A Generalization of the Sequence Convergence Problem

[benoit]

Regarding your generalisation. If s(n), s(n+1)-s(n)  are increasing  
sequences, that seems working.

Coming back to the original problem a(n)=1-(p(n-1)/p(n))*a(n-1),   I  
believe a(n)-->1/2, which can lead to conjectures related to primes.

It's easy to show that limit n-->oo (a(2n)+a(2n+1)) = 1 but that  
doesn't mean limit n-->oo a(2n)= limit n-->oo a(2n+1)= 1/2 at all.

Since : a(2n)=1/p(2n)*sum(k=1,n,p(2*k)-p(2*k-1)) ;  
a(2n+1)=1/p(2n+1)*sum(k=1,n,p(2*k+1)-p(2*k))

a(n)-->1/2 would imply   sum(k=1,n,p(2*k)-p(2*k-1)) to be asymptotic to  
(2n)*log(2n)/2 or n*log(n)

but as one can see (http://mathworld.wolfram.com/PrimeGaps.html), the  
behaviour of p(k+1)-p(k) is quite unknown.

More precisely is  :

sqrt(n)*(1/p(2*n+1)*sum(k=1,n,p(2*k+1)-p(2*k)) -  
1/p(2*n)*sum(k=1,n,p(2*k)-p(2*k-1)))  bounded ?


with (PARI)  
delta(n)=sqrt(n)*(1/prime(2*n+1)*sum(k=1,n,prime(2*k+1)-prime(2*k)) -  
1/prime(2*n)*sum(k=1,n,prime(2*k)-prime(2*k-1)))


Does limit n-->infinity delta(n) exist?  
delta(20000)=-0.5846342931598270275...

Is 1/prime(2*n+1)*sum(k=1,n,prime(2*k+1)-prime(2*k)) <  
1/prime(2*n)*sum(k=1,n,prime(2*k)-prime(2*k-1)) for all n large enough?

  If so the  sequence of n such that   
1/prime(2*n+1)*sum(k=1,n,prime(2*k+1)-prime(2*k)) >=  
1/prime(2*n)*sum(k=1,n,prime(2*k)-prime(2*k-1))

could be finite :

1,2,3,4,5,7,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,2 
9,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,5 
3,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,7 
7,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100, 
101,102,103,104,105,106,107,108,109,110,125,126,127,128,129,239,240,241, 
242,243,244,261,262,263,264,267,269,274,275,276,277,278,281,282,283,284, 
285,286,287,288,289,290,291,292,293,294,295,296,297,298,299,300,301,302, 
303,304,305,306,307,308,309,310,311,313,314

Is 314 the last term?



Benoit Cloitre
abcloitre at wanadoo.fr








Le mardi, 18 fév 2003, à 17:12 Europe/Paris, Pe Joseph-AJP070 a écrit :

> Here is a generalization of the sequence convergence problem I posted
> yesterday. (For convenience, I append
> the original problem at the end of this message.)
>
> ======================================================================= 
> =====
>
> Let s(n) be a sequence that converges to a real number K different  
> from -1.
> Define the "oscillator sequence" a(n) of s(n) by the rules:
>
>                      a(1) = 1;
>                      a(n) = 1 - (s(n-1)/s(n)) a(n-1) if n > 1.
>
> Note that the original problem below concerns the convergence of the
> oscillator sequence of s(n) = p(n).
>
> Examples of s(n) are s(n) = n, s(n) = n^2 and in general, any  
> polynomial
> function in n that is not the zero polyonomial function.
> The oscillator sequence of s(n) may converge or diverge depending on  
> s(n).
> Clearly, if s(n) = 1 is the constant sequence mapping
> each positive integer to 1, then a(n) diverges--in fact, oscillates  
> between
> 0 and 1 (hence the name "oscillator sequence").
> If s(n) = n, it is not hard to prove that a(n) converges to 1/2.
>
> Can you find conditions on s(n) that will ensure the convergence of its
> oscillator sequence? Of course, if s(n) converges, then
> lim s(n) = 1/(K + 1).
>
> ======================================================================= 
> =====
>
> J. L. Pe
>
> ======================================================================= 
> =====
>
> Original Problem:
>
> Define the sequence a(n) by: a(1) = 1; a(n) = 1-(p(n-1)/p(n))*a(n-1)  
> if n >
> 1, where p(n) denotes the n-th prime.
> It's easy to show (an exercise!) that if L = lim a(n) exists, then L =  
> 1/2.
> Can you prove the convergence of a(n) or the divergence of a(n)?
>
>
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