Talk:Limit comparison test

Comment
I think the limit comparison test can be strengthened and generalized at the same time. I dont believe we need to restrict both series so that their terms are positive. It seems to me that we need only ensure that the nth term of each series is the same sign, and that the magnitude of the nth term of one series' is appropriately related to the magnitude of the nth term of the other series. Can I get confirmation on this? 76.104.137.50 (talk) 23:56, 17 November 2011 (UTC)

'''This source says that, under certain conditions, 0 and infinity can be values of L. Can anyone else confirm this?Cmmr613 (talk) 21:15, 4 March 2012 (UTC) http://www.math.ucdavis.edu/~kouba/Math21CHWDIRECTORY/ComparisonTests.pdf'''
 * - This is true, and is also found in Briggs and Cochran's Calculus text. If L = 0 and Sum(b_k) converges, then Sum(a_k) does as well, and if L is infinity and the sum of the b_k diverges then the sum of the a_k also does. Stewart's text does not include this as part of the test, perhaps because the idea behind the limit comparison test is to find a similar series, and in these cases of L = 0 or infinity the series definitely aren't similar. It seems that different authors have different opinions on whether it's worth mentioning/actually part of test. Since some do though (and the class I'm teaching requires students to know it), it might be worth mentioning if someone who knows how to type this sort of thing up well wants to.--137.99.230.128 (talk) 21:11, 26 March 2013 (UTC)
 * - This is true, and is also found in Briggs and Cochran's Calculus text. If L = 0 and Sum(b_k) converges, then Sum(a_k) does as well, and if L is infinity and the sum of the b_k diverges then the sum of the a_k also does. Stewart's text does not include this as part of the test, perhaps because the idea behind the limit comparison test is to find a similar series, and in these cases of L = 0 or infinity the series definitely aren't similar. It seems that different authors have different opinions on whether it's worth mentioning/actually part of test. Since some do though (and the class I'm teaching requires students to know it), it might be worth mentioning if someone who knows how to type this sort of thing up well wants to.--137.99.230.128 (talk) 21:11, 26 March 2013 (UTC)

For all $$ \varepsilon $$
The proof should say "for each $$ \varepsilon > 0 $$ there is an integer $$n_0$$...". There need not be a single $$ n_0 $$ that works "for all" $$\varepsilon $$.

Tashiro (talk) 06:14, 10 July 2013 (UTC)

The example is not illuminating.
One can use direct comparison test to show the example $$ \sum_n \frac{1}{n^2+2n} $$ is convergent, because $$ \frac{1}{n^2+2n} \leq \frac{1}{n^2} $$ and the latter series converges. Maybe $$\sum_n \frac{1}{n^2-2n} $$ provides a better example to show where the theorem is actually used. — Preceding unsigned comment added by 188.158.138.52 (talk) 17:41, 2 August 2014 (UTC)