From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math.research
Subject: Re: (no subject)
Date: 30 Nov 1995 22:23:12 GMT

In article <49bm9k$9ee@news.nc.u-tokyo.ac.jp>,
Kenichiro Kashihara  <kkashiha-tky@umin.u-tokyo.ac.jp> wrote:

> Do you think the proposition below right?
><Let Pn be the n-th smallest prime number.
> Then {(P1*.....*Pn)^(1/n)}/Pn becomes 1/e if n becomes infinite.>

Yes, if I haven't gotten too sloppy in my estimates.

Taking logs, the question is whether the limit
	lim [ (1/pi(x)) theta(x)  - log(x) ]
equals  -1  as  x-->oo  ranges through the prime integers; here
pi(x) is the number of primes less than or equal to x, and
theta(x) is the sum of the logs of the primes less than or equal to  x.
I will show that in fact the above limit (taken over all real numbers  x)
really is  -1.  The proof is modelled after Landau's Primzahlen (sect. 19).

Note  pi(x)=Sum( 1 : p <= x)  [add a  1  for each prime less than  x]
	   =Sum( [theta(n)-theta(n-1)]/log n : 2 <= n <= x )
	   =Sum( theta(n) [1/log n - 1/log(n+1)] : 2 <= n <= x ) +
			theta(x)/log(floor(x)+1)
	   =Sum( theta(n) c(n) : 2 <= n <= x) + theta(x)/log(floor(x)+1),
say. We can estimate these  c(n)  to be  1/[n (log n)^2] * (1 + O(1/n) )

Thus, (log x)*pi(x) = theta(x) * c + (log x)*Sum(theta(n)*c(n))
where  c = 1 + O(1/[x log x]).

Now, it is known that  theta(n) = n + o(n), so that  theta(n)*c(n) 
is  theta(n)/[n (log n)^2] + O( 1/[n (log n)^2]), and the sum of these is
is within  O(1)  of  Sum( theta(n)/[n (log n)^2]). Also using  theta(x)=
x+o(x)  we may write the conclusion of the previous paragraph as
    (log x)*pi(x) = theta(x) + log(x)*Sum(theta(n)/[n (log n)^2]) + O(log x)

To proceed I need a little tighter estimate on the theta(n)'s. It actually
follows from using the previous equation as a "bootstrap" that
theta(x) = x + O(x/logx). Thus we can substitute into the last sum and get
Sum(1/(log n)^2) + O(Sum 1/(log n)^3). Since the summands lie within  O(1)
of the corresponding integrals, we may use integration by parts to
estimate this as  x/(log x)^2 + O(integral  1/(log t)^3); the integral
is in turn easily seen to be  O(x/(log x)^3). So the conclusion of the
preceding paragraph is now
	(log x)*pi(x) = theta(x) + x/(log x) + O(x/(log x)^2)

In particular, we see
	theta(x)/pi(x) - log x  = (-1) [x/log x]/pi(x) + O(1/(log x))
It is known that   [x/log x]/pi(x)  converges to  1  as  x-->oo, so
the limit of the expression is indeed  -1.

dave

==============================================================================

From: israel@math.ubc.ca (Robert Israel)
Newsgroups: sci.math.research
Subject: Re: (no subject)
Date: 30 Nov 1995 21:00:38 GMT

In article <49kc3l$jri@lyra.csx.cam.ac.uk>, gjm11@emu.pmms.cam.ac.uk (Gareth McCaughan) writes:
|> In article <49bm9k$9ee@news.nc.u-tokyo.ac.jp>,
|> Kenichiro Kashihara  <kkashiha-tky@umin.u-tokyo.ac.jp> wrote:
|> >  Do you think the proposition below right?
|> > <Let Pn be the n-th smallest prime number.
|> >  Then {(P1*.....*Pn)^(1/n)}/Pn becomes 1/e if n becomes infinite.>

Yes.  See below.

|> > If the same question is on An=n^k (k is a natural number),
|> > the limit is (1/e)^k.
|> > Pn/nlogn=1 (n is infinite.)
|> > Does k in this case almost 1?
|> 
|> Numerical experiments suggest that if you do this with A_n = n log n,
|> you get 0...

They may suggest it, but it's not true (except trivially because A_1 = 0).  
That is, let A_n = n log n and consider Q_n = (prod_{j=2}^n A_j)^(1/n)/A_n.
Then lim_{n -> infinity} Q_n = 1/e.

Proof: 

log Q_n = (1/n) sum_{j=2}^n (log j + log log j) - log n - log log n

now sum_{j=2}^n log j = int_1^n log t dt + O(log n) = n log n - n + O(log n)
    sum_{j=2}^n log log j = int_e^n log log t dt + O(log log n)
    (integrating by parts) = n log log n - int_e^n 1/(log t) dt + O(log log n)
                           = n log log n + o(n)
(actually the o(n) is O(n/ln(n)), but this is enough for our needs)
so log Q_n = -1 + o(1), and limit_{n -> infinity) Q_n = 1/e.

Now let F_n = (prod_{j=1}^n P_j)^(1/n)/P_n.

log F_n - log Q_n = 1/n log P_1 + 1/n sum_{j=2}^n log(P_j/A_j) + log(A_n/P_n)

and the fact that P_j/A_j -> 1 as j -> infinity implies this -> 0,
i.e. lim_{n -> infinity} P_n = 1/e.

-- 
Robert Israel                            israel@math.ubc.ca
Department of Mathematics             (604) 822-3629
University of British Columbia            fax 822-6074
Vancouver, BC, Canada V6T 1Y4