## Re: sum(1/(p(i)*log(p(i))), i=1...infinity)=Pi/2 ????

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• ... Certainly not. I believe that the sum exceeds 1.6366. To see this, do the sum over primes p
Message 1 of 4 , Jan 12, 2009
--- In primenumbers@yahoogroups.com, "AlexP" <apovolot@...> wrote:

> sum(1/(p(i)*log(p(i))), i=1...infinity)=Pi/2 ????

Certainly not. I believe that the sum exceeds 1.6366.

To see this, do the sum over primes p <= N explicitly and then use
the prime number theorem to estimate the residuum as 1/log(N).

S(N)=local(s=0);forprime(p=2,N,s=s+1/(p*log(p)));s+1/log(N);
for(k=6,9,print(S(10^k)))

1.63661864
1.63661681
1.63661652
1.63661639

Perhaps Andrey Kulsha can provide a better estimate?

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• ... Steve Finch gives the value ... in http://tinyurl.com/9x29z2 David Broadhurst ... The Open University is incorporated by Royal Charter (RC 000391), an
Message 2 of 4 , Jan 12, 2009

> do the sum over primes p <= N explicitly and then use
> the prime number theorem to estimate the residuum as 1/log(N)
...
> 1.63661639

Steve Finch gives the value

> 1.6366163233...

in http://tinyurl.com/9x29z2

----------------------------------------------------------------
The Open University is incorporated by Royal Charter (RC 000391),
an exempt charity in England and Wales and
a charity registered in Scotland (SC 038302).
• ... http://pi.lacim.uqam.ca/eng/table_en.html http://pi.lacim.uqam.ca/piDATA/plogp.txt ... Best, Andrey
Message 3 of 4 , Jan 12, 2009
> Perhaps Andrey Kulsha can provide a better estimate?

http://pi.lacim.uqam.ca/eng/table_en.html
http://pi.lacim.uqam.ca/piDATA/plogp.txt

:-)

Best,

Andrey
• ... And http://www.research.att.com/~njas/sequences/A137245 of course. Best, Andrey
Message 4 of 4 , Jan 12, 2009
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