primenumbers : Message: Re: Is this a convergent series and if so what is its sum?

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Is this a convergent series and if so what is its sum?

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Message #21083 of 21174 |

Re: Is this a convergent series and if so what is its sum?

--- In primenumbers@yahoogroups.com,
"marku606" <mark.underwood@...> wrote:

> For primes up to
> 10^3 : sum =~.166803
> 10^4 : sum =~.166329
> 10^5 : sum =~.165365
> 10^6 : sum =~.165021
> 10^7 : sum =~.165036
> 10^8 : sum =~.165025
>
> It looks like it converges as would be expected.

Here are the first 300 digits:

0.165018674700006818936682878512456426220024619244922951891979421154777156728115\
88937510036897019665460751348000455862997490922556005974900603477033766639568641\
97139814701766520550719967019598546580013612103650094400041226070750610989717700\
36230099534207416924580259659789823471141661125291034589613715

which took 174 milliseconds to compute, using the method of

Henri Cohen,
High Precision Computation of Hardy-Littlewood Constants,
http://www.math.u-bordeaux.fr/~cohen/hardylw.dvi

David

Thu Oct 29, 2009 2:26 am

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"djbroadhurst" <d.broadhurst@...>

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Message #21083 of 21174 |

Expand Messages	Author	Sort by Date
One thing I have often thought about is trying to build a quasi-alternating series out of the reciprocals of the primes, so that: - the reciprocal of every...	julienbenney	Oct 27, 2009 1:34 am
... Here's some data: For primes up to 10^3 : sum =~.166803 10^4 : sum =~.166329 10^5 : sum =~.165365 10^6 : sum =~.165021 10^7 : sum =~.165036 10^8 : sum...	marku606	Oct 27, 2009 3:19 am
... The sum is a variation on sum (9) at http://mathworld.wolfram.com/PrimeSums.html It starts (1/3)-(1/5)+(1/7)+(1/11)-(1/13)-(1/17) and equals 0.3349813253...	andrew_j_walker	Oct 28, 2009 8:47 am
... Here are the first 300 digits: ...	djbroadhurst	Oct 29, 2009 2:27 am
... Rats, my computer doesn't know what to do with a dvi file. While in my ignorance, I can't fathom how it can take merely 174 milliseconds to compute 300...	marku606	Oct 29, 2009 6:15 am
... I used only primes p < 400 {ans(A,N)=local(S=1/2.,L); forprime(p=3,A,S=S+if(p%4==1,1,-1)/p); for(s=1,N,L=if(s==1,Pi/4,if(s%2==0,zeta(s)*(1-1/2^s), ...	djbroadhurst	Oct 29, 2009 10:40 am
... To obtain, 70 good decimal digits, in 9 milliseconds, it suffices to use only the first two odd primes p = 3,5. This is so fast that, to time it...	djbroadhurst	Oct 29, 2009 5:09 pm
... Finally, here is how to obtain 70 good digits, using no odd prime at all. Observe that the formulas below involve only the numbers 2 and Pi: ...	djbroadhurst	Oct 30, 2009 3:41 pm
... In fact, polling the odd primes really helps: I found 10,000 good digits in 7 minutes, by consulting the 24 odd primes p < 100. The result is in ...	djbroadhurst	Oct 30, 2009 7:46 pm

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