Prime numbers and primality testing is a Restricted Group with 1120 members.
 Prime numbers and primality testing

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 1120 members
RE: [PrimeNumbers] primeth primes
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 I believe that your sum does not converge.
Yours is s =
1/3+1/5+1/11+1/17+1/31+1/41+1/59+1/67+1/83+1/109+1/127+1/157+1/179+1/191+1/2
11+1/241+1/277+1/283+1/331+...
Notice the difference in the denominators
2 6 14 10 18 8 16 26 16 30 22 12 20 30 36 6 50 ...
You will see that we are not guaranteed that denominators are decreasing at
a continuing rate as we are for the reciprocals of the sqauares, which
converges:
1/2^2+1/2^3+1/2^4+...
whose differences are 5 7 9 11 13 15 ... where we are guaranteed
increasingly decreasing denominators.
Milton L. Brown
> [Original Message]
> From: Werner D. Sand <Theo.3.1415@...>
> To: <primenumbers@yahoogroups.com>
> Date: 11/23/2005 10:26:45 AM
> Subject: [PrimeNumbers] primeth primes
>
> The sum of the reciprocal primeth primes p(p(n))
>
> s = 1/3 + 1/5 + 1/11 + 1/17 + 1/31 + � seems to
>
> converge to 1.
>
>
>
>
>
>
>
> Unsubscribe by an email to: primenumbersunsubscribe@yahoogroups.com
> The Prime Pages : http://www.primepages.org/
>
>
> Yahoo! Groups Links
>
>
>
>
>
>  Certainly the denominators need not decrease at a continuing rate for
convergence. Take
1/3 + 1/4 + 1/9 + 1/10 + 1/27 + 1/28 + 1/81 + 1/82 + ...
which is about 0.90406.
I believe even the Sophie Germain prime reciprocals converge  I'm
not sure of this, of course.
1/2 + 1/3 + 1/5 + 1/11 + 1/23 + ...
Asymptotically, p_k ~ k(ln k + ln ln k + 1), so p_{p_k} ~ k(ln k + ln
ln k + 1)(ln (k(ln k + ln ln k + 1)) + ln ln (k(ln k + ln ln k + 1)) +
1). In fact, this is a strict lower bound on p_k (Dusart 1999), so it
seems this could be used to prove the sum converges.
Lacking a formal proof, I'll take the most significant term k(ln k)^2.
Integrate 2...infty dk/(k(ln k)^2): with x=ln k, this is integral ln
2...infty dx/x^2 ~= 0.7213475.
Does someone want to tighten this up or check it over?
Charles Greathouse
 In primenumbers@yahoogroups.com, "Milton Brown" <miltbrown@e...>
wrote:> I believe that your sum does not converge.
1/3+1/5+1/11+1/17+1/31+1/41+1/59+1/67+1/83+1/109+1/127+1/157+1/179+1/191+1/2
>
> Yours is s =
>
> 11+1/241+1/277+1/283+1/331+...
decreasing at
>
> Notice the difference in the denominators
>
> 2 6 14 10 18 8 16 26 16 30 22 12 20 30 36 6 50 ...
>
> You will see that we are not guaranteed that denominators are
> a continuing rate as we are for the reciprocals of the sqauares, which
> converges:
>
> 1/2^2+1/2^3+1/2^4+...
>
> whose differences are 5 7 9 11 13 15 ... where we are guaranteed
> increasingly decreasing denominators.
>
> Milton L. Brown
>
> > [Original Message]
> > From: Werner D. Sand <Theo.3.1415@w...>
> > To: <primenumbers@yahoogroups.com>
> > Date: 11/23/2005 10:26:45 AM
> > Subject: [PrimeNumbers] primeth primes
> >
> > The sum of the reciprocal primeth primes p(p(n))
> > s = 1/3 + 1/5 + 1/11 + 1/17 + 1/31 + seems to
> > converge to 1.   In primenumbers@yahoogroups.com, "Milton Brown" <miltbrown@e...>
wrote:
> You will see that we are not guaranteed that denominators are
decreasing at
> a continuing rate
I really don't understand in what your above assertion has something to
do with the fact the serie actually converges or not ?
p(p(n)) ~ n*ln(n)^2 and 1/(n*ln(n)^2) serie converges so does 1/p(p(n)).
