- 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

Michael Omstead - 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

To: primenumbers@yahoogroups.com; Andy Swallow

Subject: RE: [PrimeNumbers] Prime number reciprocal sum

> From: Andy Swallow [mailto:umistphd2003@...]

Ok: you are old fashioned.

> 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...

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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