably the simplest possible" This is insanity. It is incorrect per se, we cannot put the brackets as we want (see comments "playing it fast and loose in "proofs"" above). It is also from Mid Ages. "The second one is also useful" Really? Normal math texts cannot use this proof, OMG. Using integrals to prove elementary series problem? Wow. No, everybody uses direct proof that this is not Cauchy sequence or the telescopic proof. I will add the second then. 91.76.22.132 (talk) 21:08, 15 May 2020 (UTC)
The first given proof, which, as far as we know, is also the first one in history, is indeed the simplest, and it is perfectly correct. pma 22:02, 12 October 2020 (UTC)
No, it is formally incorrect. Read Cesaro summation and Riemann series theorem. I will quote Riemann for you: "if an infinite series of real numbers is conditionally convergent, then its terms can be arranged in a permutation so that the new series converges to an arbitrary real number, or diverges [or partial sums limit will not exist at all]". And it is not obvious that for positive series we do not care about this. THAT also requires a proof. 109.252.90.66 (talk) 08:22, 16 February 2021 (UTC)
That no series of positive terms is conditionally convergent is obvious from the definition, and this kind of bracketing is not a "rearrangement" in the sense of RST (precisely: for any convergent series (conditionally or otherwise), if you bracket terms into consecutive blocks to form a new series, then the new series also converges, to the same limit).  Everything in this proof is completely correct. --JBL (talk) 11:54, 16 February 2021 (UTC)

Indeed this section has a big question mark on it. The applications given are certainly nice stories, a couple of them I think taken from Martin Gardner; we may call them metaphores, or exampla ficta, made to illustrate the facts. The worm on the rubber band is a nice little story, suitable to visualize the mathematical phenomenon, and gives to it some funny consistency and Lewis Carrol's flavor. So far, so good. But calling these stories applications is terribly misleading and diseducative. Even the length of this sections, the abundance of details, and the emphatic position it has at the beginning, are misleading. The important facts, that such an article must recall and stress, are simple, basic and powerful ideas, their genesis and their history. An easy argument like 
  
    
      
        x
        +
        1
        =
        x
      
    
    {\displaystyle x+1=x}
  
 implies 
  
    
      
        x
        =
        ±
        ∞
      
    
    {\displaystyle x=\pm \infty }
  
, in its simplicity (formally it's no more than 
  
    
      
        x
        +
        1
        >
        x
      
    
    {\displaystyle x+1>x}
  
 for any real 
  
    
      
        x
      
    
    {\displaystyle x}
  
) is fundamental in the whole treatment of the analysis of infinite. It is indeed the key point both in Bernoulli's and in Mengoli's proofs (mentioned, but hidden, in the article), and in its variation I reported. Instead of asking if a series like this converges or diverges, we give in any case a sense to its sum 
  
    
      
        S
      
    
    {\displaystyle S}
  
, and then we use an extended set of algebraic and order rules to deduce 
  
    
      
        S
        =
        +
        ∞
      
    
    {\displaystyle S=+\infty }
  
. It is the same powerful line of reasoning that pervades the development of all mathematics. I guess that the reader that find complicated 
  
    
      
        (
        x
        +
        y
        =
        x
        ,
        y
        >
        0
        )
      
    
    {\displaystyle (x+y=x,y>0)}
  
  implies 
  
    
      
        x
        =
        ±
        ∞
      
    
    {\displaystyle x=\pm \infty }
  
 will also very easily believe that mathematics is about studying crazy situations involving somebody swimming in a pool for a time larger than the age of the universe, at a speed larger than light's.  This is very bad. No surprise they close maths departments. 
Here is a tentative proposal for this section: the title of this section may be changed into Exemples or Harmonic series stories or Harmonic series in pop culture, but Applications really doesn't go. I'll try and change it into Examples; maybe you have a better idea. (I also think it should also better to shorten it and put in the bottom of the article, for the reason I explained, and expand the initial part were the various proofs are mentioned, but this has to be discussed carefully).  pma 19:01, 27 July 2019 (UTC)

