## Prime number reciprocal sum

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• By no means am I an expert on prime numbers, having only had my first taste of number theory a few semesters ago, but would it be reasonable to conjecture that
Message 1 of 9 , Sep 19 2:49 PM
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By no means am I an expert on prime numbers, having only had my first
taste of number theory a few semesters ago, but would it be
reasonable to conjecture that the sum of the reciprocals of primes
(starting at 2 and increasing thereafter) approaches and eventually
equals a finite number, i.e.

inf
(sum) 1/p = r for some r in the real numbers ??
p=2
p prime

Any proof/disproof/thoughts would be appreciated! If someone knew
the actual number (assuming it exists), that would be most
fascinating to me.

Thank you
• ... Umm, see any introductory book on number theory, and look for Merten s theorem. If you let s(x) be the sum of all reciprocals of primes p
Message 2 of 9 , Sep 19 2:54 PM
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On Fri, Sep 19, 2003 at 09:49:39PM -0000, Mike wrote:
> By no means am I an expert on prime numbers, having only had my first
> taste of number theory a few semesters ago, but would it be
> reasonable to conjecture that the sum of the reciprocals of primes
> (starting at 2 and increasing thereafter) approaches and eventually
> equals a finite number, i.e.

Umm, see any introductory book on number theory, and look for Merten's
theorem. If you let s(x) be the sum of all reciprocals of primes p<=x,
then

s(x) = loglog x + C + o(1),

where C is some constant, and o(1) denotes a term which approaches zero
as x-->oo. So no, it doesn't converge. It diverges very slowly.

Andy
• In fact, the easiest way I know to prove that the prime numbers are an infinity is that the sum of their reciprocals is divergent. But you know, if you sum the
Message 3 of 9 , Sep 19 3:06 PM
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In fact, the easiest way I know to prove that the prime numbers are an infinity is that the sum of their reciprocals is divergent.

But you know, if you sum the reciprocals of the twin primes, then the sum is convergent! And the real number that is the limit is called Brun's constant (go to the Chris Caldwell's glossary to know more).

Jose Brox

----- Original Message -----
From: Andy Swallow
Sent: Friday, September 19, 2003 11:54 PM
Subject: Re: [PrimeNumbers] Prime number reciprocal sum

On Fri, Sep 19, 2003 at 09:49:39PM -0000, Mike wrote:
> By no means am I an expert on prime numbers, having only had my first
> taste of number theory a few semesters ago, but would it be
> reasonable to conjecture that the sum of the reciprocals of primes
> (starting at 2 and increasing thereafter) approaches and eventually
> equals a finite number, i.e.

Umm, see any introductory book on number theory, and look for Merten's
theorem. If you let s(x) be the sum of all reciprocals of primes p<=x,
then

s(x) = loglog x + C + o(1),

where C is some constant, and o(1) denotes a term which approaches zero
as x-->oo. So no, it doesn't converge. It diverges very slowly.

Andy

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• ... Call me old fashioned, but surely Euclid s argument is a tad better and easier for proving the infinitude of the primes... Andy
Message 4 of 9 , Sep 19 3:13 PM
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On Sat, Sep 20, 2003 at 12:06:18AM +0200, Jose Ram?n Brox wrote:
> In fact, the easiest way I know to prove that the prime numbers are an
> infinity is that the sum of their reciprocals is divergent.

Call me old fashioned, but surely Euclid's argument is a tad better and
easier for proving the infinitude of the primes...

Andy
• ... Ok: you are old fashioned. I like this proof of the infinitude of primes. First, observe that there are an infinite number of distinct integers of the form
Message 5 of 9 , Sep 23 4:07 AM
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> From: Andy Swallow [mailto:umistphd2003@...]
> Sent: 19 September 2003 23:13
> On Sat, Sep 20, 2003 at 12:06:18AM +0200, Jose Ram?n Brox wrote:
> > In fact, the easiest way I know to prove that the prime
> numbers are an
> > infinity is that the sum of their reciprocals is divergent.
>
> Call me old fashioned, but surely Euclid's argument is a tad
> better and
> easier for proving the infinitude of the primes...

Ok: you are old fashioned.

I like this proof of the infinitude of primes.

First, observe that there are an infinite number of distinct integers
of the form F_n = 2^2^n+1 where n is an integer >= 0.

Next, observe that F_{n+1} - 2 = (2^2^{n+1} - 1) and that this
expression forms the difference between two squares. We know how to
factor such expressions.

In particular, F_{n+1} - 2 = (2^2^n + 1) * (2^2^n - 1).

But the terms in the product are just F_n and F_n - 2.

So, F_{n+1} is coprime to F_n because the latter divides F_{n+1} - 2.

We've now demonstrated an infinite series of integers, none of which
share a prime factor, so there must be an infinite number of primes.

