What is Sum(1/p log p)?
buffalo
primes and their logarithms. To explain, I'm a number theory amateur
and simply personally interested.
As well known for centuries, the harmonic series diverges; for n<=x:
Sum(1/n) = log x + gamma + O(1/x). Even more interesting, the sum of
the reciprocals of the primes diverges, too: for p<=x Sum(1/p) = log
log x + C (C=.2615...). Now I stumbled across Sum(1/n*log n) = log log
x + C, which diverges too, but with a different C, of course.
Now the main question: Whats going on with Sum(1/p*log p)? (p<=x as
above)
As Sum(1/n*log n) ~ Sum(1/p) ~ log log x one might expect something
like Sum (1/p*log p) ~ Sum (1/n*log²n); and amazingly enough this last
sum is convergent! For 1 < a < t < x Integral (1/t*log²t) dt = -1/log
x (from a to x) = 1/log a (eg., a = 2, x runs to infinity).
Is there any knowledge about the values of this special reciprocal
prime sum? I did the work to sum both series partially, resulting in
p from 2 to # of p Value
Sum (1/p*log p) 10,000 1,229 1.53
100,000 9,592 1.55
Sum (1/n*log²n) n from 2 to Value
10,000 2.00117
20,000 2.00877
40,000 2.01537
60,000 2.01885
Not a big deal, I know, but as a "zero'th approximation" ... the
sum values are comparable in size. It strikes, that the first term of
both sums makes ca. 50% of the final sum value;
1/2 ln 2 = .72..., (1.55) 1/2 ln²2 = 1.05... (2,02).
Is there a closed analytical expression for these sum values? What do
the numerical values mean? What are the constants C in both sums? And
what is the main term of the sums? I expect some interesting number
theoretical information out of that, but I have no idea about the
details. And I haven't got a clue on the error estimate in both sums.
Some background infos: As Sum(1/p) diverges, one might say, roughly,
that there are "more" prime numbers than perfect squares, as
Sum(1/n²) = Pi²/6 (very nice formula, Euler 1737). Generally
speaking, Sum(1/n^s) is the Riemann Zeta Function Zeta(s), and step by
step one choosed n to be natural, rational, irrational and finally
complex (Riemann 1859). Zeta(s) converges for Re(s)>1, and it
"codes" the prime number distribution among the natural numbers by
its zeroes. The top of the mountain is, of course, the famous Riemann
Conjecture, reading that all nontrivial zeroes are on the critical line
for Re(s)=1/2.
I tried "RTFM", but failed in Hardy/Wright ("Intro to NumTh"),
Abramowicz ("Math Tables & Formulae") and some others. Any
literature hint would be gratefully acknowledged, too.
sincerely yours, buffalo.
PS.: Sorry for the bad typing, I am not familiar in using the special
Math signs in here ... *sigh*
David C. Ullrich
>I'd like to ask a possibly dumb question about sums of reciprocal
>primes and their logarithms. To explain, I'm a number theory amateur
>and simply personally interested.
>
>As well known for centuries, the harmonic series diverges; for n<=x:
>Sum(1/n) = log x + gamma + O(1/x). Even more interesting, the sum of
>the reciprocals of the primes diverges, too: for p<=x Sum(1/p) = log
>log x + C (C=.2615...). Now I stumbled across Sum(1/n*log n) = log log
>x + C, which diverges too, but with a different C, of course.
>
>Now the main question: Whats going on with Sum(1/p*log p)? (p<=x as
>above)
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.)
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.)
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.
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.
If p is a prime and 2^n < p < 2^(n+1) then
log(p) ~~ n
and hence
(**) p*log(p) ~~ n*2^n.
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.
So the sum of 1/(p*log(p)) where p runs over _all_ primes
is
sum s_n ~~ sum(1/n^2) < infinity.
************************
David C. Ullrich
buffalo
> 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.
Steven Finch
Henri Cohen evaluates this series in an unpublished paper entitled
"High precision computation of Hardy-Littlewood constants":
http://www.ufr-mi.u-bordeaux.fr/~cohen/
See also pp. 185-188 of my book "Mathematical Constants", where
the series appears in connection with a conjecture in the following
article:
P. Erdos, Note on sequences of integers no one of
which is divisible by any other, J. London Math. Soc.
10 (1935) 126-128.
Steve Finch
http://pauillac.inria.fr/algo/bsolve/
David C. Ullrich
[...]
I've moved my reply to sci.math, where all this seems more
appropriate. (Woulda replied there to begin with except
I didn't realize that question had been posted to both
groups.)
************************
David C. Ullrich
buffalo
enough: a math forum, which kills "equals" ...! *gosh*
@David: Sorry! I ran into the same trap ...
@Steven: Thank you a lot for your info! Actually I did not expect to
attract experts like David Ullrich, Steven Finch and Henri Cohen with
my little question (and my unexperienced behaviour here ... *blush*).