I strongly dislike the section title "Examples"; "Applications" was far more accurate.  Moving it lower in the article would be fine.  --JBL (talk) 21:02, 27 July 2019 (UTC)
I agree with Joel B. Lewis: The object of the section is "Examples of use", and this is exactly what is an application. D.Lazard (talk) 08:56, 28 July 2019 (UTC)
Dear users, as I wrote, "Examples" was a tentative. I'm confident you'll find something better: please make an effort. "Applications" is inaccurate and misleading as I tried to explain to you. An "application of X" is something more than just an "example of use of X": it is an example of use of X, that motivates the introduction/creation of X. However, may I say, it seems to me that you are taking this thing a bit too much on the personal side, and this is an obstruction to accept improvements. So I will not make further edits, till this discussion reaches some conclusion. pma 21:14, 30 July 2019 (UTC)
An "application of X" is something more than just an "example of use of X": it is an example of use of X, that motivates the introduction/creation of X. This may be your belief, but it strikes me as idiosyncratic.  I also see no indication that anyone is taking anything personally. --JBL (talk) 21:55, 30 July 2019 (UTC)
I do believe something struck you. I said this article lacks serious and motivating applications, and instead sells silly stories, as applications. Which is misleading. You may call it idiosyncratic: but giving motivating applications is how good divulgation (and teaching) works. Let me repeat the question: How is that you guys are afraid that a reader may find it too difficult to understand "
  
    
      
        x
        +
        1
        =
        x
      
    
    {\displaystyle x+1=x}
  
 implies 
  
    
      
        x
        =
        ±
        ∞
      
    
    {\displaystyle x=\pm \infty }
  
", but you are not afraid that the same person may believe that the study of divergent series is well motivated by the application to silly stories? Or maybe you guys  are really convinced that divergent series are aimed to this. The model of a  hypothetical guy swimming on and off till the end of the universe, and possibly even after.
PS: What about (no irony): "Application to riddles"? At least it's accurate. pma 23:12, 30 July 2019 (UTC)
Well, I now agree with you that someone seems to be taking this personally.  Unfortunately I don't have anything constructive to say in response to your comments, since they do not seem to be responsive to anything anyone else has written. (I am a bit surprised you haven't taken advantage of the evident consensus to move the Applications section lower in the article, as you proposed; perhaps I shall do it myself.) --JBL (talk) 01:44, 31 July 2019 (UTC)
No need to be surprised - I'd like to have a constructive discussion, in order to improve the thing, and I believe my suggestions may help. There will be time for moving/changing/adding material when (if) we reach a consensus. pma 19:16, 31 July 2019 (UTC)
"Discussion". Attic Salt (talk) 23:25, 30 July 2019 (UTC)

@Attic Salt: Is this a proposed alternate title for the section? Or something else? --JBL (talk) 01:48, 31 July 2019 (UTC)
Yes. Alternative title for miscellaneous section. Attic Salt (talk) 02:00, 31 July 2019 (UTC)

Lets take 
  
    
      
        
          H
          
            n
          
        
        =
        
          ∑
          
            k
            =
            1
          
          
            n
          
        
        
          
            1
            k
          
        
        =
        1
        +
        
          
            1
            2
          
        
        +
        
          
            1
            3
          
        
        +
        
          
            1
            4
          
        
        +
        ⋯
        +
        
          
            1
            n
          
        
        .
      
    
    {\displaystyle 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 
  
    
      
        ∃
        ε
        >
        0
        :
        ∀
        k
        ∈
        
          N
        
         
        ∃
        n
        >
        k
        ,
        ∃
        p
        ∈
        
          N
        
        :
        
          |
          
            
              H
              
                n
                +
                p
              
            
            −
            
              H
              
                n
              
            
          
          |
        
        ≥
        ε
        .
      
    
    {\displaystyle \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 
  
    
      
        
          |
          
            
              H
              
                n
                +
                p
              
            
            −
            
              H
              
                n
              
            
          
          |
        
        =
        
          
            1
            
              n
              +
              1
            
          
        
        +
        ⋯
        +
        
          
            1
            
              n
              +
              p
            
          
        
        ≥
        
          
            1
            
              n
              +
              p
            
          
        
        +
        ⋯
        +
        
          
            1
            
              n
              +
              p
            
          
        
        =
        
          
            p
            
              n
              +
              p
            
          
        
        .
      
    
    {\displaystyle \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
        ≐
        n
        .
      
