## A prime limit that seems to approach a constant

Ok heres the problem:
http://i42.tinypic.com/29288yd.png
Using wolfram the first 100 results are these
heres a plot of a couple points
As you can see it doesn't seem to be approaching exactly zero, even though its very similar to 1/x (exactly the same if you replace Pn with just n)
Is there any way to prove whether this does approach 0 or some constant, or is it possible to make a program to approximate it to some extremely large n, to see if its approaching zero or some constant.

 PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor hi robnybod! welcome to pf! it's decreasing and positive, so it must have a limit to find the limit, use the usual trick of putting fn = fn-1
 Thank you! so according to that it would go to zero, correct? because than f=f(1-1/P(n)) and f goes away, so you're left with -1/P(infinity)=0, which checks

Blog Entries: 27
Recognitions:
Gold Member
Homework Help

## A prime limit that seems to approach a constant

 Quote by robnybod so according to that it would go to zero, correct?
correct!

(the reaoning isn't rigorous, but the result is ok)

 Good morning, Your infinite product ∏(1-pi-1), over all primes, does indeed converge to zero, but no fn=fn-1 trick is close to showing why. The standard elementary proof here is to rewrite your limit as (Ʃ1/n)-1 over the positive integers realizing that your limit is an euler product (google, wiki). Note btw that if you add an exponent s to all your primes, your limit equals ζ(s)-1, where ζ(s) is the Riemann zeta function, known to converge for all s>1 (and giving you non-zero limit in this case).
 Another proof would be to think of function fk as being the probability to pick a natural number that has a factor among all the prime numbers except the first k prime numbers. f0 = 1, the probability to pick a number that has a factor among all primes is 1 f1 = f0 - f0/p1, the probability to pick a number that has a factor among all prime numbers except the first prime is 1/2 at infinity this translates into lim $_{n->\infty}$fn = 0 because f$\infty$ is the same as asking what is the probability to pick a natural number that doesn't have a factor among all the prime numbers. Of course all natural numbers have a prime factor or are prime numbers therefore the answer is 0.

 Tags prime limit constant