Talk:Absolute convergence

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Why is products of series here? —The preceding unsigned comment was added by 129.7.250.116 (talk • contribs) 14:21, 31 August 2006 (UTC)

It's not absolutely, logically necessary, but it demonstrates an important application of absolute convergence. Melchoir 14:54, 31 August 2006 (UTC)[reply]

Another user

The use of the word "theorem" in the (nice) discussion that starts: ". . . this theorem can be imagined as. . ." is confusing. There is a theorem that absolute convergence implies convergence: Is that meant here? This implies the too-loose-to-be-a-theorem vebiage about "rearrangements...intuitive." But the reader has not yet met the theorem, so it is premature to call it such here.

I generally like giving intuitive grounding for math, and I think this section does a good job of doing so. I can't think of good substitute for "theorem". Might be necessary to add more words.

When I read this page I found that it didn't give enough information about the relation between absolute convergence and convergence, and especially that it conflated absolute and unconditional convergence. So I rewrote quite a bit of it.

The rewrite admittedly has a more technical feel to it: it seemed to me that to get to the bottom of absolute convergence, one should see it in proper generality, whence a section on absolute convergence for series with values in a normed abelian group. The definition of normed abelian group I use comes from Kaplansky's Set Theory and Metric Spaces (and should probably be referenced). (Apparently some people use this terminology without the symmetry axiom (iii), which means the norm might not induce a metric...) So far as I know, the main case of interest is that of a Banach space, but it seems satisfying to observe that the proofs of the easier theorems work verbatim in this level of generality. I also feel that knowing that unconditional and absolute convergence is not the same thing in general because the two fail to coincide on every infinite dimensional Banach space is an interesting and enlightening fact even if you don't really know or care much about Banach spaces (as I do not).

Probably a more careful structuring between the sections "A more general setting for absolute convergence" and "Relations with convergence" would make the latter section more palatable for students at the honors calculus / undergraduate real analysis level who do not know what a Banach space is. Perhaps someone has good ideas for this?

I started out the rewrite carrying along the parallel between series and integrals but found it tiresome, so currently there's just a short section at the bottom acknowledging that the concept makes sense for integrals. Notice that the relationship between absolute integrability and integrability depends strongly on what integration theory you're using: for the proper Riemann integral one has integrability of f implies integrability of |f| but not conversely; for the Lebesgue integral the two are equivalent by definition; for the Kurzweil-Henstock integral I think the implication goes the other way. Maybe it's better to have this as a separate article? Plclark (talk) 15:48, 29 December 2007 (UTC)Plclark[reply]

For the field of p-adic numbers, I think the following holds: Any convergent series is unconditionally convergent (though not necessarily absolutely convergent). This is because a series converges iff its terms are a zero series, and this condition remains unchanged under permutation of the terms.

On the other hand, there are quite "natural" convergent series which do not absolutely converge: e.g. the sequence 1, p,..., p, p^2, ..., p^2, p^3, ... (taking each summand p^n p^n times).

Thus, the term "unconditionally convergent" is the more "natural" in this case, IMO. --Roentgenium111 (talk) 22:27, 14 April 2008 (UTC)[reply]

With regard to the possibility of merging this article with the short article showing absolute convergence implies convergence: the latter article is written at a level accessible to a calculus student, whereas this article is much broader in scope. To me it seems reasonable to keep both articles and to keep them separate. However it might be a good idea to link to the calculus article when recalling that completeness and absolute convergence imply convergence. Plclark (talk) 16:39, 9 April 2009 (UTC)[reply]

If the accessibility of the article is a problem, the right course of action should be to mitigate that problem. I think we shouldn't address accessibility issues by maintaining separate articles. (Yes, this article could certainly use more elementary examples, like alternating series converges but not absolutely, etc. True, this is mentioned but it is mentioned in the way the article assumes readers already know such examples.) -- Taku (talk) 00:20, 10 April 2009 (UTC)[reply]

Is there anything in the WP:MOS that addresses article accessibility. andyzweb (talk) 02:57, 31 December 2009 (UTC)[reply]

The more specific WP:MSM has some information on accessibility in mathematical articles. Other than that looking at good and the few featured mathematics articles might show approaches that have worked. --JohnBlackburne^words_deeds 15:38, 12 January 2010 (UTC)[reply]

Looking to stirr some action on the merge debate I would like to settle it by Jan 21st 2010, at that point I will remove the merge discussion. andyzweb (talk) 05:46, 12 January 2010 (UTC)[reply]

• MERGE: this seems non-problematic; I'm not even sure why you need an RfC. the other article is short enough it can easily be added in as a section of this article. --Ludwigs2 23:41, 14 January 2010 (UTC)[reply]
• Agree Merge away.Doc James (talk · contribs · email) 10:28, 21 January 2010 (UTC)[reply]
• Merge. No need for RfC. ;) Regarding accessability, there speaks nothing against introducing the absolute convergence theorem as the first section in a proper way. Nageh (talk) 22:51, 31 January 2010 (UTC)[reply]

The proof is completely wrong. Am I the first to notice this?--Gallusgallus (talk) 05:19, 3 February 2010 (UTC)[reply]

Let me take a look... yep, you're right. The last two lines are completely flawed.

