Talk:Harmonic series (mathematics)

Divergence proofs

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)

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.

General Harmonic series

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)

Y 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)

playing it fast and loose in "proofs"

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)

p-series

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

$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \ .$

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)

How slowly does it diverge?

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... + 2^{n - 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:

${1\over2} + {1\over3} + {1\over4} + \cdots + {1\over n} < \int_1^n {dx \over x} < 1 + {1\over2} + {1\over3} + {1\over4} + \cdots + {1\over n -1}$

and

$\int_1^n {dx \over x} = \log_e n = {\log_{10} n \over \log_{10} e}.$

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)

Comparison Test Formula

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)

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)