Or what, am I wrong ?
JL.  As I indicated, this sum does not converge.
The comparable sum to compare it to is the reciprocal of the squares.
Consider to the denominators compared to the squares.
3 5 11 17 31 41 59 67 83 10 127 157 179 191 211 241 277 283 331 ...
1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 384 361 400 ...
After 191<196 the squares remain larger so 1/196 < 1/191
Any series like this, where you select from an infinite number of primes,
will diverge.
Milton L. Brown> [Original Message]
1/3+1/5+1/11+1/17+1/31+1/41+1/59+1/67+1/83+1/109+1/127+1/157+1/179+1/191+1/2
> From: crgreathouse <crgreathouse@...>
> To: <primenumbers@yahoogroups.com>
> Date: 11/25/2005 4:55:44 AM
> Subject: [PrimeNumbers] Re: primeth primes
>
> Certainly the denominators need not decrease at a continuing rate for
> convergence. Take
>
> 1/3 + 1/4 + 1/9 + 1/10 + 1/27 + 1/28 + 1/81 + 1/82 + ...
>
> which is about 0.90406.
>
> I believe even the Sophie Germain prime reciprocals converge  I'm
> not sure of this, of course.
>
> 1/2 + 1/3 + 1/5 + 1/11 + 1/23 + ...
>
> Asymptotically, p_k ~ k(ln k + ln ln k + 1), so p_{p_k} ~ k(ln k + ln
> ln k + 1)(ln (k(ln k + ln ln k + 1)) + ln ln (k(ln k + ln ln k + 1)) +
> 1). In fact, this is a strict lower bound on p_k (Dusart 1999), so it
> seems this could be used to prove the sum converges.
>
> Lacking a formal proof, I'll take the most significant term k(ln k)^2.
> Integrate 2...infty dk/(k(ln k)^2): with x=ln k, this is integral ln
> 2...infty dx/x^2 ~= 0.7213475.
>
> Does someone want to tighten this up or check it over?
>
> Charles Greathouse
>
>  In primenumbers@yahoogroups.com, "Milton Brown" <miltbrown@e...>
> wrote:
> > I believe that your sum does not converge.
> >
> > Yours is s =
> >
>
> > 11+1/241+1/277+1/283+1/331+...
> >
> > Notice the difference in the denominators
> >
> > 2 6 14 10 18 8 16 26 16 30 22 12 20 30 36 6 50 ...
> >
> > You will see that we are not guaranteed that denominators are
> decreasing at
> > a continuing rate as we are for the reciprocals of the sqauares, which
> > converges:
> >
> > 1/2^2+1/2^3+1/2^4+...
> >
> > whose differences are 5 7 9 11 13 15 ... where we are guaranteed
> > increasingly decreasing denominators.
> >
> > Milton L. Brown
> >
> > > [Original Message]
> > > From: Werner D. Sand <Theo.3.1415@w...>
> > > To: <primenumbers@yahoogroups.com>
> > > Date: 11/23/2005 10:26:45 AM
> > > Subject: [PrimeNumbers] primeth primes
> > >
> > > The sum of the reciprocal primeth primes p(p(n))
> > > s = 1/3 + 1/5 + 1/11 + 1/17 + 1/31 + � seems to
> > > converge to 1.
>
>
>
>
>
>
> Unsubscribe by an email to: primenumbersunsubscribe@yahoogroups.com
> The Prime Pages : http://www.primepages.org/
>
>
> Yahoo! Groups Links
>
>
>
>
>
>  Your number 0.7213475 is certainly not correct.