First, in the second-last line, the sum ${\displaystyle \sum _{i\in \{\delta _{\frac {\varepsilon }{2}}+1,...,\max X(\delta {\frac {\varepsilon }{2}})\}}a_{i}}$ must be replaced by ${\displaystyle \sum _{i\in \sigma \left(\{0,\dots ,N\}\right)\setminus \{0,...,\delta _{\frac {\varepsilon }{2}}\}}a_{i}}$.

Then, in the last line, before upper-bounding to ${\displaystyle \epsilon /2}$, the second absolute sum should be upper-bound against the sum of absolute values first, and then against the complete ${\displaystyle a_{i}}$ subset from ${\displaystyle \delta _{\frac {\varepsilon }{2}}}$ to ${\displaystyle \infty }$

I didn't double check, so there may still some slight error slipped in. Maybe some math regular can check and then update the article accordingly. Nageh (talk) 12:59, 3 February 2010 (UTC)[reply]

I am not sure about corectness in the whole, but I think some corrections and comments besides Nageh's must be done to make the proof more understandable by non-professionals:
1) "By the condition of absolute convergence:" and then something like "with the special kind of Cauchy criterion for absolutely convergent sequences".
2) Even if we know that "i" is dummy index, it is better to avoid overloading and use different index name if it runs through different domain.
3) As I see ${\displaystyle \max X(\delta _{\frac {\varepsilon }{2}})}$ must be the maximum among the set ${\displaystyle \sigma }$, not ${\displaystyle \sigma ^{-1}}$, right? Volyrkr (talk) 19:57, 9 August 2010 (UTC)[reply]

I translated this article to Japanese Wikipedia. (ja:絶対収束) In this work, I noticed that the proof is wrong. Then, I rearranged the proof. see detail below;

For any${\displaystyle \varepsilon >0}$, we can choice some ${\displaystyle \kappa _{\varepsilon },\lambda _{\varepsilon }\in \mathbb {N} }$, such that

${\displaystyle \forall N>\kappa _{\varepsilon }\ ,\sum \limits _{n=N}^{\infty }||a_{n}||<{\frac {\varepsilon }{2}}}$

and

${\displaystyle \forall N>\lambda _{\varepsilon }\ ,\left\|\sum \limits _{n=1}^{N}a_{n}-A\right\|<{\frac {\varepsilon }{2}}}$.

let,${\displaystyle N_{\varepsilon }:=\max(\kappa _{\varepsilon },\lambda _{\varepsilon })}$, ${\displaystyle M_{\sigma ,\varepsilon }:=\max \left\{\sigma ^{-1}\left(\{1,\dots ,N_{\varepsilon }\}\right)\right\}}$.

For any ${\displaystyle N>M_{\sigma ,\varepsilon }\ \ (N\in \mathbb {N} )}$, let

${\displaystyle I_{\sigma ,\varepsilon }:=\left\{1,\ldots ,N\right\}\setminus \sigma ^{-1}\left(\{1,\dots ,N_{\varepsilon }\}\right)}$,

${\displaystyle S_{\sigma ,\varepsilon }:=\min\{\sigma (k)|k\in I_{\sigma ,\varepsilon }\}}$ (note.${\displaystyle \ S_{\sigma ,\varepsilon }\geq N_{\varepsilon }+1}$), and

${\displaystyle L_{\sigma ,\varepsilon }:=\max\{\sigma (k)|k\in I_{\sigma ,\varepsilon }\}}$

then

${\displaystyle \left\|\sum \limits _{i=1}^{N}a_{\sigma (i)}-A\right\|=\left\|\sum _{i\in \sigma ^{-1}\left(\{1,\dots ,N_{\varepsilon }\}\right)}a_{\sigma (i)}-A+\sum _{i\in I_{\sigma ,\varepsilon }}a_{\sigma (i)}\right\|}$

