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

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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
Message 1 of 9 , Oct 26, 2009
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?
• ... 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
--- 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
Message 3 of 9 , Oct 28, 2009
--- 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
• ... Here are the first 300 digits:
Message 4 of 9 , Oct 28, 2009
"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
• ... 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

> 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.
• ... I used only primes p
Message 6 of 9 , Oct 29, 2009
"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
• ... 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

> 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)
• ... 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

> 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
• ... 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

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

> 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