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A prime limit that seems to approach a constantby robnybod
Tags: prime limit constant 
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#1
Feb1012, 09:49 AM

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 P_{n} 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 


#2
Feb1012, 10:10 AM

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P: 26,157

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 f_{n} = f_{n1} 


#3
Feb1012, 10:20 AM

P: 3

Thank you!
so according to that it would go to zero, correct? because than f=f(11/P(n)) and f goes away, so you're left with 1/P(infinity)=0, which checks 


#4
Feb1012, 11:08 AM

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P: 26,157

A prime limit that seems to approach a constant
(the reaoning isn't rigorous, but the result is ok) 


#5
Feb1212, 12:41 AM

P: 144

Good morning,
Your infinite product ∏(1p_{i}^{1}), over all primes, does indeed converge to zero, but no f_{n}=f_{n1} 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 nonzero limit in this case). 


#6
Mar612, 10:15 AM

P: 70

Another proof would be to think of function f^{k} as being the probability to pick a natural number that has a factor among all the prime numbers except the first k prime numbers.
f^{0} = 1, the probability to pick a number that has a factor among all primes is 1 f^{1} = f^{0}  f^{0}/p_{1}, 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 [itex]_{n>\infty}[/itex]f^{n} = 0 because f^{[itex]\infty[/itex]} 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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