From: caldwell@unix1.utm.edu (Chris Caldwell)
Newsgroups: sci.math
Subject: ln(2)+ln(3)+ln(5)+ln(7)+...+ln(p)
Date: 18 Jan 1995 00:11:27 GMT

I am looking for a pratical, accurate approximation of 

  ln(p#) = ln(2)+ln(3)+...+ln(p)

(where the sum is taken over the primes <= p).  My values 
of p are about 10^10.   Any suggestion?

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

From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: ln(2)+ln(3)+ln(5)+ln(7)+...+ln(p)
Date: 18 Jan 1995 21:55:39 GMT

In article <3fhmbf$186@martha.utk.edu>,
Chris Caldwell <caldwell@unix1.utm.edu> wrote:
>I am looking for a pratical, accurate approximation of 
>
>  ln(p#) = ln(2)+ln(3)+...+ln(p)
>
>(where the sum is taken over the primes <= p).  My values 
>of p are about 10^10.   Any suggestion?

What, no  number theorists out there? This is a classical function, a
careful analysis of which leads to the prime number theorem. Here's
Landau writing in 1909:
section 17: sets \theta(x) = \Sum_{p\le x} \log(p)
section 19: shows \theta(x)/x and \pi(x)/(x/log(x)) have the same limit
section 24-25: shows that limit is 1.

So  ln(p#) = p + o(p).

As I recall, the better you do at estimating the error in the Prime
Number Theorem, the better you can estimate the error term  o(p); the
correspondence is too cumbersome for me to remember but it's not deep;
in fact I believe this is the route used in "elementary" proofs of
the PNT.

For what it's worth, a sum over consecutive primes can often be rewritten
as an ordinary sum using the approximation that the n-th prime is roughly
n log(n), and the sum up to  N  is a sum over the first  N/log(N) primes.
Of course the error terms in these approximations can kill you but this
is good enough to discover that the dominant term in  ln(p#) is indeed  p.

dave