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

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  • julienbenney
    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
    Message 1 of 9 , Oct 26, 2009
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      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 prime of the form 4n+1 is a positive term
      - the reciprocal of every prime of the form 4n+3 is a negative term
      - the reciprocal of 2 is a positive term, both because it is like primes of the form 4n+1 in being expressible as two squares and because such a result I imagine as necessary to give a positive sum to the whole series

      The first fifty terms of the series, to give you the idea, would be:

      (1/2)-(1/3)+(1/5)-(1/7)-(1/11)+(1/13)+(1/17)-(1/19)-(1/23)+(1/29)-(1/31)+(1/37)+(1/41)-(1/43)-(1/47)+(1/53)-(1/59)+(1/61)-(1/67)-(1/71)+(1/73)-(1/79)-(1/83)+(1/89)+(1/97)+(1/101)-(1/103)-(1/107)+(1/109)+(1/113)-(1/127)-(1/131)+(1/137)-(1/139)+(1/149)-(1/151)+(1/157)-(1/163)-(1/167)+(1/173)-(1/179)+(1/181)-(1/191)+(1/193)+(1/197)-(1/199)-(1/211)-(1/223)-(1/227)+(1/229)

      I have summed this series using a hand calculator as far as 2500 (the first 365 primes) and found that it appears to ultimately converge to some number between 0.165 and 0.170, though sequences of positive and negative terms can be found if you look above at the early terms and make the ultimate sum hard to estimate without software which I cannot access.

      By analogy with the alternating harmonic series and the fact that primes of form 4n+1 and 4n+3 will be equal in number, I have little doubt this series must (conditionally) converge. I have long wondered if there is a formula for the sum. I have often imagined the sum in terms of ln(lnk) but realise this is logically unlikely.

      Can anyone confirm that this quasi-alternating series really does converge and provide some information about the sum?
    • marku606
      ... 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
      Message 2 of 9 , Oct 26, 2009
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        --- In primenumbers@yahoogroups.com, "julienbenney" <jpbenney@...> wrote:
        >
        > Can anyone confirm that this quasi-alternating series really does converge and provide some information about the sum?
        >

        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 =~.165025

        It looks like it converges as would be expected. That's about all I can tell.
        Mark
      • andrew_j_walker
        ... 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
        Message 3 of 9 , Oct 28, 2009
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          --- In primenumbers@yahoogroups.com, "marku606" <mark.underwood@...> wrote:
          >
          >
          > --- In primenumbers@yahoogroups.com, "julienbenney" <jpbenney@> wrote:
          > >
          > > Can anyone confirm that this quasi-alternating series really does converge and provide some information about the sum?
          > >
          >
          > 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 =~.165025
          >
          > It looks like it converges as would be expected. That's about all I can tell.
          > Mark
          >

          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 (ie your sum and this sum add to 0.5)

          Andrew
        • djbroadhurst
          ... Here are the first 300 digits:
          Message 4 of 9 , Oct 28, 2009
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            --- 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:

            0.165018674700006818936682878512456426220024619244922951891979421154777156728115889375100368970196654607513480004558629974909225560059749006034770337666395686419713981470176652055071996701959854658001361210365009440004122607075061098971770036230099534207416924580259659789823471141661125291034589613715

            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
          • marku606
            ... 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
            Message 5 of 9 , Oct 28, 2009
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              --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:

              > Here are the first 300 digits:
              >
              > 0.165018674700006818936682878512456426220024619244922951891979421154777156728115889375100368970196654607513480004558629974909225560059749006034770337666395686419713981470176652055071996701959854658001361210365009440004122607075061098971770036230099534207416924580259659789823471141661125291034589613715
              >
              > 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
              >

              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 digits of this number. Surely that requires an ungodly number of primes to go through, which I can't reconcile with 174 milliseconds. Clearly, I'm missing something.
            • djbroadhurst
              ... I used only primes p
              Message 6 of 9 , Oct 29, 2009
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                --- In primenumbers@yahoogroups.com,
                "marku606" <mark.underwood@...> wrote:

                > I can't fathom how it can take merely 174 milliseconds to
                > compute 300 digits of this number. Surely that requires an
                > ungodly number of primes to go through, which I can't reconcile
                > with 174 milliseconds.

                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),
                sumalt(k=0,(-1)^k/(2*k+1)^s)));
                forprime(p=3,A,L=L*(1-(if(p%4==1,1,-1)/p)^s));
                S=S+moebius(s)/s*log(L));S}

                \p300
                S=ans(400,120)
                print(gettime" milliseconds")

                ....
                149 milliseconds

                David
              • djbroadhurst
                ... 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
                Message 7 of 9 , Oct 29, 2009
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                  --- In primenumbers@yahoogroups.com,
                  "djbroadhurst" <d.broadhurst@...> wrote:

                  > I used only primes p < 400

                  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 accurately, I performed
                  1000 repetitions of the same calculation:

                  {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,if(issquarefree(s),
                  L=if(s==1,Pi/4,if(s%2==0,zeta(s)*(1-1/2^s),
                  sumalt(k=0,(-1)^k/(2*k+1)^s)));
                  forprime(p=3,A,L=L*(1-(if(p%4==1,1,-1)/p)^s));
                  S=S+moebius(s)/s*log(L)));S}

                  \p70
                  repeat=1000;gettime;
                  {for(n=1,repeat,S=ans(5,100);if(n==repeat,print(S);
                  print(ceil(gettime/repeat)" milliseconds")))}

                  0.1650186747000068189366828785124564262200246192449229518919794211547772
                  9 milliseconds

                  It is to be regretted that Henri Cohen never published his
                  wonderful preprint, which appears to date back to 1991.

                  Many have mined it, subsequently.

                  > Full many a gem of purest ray serene
                  > The dark unfathom'd caves of ocean bear.

                  David (in gratitude to Henri, for this and so much else)
                • djbroadhurst
                  ... 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:
                  Message 8 of 9 , Oct 30, 2009
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                    --- In primenumbers@yahoogroups.com,
                    "djbroadhurst" <d.broadhurst@...> wrote:

                    > To obtain 70 good decimal digits, in 9 milliseconds,
                    > it suffices to use only the first two odd primes p = 3,5.

                    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:

                    f(t)=(t/2/cos(t/2)+1-t/tan(t))/2;
                    default(seriesprecision,150);V=Vec(f(t));
                    L(s)=Pi^s*V[s]*(1-(s%2==0)/2^s);
                    \p70
                    print(1/2+sum(s=1,#V,moebius(s)/s*log(L(s))));

                    0.1650186747000068189366828785124564262200246192449229518919794211547772

                    David
                  • djbroadhurst
                    ... In fact, polling the odd primes really helps: I found 10,000 good digits in 7 minutes, by consulting the 24 odd primes p
                    Message 9 of 9 , Oct 30, 2009
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                      --- In primenumbers@yahoogroups.com,
                      "djbroadhurst" <d.broadhurst@...> wrote:

                      > how to obtain 70 good digits, using no odd prime at all

                      Off-list, Maximilian Hasler asked, amusingly:

                      > Can this method also be used to produce reliable
                      > statistics without using any data?

                      > If you can generalize your method, there will be
                      > no more polls needed at all...

                      > Do you realize that you are working towards a
                      > democracy where the government will know what the
                      > people wants even without them having to vote?

                      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
                      http://physics.open.ac.uk/~dbroadhu/cert/cohenbig.out

                      David (campaign for a discriminating democracy)
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