The sum of the first 11 terms
1/3+1/5+1/11+1/17+1/31+1/41+1/59+1/67+1/83+1/109+1/127
is 0.80071164
Milton L. Brown
> [Original Message]
1/3+1/5+1/11+1/17+1/31+1/41+1/59+1/67+1/83+1/109+1/127+1/157+1/179+1/191+1/2
> From: crgreathouse <crgreathouse@...>
> To: <primenumbers@yahoogroups.com>
> Date: 11/25/2005 4:55:44 AM
> Subject: [PrimeNumbers] Re: primeth primes
>
> Certainly the denominators need not decrease at a continuing rate for
> convergence. Take
>
> 1/3 + 1/4 + 1/9 + 1/10 + 1/27 + 1/28 + 1/81 + 1/82 + ...
>
> which is about 0.90406.
>
> I believe even the Sophie Germain prime reciprocals converge  I'm
> not sure of this, of course.
>
> 1/2 + 1/3 + 1/5 + 1/11 + 1/23 + ...
>
> Asymptotically, p_k ~ k(ln k + ln ln k + 1), so p_{p_k} ~ k(ln k + ln
> ln k + 1)(ln (k(ln k + ln ln k + 1)) + ln ln (k(ln k + ln ln k + 1)) +
> 1). In fact, this is a strict lower bound on p_k (Dusart 1999), so it
> seems this could be used to prove the sum converges.
>
> Lacking a formal proof, I'll take the most significant term k(ln k)^2.
> Integrate 2...infty dk/(k(ln k)^2): with x=ln k, this is integral ln
> 2...infty dx/x^2 ~= 0.7213475.
>
> Does someone want to tighten this up or check it over?
>
> Charles Greathouse
>
>  In primenumbers@yahoogroups.com, "Milton Brown" <miltbrown@e...>
> wrote:
> > I believe that your sum does not converge.
> >
> > Yours is s =
> >
>
> > 11+1/241+1/277+1/283+1/331+...
> >
> > Notice the difference in the denominators
> >
> > 2 6 14 10 18 8 16 26 16 30 22 12 20 30 36 6 50 ...
> >
> > You will see that we are not guaranteed that denominators are
> decreasing at
> > a continuing rate as we are for the reciprocals of the sqauares, which
> > converges:
> >
> > 1/2^2+1/2^3+1/2^4+...
> >
> > whose differences are 5 7 9 11 13 15 ... where we are guaranteed
> > increasingly decreasing denominators.
> >
> > Milton L. Brown
> >
> > > [Original Message]
> > > From: Werner D. Sand <Theo.3.1415@w...>
> > > To: <primenumbers@yahoogroups.com>
> > > Date: 11/23/2005 10:26:45 AM
> > > Subject: [PrimeNumbers] primeth primes
> > >
> > > The sum of the reciprocal primeth primes p(p(n))
> > > s = 1/3 + 1/5 + 1/11 + 1/17 + 1/31 + � seems to
> > > converge to 1.
>
>
>
>
>
>
> Unsubscribe by an email to: primenumbersunsubscribe@yahoogroups.com
> The Prime Pages : http://www.primepages.org/
>
>
> Yahoo! Groups Links
>
>
>
>
>
>  What has to be neatened up? p(p(n)) is on the order of n*log(n)^2
and the integral test shows 1/[nlog(n)^2] convereges so sum 1/p(p
(n)) converegs.
Adam
 In primenumbers@yahoogroups.com, "crgreathouse"
<crgreathouse@g...> wrote:>
for
> Certainly the denominators need not decrease at a continuing rate
> convergence. Take
ln
>
> 1/3 + 1/4 + 1/9 + 1/10 + 1/27 + 1/28 + 1/81 + 1/82 + ...
>
> which is about 0.90406.
>
> I believe even the Sophie Germain prime reciprocals converge  I'm
> not sure of this, of course.
>
> 1/2 + 1/3 + 1/5 + 1/11 + 1/23 + ...