Take a bow, M. Pierre Fermat.

Paul
• In a message dated 23/09/03 12:09:28 GMT Daylight Time, ... Have you? F_{n+1} is coprime to F_n, and F_{n+2} is coprime to F_{n+1), but is F_{n+2} necessarily
Message 6 of 9 , Sep 23 4:29 AM
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In a message dated 23/09/03 12:09:28 GMT Daylight Time,
pleyland@... writes:

>So, F_{n+1} is coprime to F_n because the latter divides F_{n+1} - 2.
>
>We've now demonstrated an infinite series of integers, none of which
>share a prime factor, so there must be an infinite number of primes.

Have you?
F_{n+1} is coprime to F_n, and F_{n+2} is coprime to F_{n+1), but is F_{n+2}
necessarily coprime to F_n?

Mike

[Non-text portions of this message have been removed]
• ... Rather amusing really. But of course, the proof that Paul presented can be more properly written out to show that F_n and F_{n+k} have no common factors
Message 7 of 9 , Sep 23 4:58 AM
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> >So, F_{n+1} is coprime to F_n because the latter divides F_{n+1} - 2.
> >
> >We've now demonstrated an infinite series of integers, none of which
> >share a prime factor, so there must be an infinite number of primes.
>
> Have you?
> F_{n+1} is coprime to F_n, and F_{n+2} is coprime to F_{n+1), but is F_{n+2}
> necessarily coprime to F_n?

Rather amusing really. But of course, the proof that Paul presented can
be more properly written out to show that F_n and F_{n+k} have no common
factors for any k>0.

So that a proof using Fermat numbers. Both that method and Euclid's can
be used to estimate the number of primes up to a certain point. And both
estimates are terrible when compared to more 'modern' methods. But just
for the record, the estimate given by Euclid's argument is a lot closer
to the truth than that given by considering Fermat numbers...

Andy
• Yes it is. Think about it. To get you started, try factoring one of the terms in the product I gave you. Paul ________________________________ From:
Message 8 of 9 , Sep 23 4:59 AM
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Yes it is. Think about it.

To get you started, try factoring one of the terms in the product I gave you.

Paul

________________________________

From: Mikeoakes2@... [mailto:Mikeoakes2@...]
Sent: 23 September 2003 12:29
To: Paul Leyland
Subject: Re: [PrimeNumbers] Prime number reciprocal sum

In a message dated 23/09/03 12:09:28 GMT Daylight Time, pleyland@... writes:

>So, F_{n+1} is coprime to F_n because the latter divides F_{n+1} - 2.
>
>We've now demonstrated an infinite series of integers, none of which
>share a prime factor, so there must be an infinite number of primes.

Have you?
F_{n+1} is coprime to F_n, and F_{n+2} is coprime to F_{n+1), but is F_{n+2} necessarily coprime to F_n?

Mike

[Non-text portions of this message have been removed]
• Just wondering. Is the following argument right? Consider a continuous list of primes starting 2 ie 2,3,5,7..............p 1. we can define composites and
Message 9 of 9 , Sep 23 6:59 AM
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Just wondering. Is the following argument right?

Consider a continuous list of primes starting 2 ie 2,3,5,7..............p

1. we can define composites and primes in the range >p <p^2

2. there will always be at least one prime in the range >p <p^2.
Therefore, there must be an infinitude of prime numbers.

Navid

-----Original Message-----
From: Paul Leyland [mailto:pleyland@...]
Sent: 23 September 2003 12:08
Subject: RE: [PrimeNumbers] Prime number reciprocal sum

> From: Andy Swallow [mailto:umistphd2003@...]
> Sent: 19 September 2003 23:13
> On Sat, Sep 20, 2003 at 12:06:18AM +0200, Jose Ram?n Brox wrote:
> > In fact, the easiest way I know to prove that the prime
> numbers are an
> > infinity is that the sum of their reciprocals is divergent.
>
> Call me old fashioned, but surely Euclid's argument is a tad
> better and
> easier for proving the infinitude of the primes...

Ok: you are old fashioned.

I like this proof of the infinitude of primes.

First, observe that there are an infinite number of distinct integers
of the form F_n = 2^2^n+1 where n is an integer >= 0.

Next, observe that F_{n+1} - 2 = (2^2^{n+1} - 1) and that this
expression forms the difference between two squares. We know how to
factor such expressions.

In particular, F_{n+1} - 2 = (2^2^n + 1) * (2^2^n - 1).

But the terms in the product are just F_n and F_n - 2.

So, F_{n+1} is coprime to F_n because the latter divides F_{n+1} - 2.

We've now demonstrated an infinite series of integers, none of which
share a prime factor, so there must be an infinite number of primes.

Take a bow, M. Pierre Fermat.

Paul

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