# A prime limit that seems to approach a constant

by robnybod
Tags: prime limit constant
 P: 3 Ok heres the problem: 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. Thanks in advance, and sorry if the answer is obvious
 Sci Advisor HW Helper Thanks P: 26,160 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
 P: 3 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
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P: 26,160
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)
 P: 144 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).
 P: 70 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.

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