- --- 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?

David Broadhurst

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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). - --- In primenumbers@yahoogroups.com, "David Broadhurst"

<d.broadhurst@...> wrote:

> 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

David Broadhurst

----------------------------------------------------------------

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). > 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> http://pi.lacim.uqam.ca/eng/table_en.html

And

> http://pi.lacim.uqam.ca/piDATA/plogp.txt

http://www.research.att.com/~njas/sequences/A137245

of course.

Best,

Andrey