    
    {\displaystyle p\doteq n.}
  
 Then 
  
    
      
        ∀
        n
        ∈
        
          N
        
        :
        
          |
          
            
              H
              
                2
                n
              
            
            −
            
              H
              
                n
              
            
          
          |
        
        ≥
        
          
            1
            2
          
        
        .
      
    
    {\displaystyle \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)

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 proof – finding 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)
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)
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)
I agree with Deacon Vorbis. --JBL (talk) 22:23, 15 May 2020 (UTC)
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)
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)
"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)

The current version of section "Comparison test" uses sloppy reasoning which gives the reader a wrong impression about calculating with infinite series. 
In the first indent (the inequation), the Direct comparison test is used - it is mentioned later (i.e., in the wrong place). 
The second indent regroups an infinite series without even mentioning the rearrangement theorem (absolute convergence#Rearrangements and unconditional convergence). A naive reader will get the impression that there is no problem with rearranging any series.
While all this could be fixed, it is unnecessary. I made a suggestion for a shorter argument here, but was reverted by Deacon Vorbis. I'd be willing to fix my messing up of the formula alignment, if this was the only reason of reversion. One aspect would of course be the original version of Oresme; however, I couldn't yet find it. - Jochen Burghardt (talk) 15:49, 2 October 2020 (UTC)

Meh. A short note about grouping terms would be okay and should be enough here.  The terms in the series are all positive, so there are no issues anyway.  And besides, the grouping isn't even strictly necessary; it's mainly just there for clarity to the reader that we're getting runs of terms that all sum to 1/2. –Deacon Vorbis (carbon • videos) 16:13, 2 October 2020 (UTC)
I agree with DV about the lack of a problem with the longer form.  The shorter version packages exactly the same idea in a more technical (so harder-to-grasp) formulation, without any real advantage.  (I have no idea what Oresme actually wrote, but I'd bet actual money it looked more like the long version than the short one.) --JBL (talk) 00:13, 3 October 2020 (UTC)
You'd lose it. Here's Oresme's proof:
I argue that the whole will become infinite. This is proved by this way: there exist an infinite number of parts each of which being greater than one half, and therefore the whole will become infinite. The first is evident, because the fourth part and the third part are more than one half, the same with the fifth up to the eigth part and with the ninth to the sixteenth part, &c, in infinitum.
pma 21:45, 12 October 2020 (UTC)
??? That ... looks much more like the long version than the short version, as I said. --JBL (talk) 22:10, 12 October 2020 (UTC)
Maybe because the short and the long are also quite similar to each other. But the above short version by Jochen Burghardt turns out to be the exact translation into formulas of Oresme's words. The present long version have some more details, that seems to me avoidable. In fact the short version seems superior to me. Anyway, the original Oresme's proof may be worth a footnote, like other quotations from Mengoli and the Bernoulli. I found in Sourcebook in the Mathematics of Medieval Europe and North Africa; unfortunately, I wasn't able to find the original Latin text.pma 22:53, 12 October 2020 (UTC)
Your comments appear to miss entirely the point of the differences between the proofs.  The Oresme proof you've quoted, like the longer version, avoids being over-burdened by the pedantic and unnecessary technical details. --JBL (talk) 23:21, 12 October 2020 (UTC)
Dear JLB, to me both versions are OK, (with a light preference to the shorter, as I said, and maybe the original is even better). We all want to avoid pedantic and unnecessary technical details, but I am honestly curious about your reasoning. Would you clarify your last sentence? How is that the shorter version is more burden by details than the longer? What details do you see in the shorter that are not in the longer? Thanks pma 08:41, 13 October 2020 (UTC)
The long version was understandable a few centuries before Newton, and should be understandable today by any moderately bright 12 year-old.  The short version takes this simple idea and wraps it in a modern jargon (an arbitrary positive integer k, switching between infinite sums and finite sums, most especially "sufficiently large") that will be unfamiliar to and not decipherable by anyone who has not taken a first class in proof-based calculus.  This modern machinery is great and is useful for many things, but it is not necessary here and it interferes with clearly communicating the central idea of the proof. --JBL (talk) 13:04, 13 October 2020 (UTC)
I see, you refer to "Therefore, the partial sums grow beyond every number, if k is chosen sufficiently large". This is also alluded in the inequality reported in the long version, which is the key point of the proof. Note however that what you call "modern machinery" are ideas that were perfectly clear to the Hellenistic mathematics, to whom we owe them: avoid talking about "infinite processes", like "a sum of infinite objects", which is not an idea that simple and clear, and introducing instead some quantifiers. Anyway, I like the present version as well. pma 13:40, 13 October 2020 (UTC)

