Hauke Worpel
Posted on Friday, 10 January, 2003 - 01:22 pm:
I have a conjecture that is totally counterintuitive concerning divergent series. But I can't find a counterexample, can't think of a simple way to disprove it so now I am beginning to think it is actually true.

Let {a(n)} be a sequence of ascending positive integers such that
¥
å
n=1 
 a(n)

diverges. Then
¥
å
n=1 
 a(a(n))

also diverges.


It might be hard to understand what I mean, so imagine we're summing the reciprocals of the 2nd, 3rd, 5th, 7th etc prime numbers ie. 1/3+1/5+1/11+1/17,...

I have proved that it is true for an =cn+d, which is a very, very elementary result.

Any ideas on how this can best be tackled in general?

Peter Conlon
Posted on Friday, 10 January, 2003 - 01:56 pm:
Surely if {a(n)} is a sequence of ascending positive integers, then its sum to infinity is bound to diverge? or am i missing something...

peter
Hauke Worpel
Posted on Friday, 10 January, 2003 - 03:16 pm:
Should be 1/a(n), sorry. Bother cut-and-paste!
Brad Rodgers
Posted on Friday, 10 January, 2003 - 08:03 pm:
I think the sum of the 2nd, 3rd, 5th, 7th, etc., prime number will converge - in fact, I think it'll follow directly from the prime number theorem.

There may be an easier method to prove this, though; I'll look for it, but if I can't find anything I'll probably post by tomorrow with the proof using the prime number theorem.

Brad
Michael Doré
Posted on Friday, 10 January, 2003 - 08:33 pm:
Here's a hint: an = n log n has the desired property except that an aren't integers. Can you construct a counter-example from here?
Hauke Worpel
Posted on Saturday, 11 January, 2003 - 12:37 am:
If you're asking me to see whether ò1¥ 1/(n log n log(n log n))dn goes to infinity, that is a bit above me. I had the same thought a few days ago and couldn't solve the integral.

I see what you're getting at though. I graphed the two functions, and the second appears to grow much faster ie. its reciprocals approach zero much faster.

That's hardly a proof though, and even if I could prove it, constructing a series of only integers is something else entirely.
Michael Doré
Posted on Saturday, 11 January, 2003 - 01:13 am:
Hmmm, just realised that Brad's answer and my answer are effectively the same because the nth prime is asymptotic to n log n by the Prime Number Theorem.

However you don't need the prime number theorem to solve your problem - because you can just let an be any integer sequence which is asymptotic to n log n. For example set an=[n log n] where [] denotes integer part. Then an is obviously strictly increasing because (n+1)log(n+1)-n log n ³ 1. It is well known and easy to prove that if xn and yn are positive sequences with xn ~ yn then
å
xn

converges iff
å
yn

converges. So it suffices to show that
å
1/(n log n)

diverges and
å
1/(n log n log(n log n ))

converges.

Well possibly the easiest way to do this is to use the fact that if xn is a decreasing positive sequence then
å
xn

converges iff
å
2n x2n

converges. Can you see how to prove this? (Hint: think of the standard proof that
å
1/n

diverges; obtain estimates for the sum in blocks which are powers of two.) The result is now clear.

Alternatively as you suggest you can use the integral test. You want to show that ò1¥1/(x log x) dx diverges and ò1¥ 1/(x log x log(x log x)) dx converges. In the first just substitute x=log t. In the latter you can simplify things by noting 1/(x log x log(x log x)) £ 1/ (x log x log(x))=1/(x (log x)2) and proceed similarly.

Let me know if you need any further hints.

Michael

Hauke Worpel
Posted on Saturday, 11 January, 2003 - 02:45 am:
Duh! I am the biggest idiot of all time. I should have seen that straight away.

So now we know that the 'primeth primes' harmonic series converges, which begs the question 'what does it converge to?', but I doubt that it will be very interesting.
Ben Tormey
Posted on Saturday, 11 January, 2003 - 08:57 pm:
Alternatively, use Raabe's test .