>
> Asymptotically, p_k ~ k(ln k + ln ln k + 1), so p_{p_k} ~ k(ln k +
> ln k + 1)(ln (k(ln k + ln ln k + 1)) + ln ln (k(ln k + ln ln k +
1)) +
> 1). In fact, this is a strict lower bound on p_k (Dusart 1999),
so it
> seems this could be used to prove the sum converges.
^2.
>
> Lacking a formal proof, I'll take the most significant term k(ln k)
> Integrate 2...infty dk/(k(ln k)^2): with x=ln k, this is integral
ln
> 2...infty dx/x^2 ~= 0.7213475.
>
> Does someone want to tighten this up or check it over?
>
> Charles Greathouse  Message 17230 of 17232 indicates that pn>=n(log(n)+log(log(n))+1) so
we conclude that pn>n*log(n) and then "compose this with itself" to
get p(pn)>pn*log(pn)>n*log(n)*log[n*log(n)]>n*log(n)^2 so 1/p(pn)<1/
[n*log(n)^2].
The tail end of the sum sum(1/p(pn),n=k+1..infinity) is bounded above
by sum(1/[n*log(n)^2],n=k+1..infinity) which is bounded above by
integral(1/[x*log(x)^2],x=k..infinity) which is 1/[3*log(k)^3].
You add up the first four terms 1/p(p1)+1/p(p2)+1/p(p3)+1/p(p4)=1/p(2)
+1/p(3)+1/p(5)+1/p(7)=1/3+1/5+1/11+1/17<.68306596 and then you have
the error term of int(1/[x*log(x)^2],x=3..infinity)=1/[3*log(3)^3]
<.25138849 so the whole sum is <.68306596+.25138849=.93445445.
For n=7 I get an upper bound of .814611570570018210618938079323
Adam
 In primenumbers@yahoogroups.com, "Werner D. Sand"
<Theo.3.1415@w...> wrote:>
> The sum of the reciprocal primeth primes p(p(n))
>
> s = 1/3 + 1/5 + 1/11 + 1/17 + 1/31 + seems to
>
> converge to 1.
>  Hi, All!
>
Sum_{n=1}^{1000}Prime(Prime(n))=0.9416291650691093498004588771001232327
> You add up the first four terms 1/p(p1)+1/p(p2)+1/p(p3)+1/p(p4)=1/p(2)
> +1/p(3)+1/p(5)+1/p(7)=1/3+1/5+1/11+1/17<.68306596 and then you have
> the error term of int(1/[x*log(x)^2],x=3..infinity)=1/[3*log(3)^3]
> <.25138849 so the whole sum is <.68306596+.25138849=.93445445.
>
> For n=7 I get an upper bound of .814611570570018210618938079323
>
Sum_{n=1}^{100000}Prime(Prime(n))=0.97723454001125247632387261754663756

Best regards. Someone.
http://www.someoneltd.boom.ru/ http://home.tula.net/frazez/  Hello,
try to plot some trends and you'll see that this sum reaches 1 for p(p(n)) near exp(25) (so n is about 145e6), and converges to the value of 1.04...
