Wikipedia:Reference desk/Archives/Mathematics/2010 August 22

= August 22 =

arithmetic progressions and primes
I want to show that any arithmetic progression if continued long enough wouldn't contain primes. How should I proceed. Thanks-Shahab (talk) 07:46, 22 August 2010 (UTC)


 * [personal attack removed] If it contained primes over and over you'd clearly get a contradiction. Think about it and maybe you'll see it —Preceding unsigned comment added by 114.72.236.126 (talk) 08:10, 22 August 2010 (UTC)
 * Do you mean that any arithmetic progression eventually contains a number which is not prime? This follows from the prime number theorem. -- Meni Rosenfeld (talk) 08:53, 22 August 2010 (UTC)

Ya don't need any [profanity removed] prime number theory to see it. Besides, it doesn't follow from th eprime number theorem since ya could get primes appearing in the sequence over and over but less frequently as n goes to INFINITY WITHOUT contradicting any bullcrap prime number theory. So no, think harder Meni Rosenfeld and Shahb. —Preceding unsigned comment added by 114.72.192.74 (talk) 09:02, 22 August 2010 (UTC)
 * Please be polite. Using words such as "bullcrap" is not acceptable, especially in the context of a public forum. I think that you need to think harder; see the post below by Meni Rosenfeld. PS  T  13:58, 22 August 2010 (UTC)


 * Yes I did mean so Meni, like the AP 5,11,17,23,29 has primes only but then the next term is non-prime. Uptil now the biggest such prime AP that has been found has around 20 terms. Not sure how the prime number theorem applies. -Shahab (talk) 09:05, 22 August 2010 (UTC)
 * Prove by contradiction. If there was a sequence with only primes, what would a lower bound for $$\lim_{n\to\infty}\frac{\pi(n)}{n}$$ be? What is the limit in light of the PNT? -- Meni Rosenfeld (talk) 09:19, 22 August 2010 (UTC)

[irrelevant vulgarities and personal attack removed] —Preceding unsigned comment added by 110.20.58.220 (talk) 10:25, 22 August 2010 (UTC)

[personal attacks removed] —Preceding unsigned comment added by 114.72.192.74 (talk) 09:09, 22 August 2010 (UTC)


 * Because if an arithmetic progressions with difference d contained only primes then the asymptotic density of primes would be at least 1/d, which contradicts the prime number theorem. Gandalf61 (talk) 09:19, 22 August 2010 (UTC)


 * For a constructive proof, recall that an arithmetic progression looks like $$a + bn$$ for some choice of $$a$$ and $$b$$. You want a choice of $$n$$ that will make this be obviously non-prime.  Since all you have to work with are $$a$$ and $$b$$, try choosing one of those for $$n$$ and seeing what you can get.  —Preceding unsigned comment added by 203.97.79.114 (talk) 09:20, 22 August 2010 (UTC)


 * ... but make sure your proof covers the case a = 1. Gandalf61 (talk) 09:30, 22 August 2010 (UTC)


 * Thanks. I selected n=b^2 for a=1 and n=a otherwise. This seems so much easier then PNT.-Shahab (talk) 12:31, 22 August 2010 (UTC)


 * See also Primes in arithmetic progression (created by me) whics says: "an AP with common difference a cannot contain more consecutive prime terms than the value of the smallest prime that does not divide a." PrimeHunter (talk) 01:46, 24 August 2010 (UTC)

Divergent Series
OK, so one of the most important lemmas on convergent series sequences is that if a series sequence converges to a limit then any subsequence also converges to that limit. This is not true for divergent series sequences, since you can take the sum of the reciprocals of the natural numbers, which diverges, and, as a subsequence, the sum of the reciprocals of the squares, which converges. My question is, if you have a sequence and you're investigating convergence, does finding a divergent subsequence imply that the sequence itself is divergent? Thanks asyndeton   talk  11:38, 22 August 2010 (UTC)
 * You seem to be confusing series with sequences. For sequences, if the sequence converges, then every subsequence converges to the same limit. Hence if any subsequence diverges, then the sequence itself diverges. No such results hold for series. Algebraist 12:05, 22 August 2010 (UTC)
 * Well, if a series absolutely converges, then any subseries does as well, though not necessarily to the same value. -- 1.46.157.19 (talk) 12:17, 22 August 2010 (UTC)
 * Sorry for being so casual on naming; I've amended my question for posterity. And thanks for your answer, it's just what I needed. asyndeton   talk  12:22, 22 August 2010 (UTC)
 * Let me point out that you actually answered your own question. "If a sequence converges, every subsequence converges" and "If a subsequence diverges, the sequence diverges" are contrapositives.  —Preceding unsigned comment added by 203.97.79.114 (talk) 12:30, 22 August 2010 (UTC)

The amended question is still confused about sequences versus series. The question about the sum of the reciprocals is about series, not sequences, whereas the lemma you mention is about sequences, not series. Michael Hardy (talk) 19:25, 22 August 2010 (UTC)

Every sequence is a series and every series is a sequence. If an is a sequence and bn = an - an-1 and b1 = a1 then bn is a series and an can be a series. If bn is a series then its partial sums form a sequence. Every theory about sequence has one theory about series. And Meni Rosenfeld, don't dob on me mate. If I can't correct someone for the good of humanity, what can I do? And please don't say I have attacked anyone. I just asked Meni to not "dob on me mate". That's not a personal attack. I just want to better understand what I've done wrong and unless you tell me, I can't understand that. Blocking me won't help me. —Preceding unsigned comment added by 110.20.6.240 (talk) 22:57, 22 August 2010 (UTC)
 * Yes, sequence and series are related via partial sums, but care must be taken not to conflate the two. The lemma mentioned by the OP was of the convergence of subsequences, where an arbitrary subseries is guaranteed convergence only if the original series is absolutely convergent.  With a conditionally convergent series of real values, you may choose any arbitrary real number and there will be a subseries which converges to that value.  (Actually, there are infinitely many such subseries for every real number.)  The Riemann series theorem addresses permutations of conditionally convergent series; is there a name for the corresponding statement regarding subseries? -- 1.46.93.58 (talk) 00:22, 23 August 2010 (UTC)


 * Re the other issue, please see Wikipedia's policy on civility. Editors intelligent enough to contribute answers at the mathematics reference desk should also be able to understand the civility policy and follow it without a lot of additional explanation from other users.  Therefore, any failure to do this should be treated as presumptive trolling and result in blocks. 67.122.209.167 (talk) 01:59, 23 August 2010 (UTC)

Sorry it won't happen again. I've removed "bulltwang" from my posts. Sorry again.