${\displaystyle \leq \left\|\sum _{j=1}^{N_{\varepsilon }}a_{j}-A\right\|+\left\|\sum _{i\in I_{\sigma ,\varepsilon }}a_{\sigma (i)}\right\|\leq \left\|\sum _{j=1}^{N_{\varepsilon }}a_{j}-A\right\|+\sum _{i\in I_{\sigma ,\varepsilon }}\|a_{\sigma (i)}\|}$

${\displaystyle \leq \left\|\sum _{j=1}^{N_{\varepsilon }}a_{j}-A\right\|+\sum _{j=S_{\sigma ,\varepsilon }}^{L_{\sigma ,\varepsilon }}\|a_{j}\|\leq \left\|\sum _{j=1}^{N_{\varepsilon }}a_{j}-A\right\|+\sum _{j=N_{\varepsilon }+1}^{\infty }\|a_{j}\|<\varepsilon }$

therefore

${\displaystyle \forall \varepsilon >0\ ,\exists M_{\sigma ,\varepsilon }\,\forall N>M_{\sigma ,\varepsilon }\,\,,\left\|\sum \limits _{i=1}^{N}a_{\sigma (i)}-A\right\|<\varepsilon }$

then ${\displaystyle \sum \limits _{i=1}^{\infty }a_{\sigma (i)}=A}$ Q.E.D.

Please do double check of this proof. Loasa (talk) 13:31, 9 January 2011 (UTC)[reply]

Oh thank the heavens, I'm not hallucinating and the proof is wrong! I looked over Loasa's proof, and I believe it is correct as written -- I'll replace the old proof in the main article with this one.

Volyrkr, I believe using ${\displaystyle \sigma ^{-1}}$ is correct: to find where the ${\displaystyle N}$th term of ${\displaystyle a_{i}}$ is in the rearrangement, you use ${\displaystyle \sigma ^{-1}(N)}$. Blue Wizard (talk) 19:39, 13 March 2011 (UTC)[reply]

This WLOG assumption hides a lot! This is only valid if the sum of |a_i| converges iff the sum of |Re(a_i)| and sum of |Im(a_i)| converges. While this may be true, it is disingenuous to say that the proof is complete without it. The argument I give, expanded from Thm. 3.45 of Rudin in clean and indisputable. I don't see why the original proof should be kept.

Alsosaid1987 (talk) 21:58, 29 January 2017 (UTC)[reply]

In particular what needs to be proved is the following: if \sum a_i and \sum b_i converge, where a_i, b_i>0, then \sum \sqrt{a_i^2+b_i^2} converges. Please let me know if there is a quick proof of this. Alsosaid1987 (talk) 22:28, 29 January 2017 (UTC)[reply]

Sorry. My bad, I guess only the reverse implication is needed: \sum |a_i| converging is equivalent to \sum \sqrt{Re(a_i)^2+Im(a_i)^2} converging. This needs to imply that \sum |Re(a_i)| and \sum |Im(a_i)| both converge, since this allows us to argue sequences of real numbers. This is true since \sqrt{Re(a_i)^2+Im(a_i)^2}\geq\sqrt{Re(a_i)^2}=|Re(a_i)| and similarly, \sqrt{Re(a_i)^2+Im(a_i)^2}\geq |Im(a_i)|, so we get what we want by the comparison test. I believe this should probably be made explicit in the text, unless you believe it is too obvious. Alsosaid1987 (talk) 22:41, 29 January 2017 (UTC)[reply]

This section is a mess. Most of the statements made are probably true, as far as my knowledge of this subject goes, and I have taken the liberty of restating and restructuring them somewhat, but as a whole, this section seems to lack focus and require 1) addition of reference(s) and 2) the attention of someone with expertise in integration theory.

-Jimmy Alsosaid1987 (talk) 05:06, 30 January 2017 (UTC)[reply]

I just added a section on the convergence of sums over sets, largely taken from Tao's analysis book. However, Tao seems to use a definition of "absolute convergence" which is not the same definition as the one used by other authors for the absolute convergence of an iterated sum. Tao proves that if a series is absolutely convergent in his sense, then the iterated sum equals the sum over the set of all pairs. But we don't get a proof that a series is absolutely convergent (in Tao's sense) if and only if the iterated sum is convergent, leaving it open that perhaps the iterated sum could converge even though the sum is not absolutely convergent. I'm sure that's not true, but I don't know of a reference which establishes the consistency of Tao's definition with that of other authors. Would love it if someone knew of a reference or otherwise could improve upon what I've written. Addemf (talk) 16:11, 4 August 2021 (UTC)[reply]

The example of 1 - 1 + 1 - 1 + ... given in the section background does not converge and therefore is not really related to the subject. A better example would be given by the alternating harmonic series. 45.166.163.85 (talk) 19:12, 1 December 2021 (UTC)[reply]