Talk:Direct comparison test

message: "if an <= bn for all n, and suppose that  sum (from n= 0 to infinty) bn is convergent. Then sum an is convergent." 
(Craw, 2002)
Notice "for all n". Wiki says that "for sufficiently large n".
Also, please explain why comparing the ratios of a series to another known series is betten than comparing the ratio to 1, as in the d'Alembert test. I'm sure there must be a reason, but I can't figure it.
Craw, I., 2002, The Comparison Test, The University of Aberdeen, Available from
http://www.maths.abdn.ac.uk/~igc/tch/ma2001/notes/node50.html
212.159.75.167 17:33, 10 December 2006 (UTC)
thread_title:  Questions 

message: In my opinion the described comparison test of the second kind is plain wrong. As a counterexample you can take a series b_n=s^n with s<1 and 
a series a_n=r^n with r>1 (i.e. a converging and a non-converging geometric series) and obviously the condition is fulfilled with C=r/s, but as obivously a_n is not converging while b_n is. This could be fixed by requiring C<=1, but then I could directly use the ratio test (at least, I can't see a point in comparing to another series in that case). So, I would vote for removal of this passage and any references to it. 134.169.77.186 10:29, 29 August 2007 (UTC)
replies: text: The material on the "comparison test of the second kind" was completely incorrect, and has now been deleted. Jim 00:03, 5 September 2007 (UTC)
replies: text:  Well... not completely incorrect. Just incorrect. [w] I've re-added a similar so-called "ratio comparison test" from Buck's Advanced Calculus. BTW, Buck uses the RCT to prove both the ratio test and Raabe's test, so I assume this means the RCT is somewhat useful in and of itself. - dcljr (talk) 05:18, 3 May 2012 (UTC)
thread_title:  Comparison test of the second kind 

message: The test is also taught as the "direct comparison test", to separate it from the "limit comparison test", at least at my university.  —Preceding unsigned comment added by ThomasOwens (talk • contribs) 16:15, 21 October 2007 (UTC)
thread_title:  Name of Test 

message: In the proof it assumes that 
S_n <= T_n
but a_1 could be a_1 = 999.000.000, what I mean it could be really big as the condition a_n < b_n only is required for very large n, so that asumption I think is incorrect.  — Preceding unsigned comment added by 186.18.76.220 (talk) 00:52, 14 November 2011 (UTC)
thread_title:  Proof flawed? 

message: They are often confused, so I mentioned they are different at the beginning of the article.
Why is the direct comparison test called CQT?  — Preceding unsigned comment added by Arathron (talk • contribs) 00:10, 15 November 2011 (UTC)
replies: text: I couldn't find a reference for "CQT" using Google that wasn't just quoting this article, so I've removed it. If someone has a reference, they can put it back with a citation. - dcljr (talk) 22:18, 1 May 2012 (UTC)
thread_title:  relation with [[limit comparison test]] 

message: Substituted at 02:00, 5 May 2016 (UTC)
thread_title: Assessment comment

message: In  the DCT is stated in two variants:
# once for convergence, and
# again for absolute convergence.  
In each case, the test provides two statements:
# that a series dominated by a [or, an absolutely] convergent series also converges [absolutely];
# that a series that dominates a [or, an absolutely] divergent series also diverges [absolutely].
This is all fine.  Now we come to  (and resp. ), which doesn't mention absolute convergence or divergence of the integrals (resp. real-valued series).  Nor does the RCT mention divergence at all.
Questions: Are analogues of all four statements for the DCT for series also available for integrals?  And for the RCT?  If so, shouldn't the article also cover them?  And if not – which would be surprising! – shouldn't the article explain why not?
yoyo (talk) 08:57, 1 October 2018 (UTC)
thread_title:  More questions