From: "Paul Pollack" <paulp@arches.uga.edu>
Subject: Dirichlet density vs. natural density
Date: Tue, 23 Oct 2001 19:54:03 -0700
Newsgroups: sci.math
Summary: Comparison of notions of densities of sets of integers

Let P denote a set of (positive) rational primes. One defines the Dirichlet
density of P to be

  DD(P) = lim [s->1+] sum(1/p^s, p in P)/sum(1/p^s, all primes p),

and the natural density of P by

  ND(P) = lim [x-> oo] #{p<= x: p in P}/#{p prime <= x}.

Then it is well-known that if ND(P) exists, then DD(P) exists and has the
same value. The converse is false; Serre's _Course in Arithmetic_ gives the
example Q = {p prime : p has leading decimal digit 1}, for which DD(Q) =
log(2)/log(10) while ND(Q) fails to exist.

Now showing ND(Q) fails to exist is relatively simple: the function f(x) =
#{p <= x: p in Q} is constant on, say, [2*10^k,9*10^k] (for each k >= 1),
whereas there is a constant A > 1 for which pi(9*10^k)/pi(2*10^k) >= A > 1
for all k sufficiently large. (By the prime number theorem we may take A
anything smaller than 9/2, but even Chebyshev's theorem with moderately
precise upper and lower bounds would yield a sufficient A.) This implies
f(x)/pi(x) cannot tend to a limit as x -> oo.

On the other hand, there doesn't seem to be an easy proof that DD(Q) exists.
Partial summation shows that if P is any set of primes for which sum(1/p, p
in P and p<=x) ~ c*sum(1/p, p<=x) [as x->oo], then DD(P) = c, and I have
heard there is a deep, much more difficult theorem that says a converse of
this holds. Unfortunately, if we attempt to estimate sum(1/p, p in Q, p<=x)
by summing over intervals and using the elementary estimate

 sum(1/p, p<=x) = log log x + C + O(1/log x),

we get the expected main term but with an error term of the same order of
magnitude, so this is no good. The prime number theorem with a relatively
weak error term -- namely pi(x) = li(x) + O(x/log^3 x) -- yields the
stronger estimate

 sum(1/p, p<=x) = log log x + C + O(1/log^2 x),

which gives the needed result -- unfortunately this is not what I consider
elementary. Which leads to my question: Is there an elementary example of a
set of primes for which the Dirichlet density exists but the natural density
does not? By an elementary example, I mean something that can be shown to be
an example by methods on the level of Chebyshev's and Mertens' theorems.

Thanks,
Paul
==============================================================================

From: hrubin@odds.stat.purdue.edu (Herman Rubin)
Subject: Re: Dirichlet density vs. natural density
Date: 24 Oct 2001 09:21:43 -0500
Newsgroups: sci.math

In article <9r50h6$r4h$1@cronkite.cc.uga.edu>,
Paul Pollack <paulp@arches.uga.edu> wrote:

[quote of most of previous message deleted --djr]

For this example, it is not very deep.  The prime number theorem
gives the number of primes in any sufficiently far out interval
of the form (c, kc), k>1, is approximately the integral of 
1/ln(x) over that interval, and this is between (k-1)c/ln(c) and
(k-1)c/ln(kc).  One can get use this to get asymptotic bounds
on the sums of 1/p^s in such intervals.  The upshot of this is
that one is looking, for s near 1, at the integral of 1/(x^s ln(x))
over the intervals in which primes are taken compared to that 
over all x larger than some fixed value.  The slow variation of
ln(x) enables the proof to be finished.

-- 
This address is for information only.  I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
hrubin@stat.purdue.edu         Phone: (765)494-6054   FAX: (765)494-0558