## RE: [PrimeNumbers] 1/p = p, always, especially whan p is prime

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• I would comment, except for the site seems to have disappeared. http://algo.inria.fr/banderier/Seminar/Vardi/index.html ... i.e. we sum 1/p for p in ak+b, i.e.
Message 1 of 3 , May 1, 2002
I would comment, except for the site seems to have disappeared.

http://algo.inria.fr/banderier/Seminar/Vardi/index.html

>I meant sum over the primes.

i.e. we sum 1/p for p in ak+b, i.e. p over ak+b.

>Your second error was to believe that INRIA can claim 'the sum
>of an infinite numbers of positive increasing values is equal to
>infinity'.

My third error must therefore have been to ....

While I have no problem believing that every sum of inverted primes from an
AP is infinite, I do not see how the RHS implies the LHS without further and
extensive work.

>At first glance, it doesn't look trivial

It's not. This would say a great deal about the distribution of primes in
AP's, and perhaps more importantly, offer insight as to the 'cut-off' point
that determines whether a sequence is finite or infinite.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com

-----Original Message-----
From: Marcel Martin [mailto:znz@...]
Sent: 30 April 2002 22:31
Subject: Re: [PrimeNumbers] 1/p = p, always, especially whan p is prime

>I meant sum over the primes.

Yes, I had understood. And that's one of your two errors. What is
summed is the inverses of the primes, not the primes themselves.
Your second error was to believe that INRIA can claim 'the sum
of an infinite numbers of positive increasing values is equal to
infinity'.
The INRIA (very roughly, "National Institut for Research in Computer
Science") is not a bozo club. When one finds what looks like an error
in what they published, there are two possibilities:
1) this is a typo;
2) the reader doesn't quite understand what they wrote.

>But how does the statement that ak+b contains an infinite number
>of primes imply that sum(1/p) is infinite?

That's precisely what they wrote, a proof. At first glance, it
doesn't look trivial. And, always at first glance, I am not sure
to be able to understand it. I think that my knowledge about
the tools they used could be written on a postage stamp. A small
one.

Marcel Martin

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• DOES THE SUM OF THESE PRIMES MEAN THE SUM OF THE INVERSES ? That s precisely why you ironically wrote the immortal statement , because if we sum the
Message 2 of 3 , May 2, 2002
DOES "THE SUM OF THESE PRIMES" MEAN "THE SUM OF THE INVERSES"?
That's precisely why you ironically wrote 'the immortal statement',
because if we sum the primes, in that case, of course, the sum is
trivially infinite.

It was a small, and fairly self-correcting typo, which I did correct in a
following post.

As for the RHS implying the LHS, I have yet to comprehend its subtleties.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com

-----Original Message-----
From: Marcel Martin [mailto:znz@...]
Sent: 01 May 2002 23:32
Subject: Re: [PrimeNumbers] 1/p = p, always, especially whan p is prime

>>I meant sum over the primes.
>i.e. we sum 1/p for p in ak+b, i.e. p over ak+b.

That's incredible. In your first post, you wrote this

>Apparently, the fact that there are an infinite number of primes in a+kq
>implies the sum of these primes is infinite.
^^^^^^^^^^^^^^^^^^^^^^^

DOES "THE SUM OF THESE PRIMES" MEAN "THE SUM OF THE INVERSES"?
That's precisely why you ironically wrote 'the immortal statement',
because if we sum the primes, in that case, of course, the sum is
trivially infinite.

You was wrong. Period. And now, as usual, you are trying to get
out of the mess by trying to change what you said. You did that
with me about Wilson theorem, you did that with Jud McCranie about
Godel theorem. You always do that.
Is it so difficult to say 'Ok, I was wrong. Next chapter.'?

Marcel Martin

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