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)

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)

This review is transcluded from Talk:Harmonic series (mathematics)/GA1. The edit link for this section can be used to add comments to the review.
Reviewer: XOR'easter (talk · contribs) 21:01, 12 March 2022 (UTC)

I find the article well-written, in an encyclopedic rather than a textbook-like way. It is pitched to an appropriate audience level given the material, and the least technical parts are pushed to the forefront.  Approved
The article is verifiable with no original research. It presents standard material without novel synthesis. Citations are properly and consistently formatted. I have a few minor points that I will detail below.  GA on hold
The coverage addresses the main topics without drowning in detail. There is doubtless room for expansion, given how many places in mathematics a subject like this will pop up, but we're not going for FA-level comprehensiveness here.  Approved
I find no editorial bias.  Approved
No signs of edit-warring or content disputes.  Approved
Images are appropriate; no licensing issues.  Approved
Now, for a few things that look like easy fixes:
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).

Replaced with new Dunham source, and reordered to clarify that the first publication was Jacob's but that in it he credited Johann with the proof. —David Eppstein (talk) 02:39, 13 March 2022 (UTC)
Reference 14, Havil (2003), appears not to actually use the term harmonic number — it just uses 
  
    
      
        
          H
          
            n
          
        
      
    
    {\displaystyle H_{n}}
  
, in all the instances I can find. "No property is more unexpected than 
  
    
      
        
          H
          
            n
          
        
      
    
    {\displaystyle 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. I'm not convinced that a citation explicitly introducing the term is obligatory, given that we have a whole article on it (it's unlikely to be challenged), but maybe there's a secondary source that reports the first known use of the term, or something like that.

I don't know about secondary, but I appear to have found the primary source with the first use of the term: Knuth's Art of Computer Programming (1968). —David Eppstein (talk) 03:11, 13 March 2022 (UTC)
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.

Ok, added a gloss for this unit. With the source's definition of a double pace as 5 feet, it's closer to 1.4 miles, but I used km as the primary unit and miles converted from that. —David Eppstein (talk) 08:02, 13 March 2022 (UTC)
Euler's conclusion that the partial sums of reciprocals of primes grow as a double logarithm of the number of terms has been confirmed by later mathematicians as one of Mertens' theorems, and can be seen as a precursor to the prime number theorem. The statement looks right, but the source provided does not include a specific mention of Mertens' theorems. Maybe it should be supplemented?

Added another reference (Pollack). —David Eppstein (talk) 08:27, 13 March 2022 (UTC)
These are all very low-grade issues, but they might get slightly in the way of a student's understanding. XOR'easter (talk) 21:25, 12 March 2022 (UTC)

Ok, I think I've handled them all. Time for another look? —David Eppstein (talk) 08:27, 13 March 2022 (UTC)
Looks good to me! XOR'easter (talk) 18:50, 13 March 2022 (UTC)

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)

Convergence_tests#p-series_test also points here. 66.44.49.56 (talk) 02:18, 1 August 2022 (UTC)
So it happened in this edit by David Eppstein, 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)

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)

The lead says The first 
  
    
      
        n
      
    
    {\displaystyle n}
  
 terms of the series sum to approximately 
  
    
      
        ln
        ⁡
        n
        +
        γ
      
    
    {\displaystyle \ln n+\gamma }
  
. This is correct, since the difference between the sum of the n first terms and 
  
    
      
        ln
        ⁡
        n
        +
        γ
      
    
    {\displaystyle \ln n+\gamma }
  
 is less than 
  
    
      
        1
        
          /
        
        n
      
    
    {\displaystyle 1/n}
  
 and tends to zero when 
  
    
      
        n
      
    
    {\displaystyle n}
  
 tends to infinity. However, the term 1/(2n) is important in some contexts, since it reduces the approximation error from less than 1/(2n) to less than 1/(8n2), which is much smaller. See § Growth rate. D.Lazard (talk) 11:05, 17 March 2024 (UTC)