Re: What is Sum(1/p log p)?

Comment on notation: When you write 1/a*b you seem to mean 1/(a*b).
If so you should include the parentheses; 1/a*b is actually (1/a)*b
(or at least could be read that way.)

Correct. Again I'm sorry for lacking accuracy of notation.

Anyway, I don't know this second whether that series converges,
but it can be figured out using the Prime Number Theorem: If
pi(n) is the number of primes less than n then
pi(n) ~ n/log(n)
as n -> infinity (here a ~ b means that a/b tends to 1.)

I know this as "asymptotically equal" ("Landau notation"). And
I know PNT.

An easy way to figure out a lot of things like this is to
split the sum according to powers of 2 (see below for what
I mean by that). There are doubtless more elegant ways to
give the argument below, whatever it turns out to be,
but just splitting on powers of 2 gives a version that
requires a minumum of theoretical justification.

Beautiful idea. As I'm only an amateur, I had no idea how to settle
this.

Let's write a ~~ b to mean that a/b and b/a are both bounded
(or are bounded at least when n is large). Now PNT shows that
pi(2^n) ~~ 2^n/n,
and it follows that
(*) pi(2^(n+1)) - pi(2^n) ~~ 2^n/n.

I think I got that: pi(2^(n+1)) ~~ 2^(n+1)/(n+1), hence
pi(2^(n+1) - pi(2^n) ~~ 2^(n+1)/((n+1) - 2^n/n = 2^n*(n-1)/(n(n+1))
~~ 2^n/n as n tends to infinity (the +/- 1 is irrelevant for large n).
This means, the difference of number of primes between two consecutive
powers of 2 is of the same order as the number of primes below the
smaller power of 2. Roughly speaking, there are asymptotically equal
many primes between two consecutive powers of 2 and below the smaller
power of two.

If p is a prime and 2^n < p < 2^(n+1) then
log(p) ~~ n

Yep. That's again PNT, correct?

and hence
(**) p*log(p) ~~ n*2^n.

Got it.

Say s_n is the sum of 1/(p*log(p)), where p runs over
all the primes between 2^n and 2^(n+1). Then combining
(*) and (**) shows that
s_n ~~ (2^n/n)/(n*2^n) 1/n^2.

You seem to have missed an "="-sign. S_n ~~ (2^n/n)/(n*2^n) =
1/n^2. But the idea is clear: Show that s_n is asymptotically equal to
1/n^2 for 2^n < p < 2^(n+1), this is valid for all n and therefore the
sums are asymptotically equal for all n to infinity. Wonderful. Plainly
brillant!

So the sum of 1/(p*log(p)) where p runs over _all_ primes is
sum s_n ~~ sum(1/n^2) < infinity.

Excellent. Thank you a lot for this marvellous and short proof.

Hum ... numerically, would you agree to 1.442695... = 1/(ln 2) ~~ s_n
~~ pi²/6 = 1.6449341..., ie., 1/(ln 2) is the "leading term" of
the sum in question? There's only a tiny 0.2<something> missing ...

Thanks a lot again, sincerely yours, Buffalo.


.

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