Wikipedia:Reference desk/Archives/Mathematics/2020 April 7

= April 7 =

Series that seems to converge actually diverges
What is the best series to use to make sure students know that a series that sometimes seems to converge actually diverges?? A good answer would probably be $$ n!/(1000^n) $$; it initially looks as if it converges to 1.001002006024... but if you find out what happens to the ratios of the terms you'll notice that the terms will start increasing faster and faster after only 1000 terms. Thus, it diverges; the terms will have ratios above a googol when the number of terms goes past 10^103. Are there any even more interesting series that seem to converge but that actually diverge?? Georgia guy (talk) 15:21, 7 April 2020 (UTC)
 * $$\sum \frac{1}{n}$$? $$\sum \frac{1}{n \log n}$$? $$\sum \frac{1}{n \log n \log \log n}$$? And so on? (That these all converge diverge is a consequence of the Cauchy condensation test.) Double sharp (talk) 15:31, 7 April 2020 (UTC)
 * You mean, that they all diverge... Burzuchius (talk) 21:26, 7 April 2020 (UTC)
 * Yes indeed. Edited. Double sharp (talk) 02:42, 8 April 2020 (UTC)
 * The first of those is the harmonic series, which diverges. Georgia guy (talk) 15:39, 7 April 2020 (UTC)
 * But slowly, like $$\log x$$. ^_^ Double sharp (talk) 16:15, 7 April 2020 (UTC)
 * Also a consequence of the integral test, which students would likely see first I'd think. If you just want to throw some cool examples at them, you can also use the divergence of the sum of the reciprocals of the primes: $$\textstyle\sum_n \frac{1}{p_n} = \infty.$$  I'm not sure if there's a good way to show this to first-year Calculus students, but it's a nice example anyway. –Deacon Vorbis (carbon &bull; videos) 16:51, 7 April 2020 (UTC)
 * They all diverge. --Lambiam 17:14, 7 April 2020 (UTC)
 * But Double sharp initially said they all converge. Georgia guy (talk) 17:43, 7 April 2020 (UTC)
 * Ridiculous typo on my part, sorry. Double sharp (talk) 02:42, 8 April 2020 (UTC)
 * If you want convergent ones that diverge really slowly, take $$\sum \frac{1}{n^p}$$, $$\sum \frac{1}{n (\log n)^p}$$, $$\sum \frac{1}{n \log n (\log \log n)^p}$$, etc. for $$p$$ just greater than 1. Double sharp (talk) 02:43, 8 April 2020 (UTC)
 * I think the adaptation of Euler's intuition in the section Divergence of the sum of the reciprocals of the primes may be a good way to show this result. Double sharp (talk) 02:46, 8 April 2020 (UTC)
 * Asymptotic expansions of functions "usually" diverge, but partial sums of them provide useful approximations. So that's a whole subject devoted to such series, more or less.John Z (talk) 18:43, 7 April 2020 (UTC)


 * (I have no idea how interesting this is.) $$\sum_{n=1}^\infty \frac{x^n}{n^3}$$ diverges for all $$x > 1$$, but when $$x = 1.01$$, the numerical values of the first 301 partial sums look as if the series might converge. But then the 302nd term is larger than the 301st. It may also be interesting for your students (or not) to learn that $$\sum_{n=1}^\infty n^{{-}s}$$ diverges when $$s = 1$$, but converges for all $$s > 1$$. --Lambiam 19:35, 7 April 2020 (UTC)


 * Let S be the set of divergent series consisting of elements that can be formulated with less than 20 symbols. Which element of S diverges the slowest? Count Iblis (talk) 07:38, 8 April 2020 (UTC)
 * It is not necessarily the case that some element is as slow as, or slower than, all other elements. Also, depending on the language used for the formulas, convergence may be an undecidable property. --Lambiam 17:43, 8 April 2020 (UTC)