Talk:Direct comparison test

Questions
"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)Timbo

Comparison test of the second kind
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) (ezander)
 * 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)
 * 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)

Name of Test
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)

Proof flawed?
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)

relation with limit comparison test
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)
 * 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)

Move

 * The following discussion is an archived discussion of a requested move. Please do not modify it. Subsequent comments should be made in a new section on the talk page. Editors desiring to contest the closing decision should consider a move review. No further edits should be made to this section. 

The result of the move request was: moved. Andrewa (talk) 06:19, 8 February 2013 (UTC)

Comparison test → direct comparison test – This current name should be a disambiguation for DCT and limit comparison test. As mentioned, the term can refer to both. Ranze (talk) 14:54, 31 January 2013 (UTC)
 * Support Seems reasonable. This is probably one of those rare cases where a dab with only two entries is appropriate, give that "comparison test" is vague. --BDD (talk) 17:53, 7 February 2013 (UTC)
 * The above discussion is preserved as an archive of a requested move. Please do not modify it. Subsequent comments should be made in a new section on this talk page or in a move review. No further edits should be made to this section.

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

More questions
In the DCT is stated in two variants:
 * 1) once for convergence, and
 * 2) again for absolute convergence.

In each case, the test provides two statements:
 * 1) that a series dominated by a [or, an absolutely] convergent series also converges [absolutely];
 * 2) 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)