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A prime limit that seems to approach a constant

by robnybod
Tags: prime limit constant
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robnybod
#1
Feb10-12, 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 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
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tiny-tim
#2
Feb10-12, 10:10 AM
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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
robnybod
#3
Feb10-12, 10:20 AM
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

tiny-tim
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Feb10-12, 11:08 AM
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A prime limit that seems to approach a constant

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

(the reaoning isn't rigorous, but the result is ok)
Norwegian
#5
Feb12-12, 12:41 AM
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).
atomthick
#6
Mar6-12, 10:15 AM
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 [itex]_{n->\infty}[/itex]fn = 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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