I'll have a look in your book. - Unfortunately I cannot read the
.dvi-file by Henri Cohen. Seems to be a 70-kB-Video ... anyway, I'll
try to get the appropriate software for it. And I'll try to catch
Erdos' article (will be difficult, as I do not have access to the
university library in Dortmund) - btw, did he use this sum in his
elementary proof of the PNT 1948/9? Last question: Is your book
available in the net (.pdf, .doc or something)?
Thank you all for your help, info and answers.
PS.: Seems to be normal to post with full real name here ... I thought,
one _must_ choose a nick, as is usual in other forums etc. Anyway, my
name is Wolfgang Rave, I'm from Germany (Hagen), and I've been NTh
amateur for some years now ...
Sincerely yours, Wolfgang.
David C. Ullrich
"research", just for the record:
On 13 Feb 2006 23:45:33 -0800, "buffalo" <in...@rave-edv.de> wrote:
>_We_ do not miss the "equal"-signs, its the newsgroup!! Amazingly
>enough: a math forum, which kills "equals" ...! *gosh*
I doubt that the typical reader has any idea what you're
talking about here. I know I don't.
>@David: Sorry! I ran into the same trap ...
Similarly here. I suspect you don't realize that people
reading this are not aware of some email or something
that's sitting in front of you...
>@Steven: Thank you a lot for your info! Actually I did not expect to
>attract experts like David Ullrich, Steven Finch and Henri Cohen
But just for the record, I'm far from being an expert on such
matters - if there was a point to my reply it was just to show
that one didn't really need to be an expert to see whether that
series converged, fairly standard convergence-testing techniques
suffice (the standard technique analogous to the argument I
gave is the "Cauchy Condensation Principle".)
> with
>my little question (and my unexperienced behaviour here ... *blush*).
>I'll have a look in your book. - Unfortunately I cannot read the
>.dvi-file by Henri Cohen. Seems to be a 70-kB-Video ... anyway, I'll
>try to get the appropriate software for it. And I'll try to catch
>Erdos' article (will be difficult, as I do not have access to the
>university library in Dortmund) - btw, did he use this sum in his
>elementary proof of the PNT 1948/9? Last question: Is your book
>available in the net (.pdf, .doc or something)?
>
>Thank you all for your help, info and answers.
>
>PS.: Seems to be normal to post with full real name here ... I thought,
>one _must_ choose a nick, as is usual in other forums etc. Anyway, my
>name is Wolfgang Rave, I'm from Germany (Hagen), and I've been NTh
>amateur for some years now ...
>
>Sincerely yours, Wolfgang.
************************
David C. Ullrich
Vladimir Drobot
This is perhaps beating on a dead horse, but here is even simpler proof that
the series Sum (1/(p ln(p)) over the primes converges:
First of all, the series Sum (1/(n (ln n)^2) converges, by applying the
standard integral test: The antiderivative of 1/(x (ln x)^2) is
-1/(ln x) which vanishes at infinity. Next, the nth prime p_n is about n
ln(n) This follows from the prime number theorem, of course, but is much
weaker.
Actually all that's needed is that p_n > 0.5 n ln(n), or any positive
constant replacing 0.5. There are many simple elementary proofs of that.
Since p_n > n we get p_n ln(p_n) > 0.5 n (ln (n)) (ln (n) = 0.5 n (ln n)^2
and the convergence follows from the standard comparison test.
Vladimir Drobot
Wolfgang Rave
@David: I'm awfully sorry, but I cannot see the equal-signs on my
screen. They're gone. In no thread here I could detect a single
"equal". Believe it or not, but it's true.
================================================================
The paper by Henri Cohen, recommended by Steven Finch (thanx again!),
reads roughly as follows. They compute the sum(1/(p^m log p) by
I(m) = Integral log(Zeta(t)) dt [from m to infty].
This integral may be computed by means of Euler-MacLaurin summation.
However, this method is not valid for m=1, which is the case in
question here, as f(t)=log(Zeta(t)) and its derivatives are not defined
for t=1. Instead, they use
f_2(t) = Sum[(-1)^n n^(-t)] = (1-2^(1-t))Zeta(t)
which is welldefined for t>0 and can be computed via Euler-MacLaurin.
So they end up with
I(1) = Zeta(2)/ln 2 + Integral f_2(t) dt [1..infty]
Now the numerical values.
Sum(1/(p ln p)) = 1.636616323351260...
(which is very close to Zeta(2) = Pi²/6 = 1,6449340...).
Pi²/(6 ln 2) = 2.373138220846...
I(1) = 1.79756995862873940...
Their difference, Integral(f_2(t))dt [1...oo], is almost exactly Gamma;
Zeta(2)/ln 2 - I(1) = 0.575..., there's only a tiny 0.002 missing.
Is this mere coincidence or a mathematical relation?
Wolfgang.