* * * * * * *
n p(n) p(p(n)) sum(1/p(p(n)))
125791 1666589 26717767 0.978334479698888415
251582 3521831 59278873 0.981433198741145300
377373 5446943 94216519 0.983104909458966484
503164 7419067 130765361 0.984234195530853506
628955 9425137 168525023 0.985079781142629655
754746 11458679 207263417 0.985751901454822645
880537 13515049 246824987 0.986307552426211293
1006328 15589333 287075561 0.986779780231421793
1132119 17679553 327922403 0.987189515870527350
1257910 19787381 369389123 0.987550778379484510
1383701 21906617 411305143 0.987873375985004616
1509492 24040117 453728633 0.988164463242098489
1635283 26183929 496554677 0.988429408082470812
1761074 28335751 539744393 0.988672325751093269
1886865 30498953 583324493 0.988896453017897082
2012656 32670817 627239161 0.989104375276345374
2138447 34856453 671587723 0.989298165912155787
2264238 37045069 716146609 0.989479528649857192
2390029 39237901 760935299 0.989649909311663483
2515820 41439943 806020277 0.989810516997660378
2641611 43651747 851437423 0.989962345562505603
2767402 45865223 897028177 0.990106265471564048
2893193 48090131 942948289 0.990243031050776810
3018984 50316947 989010599 0.990373278036306816
3144775 52550809 1035324749 0.990497581470033747
3270566 54792151 1081900493 0.990616430063422676
3396357 57034421 1128583543 0.990730261584076278
3522148 59284571 1175542889 0.990839466215913510
3647939 61539047 1222679863 0.990944385446801957
3773730 63796961 1269963341 0.991045326656961421
3899521 66060539 1317466019 0.991142570398026806
4025312 68324761 1365036661 0.991236368302043147
4151103 70595879 1412867293 0.991326942344998715
4276894 72870559 1460826767 0.991414496757487683
4402685 75152381 1508994661 0.991499217767051643
4528476 77434891 1557273923 0.991581274645749774
4654267 79723759 1605760991 0.991660818813572631
4780058 82011379 1654295431 0.991737995382618730
4905849 84307417 1703069941 0.991812935530606868
5031640 86606083 1751959211 0.991885756865868117
5157431 88906457 1800937471 0.991956571649872347
5283222 91211459 1850110027 0.992025482328520660
5409013 93519659 1899398981 0.992092583877380152
5534804 95828917 1948774811 0.992157964574395842
5660595 98141369 1998261647 0.992221708006590970
5786386 100459397 2047923487 0.992283889006131001
5912177 102776371 2097628613 0.992344578739656232
6037968 105097561 2147483563 0.992403845704372124
It would be interesting to compute the first n with sum > 1 :)
WBR,
Andrey
[Nontext portions of this message have been removed]  It's
p_k > k (log k + log log k  1)
for k >= 2.
You changed the  to a +.
Charles Greathouse
 In primenumbers@yahoogroups.com, "Adam" <a_math_guy@y...> wrote:
> Message 17230 of 17232 indicates that pn>=n(log(n)+log(log(n))+1) so
> we conclude that pn>n*log(n) and then "compose this with itself" to
> get p(pn)>pn*log(pn)>n*log(n)*log[n*log(n)]>n*log(n)^2 so 1/p(pn)<1/
> [n*log(n)^2].
>
> The tail end of the sum sum(1/p(pn),n=k+1..infinity) is bounded above
> by sum(1/[n*log(n)^2],n=k+1..infinity) which is bounded above by
> integral(1/[x*log(x)^2],x=k..infinity) which is 1/[3*log(k)^3].
>
> You add up the first four terms 1/p(p1)+1/p(p2)+1/p(p3)+1/p(p4)=1/p(2)
> +1/p(3)+1/p(5)+1/p(7)=1/3+1/5+1/11+1/17<.68306596 and then you have
> the error term of int(1/[x*log(x)^2],x=3..infinity)=1/[3*log(3)^3]
> <.25138849 so the whole sum is <.68306596+.25138849=.93445445.
>
> For n=7 I get an upper bound of .814611570570018210618938079323
>
> Adam
>
>  In primenumbers@yahoogroups.com, "Werner D. Sand"
> <Theo.3.1415@w...> wrote:
> >
> > The sum of the reciprocal primeth primes p(p(n))
> >
> > s = 1/3 + 1/5 + 1/11 + 1/17 + 1/31 + seems to
> >
> > converge to 1.
> >
> > Any series like this, where you select from an infinite number of
Certainly you concede that the reciprocal sum of the 1st, 4th, 9th,
> primes, will diverge.
... n^2th, ... primes converges?
What about the reciprocals of the twin primes
B=(1/3+1/5)+(1/5+1/7)+(1/(11)+1/(13))+(1/(17)+1/(19))+....
which converges to about 1.902160583?
Charles Greathouse
 In primenumbers@yahoogroups.com, "Milton Brown" <miltbrown@e...>
wrote:> As I indicated, this sum does not converge.
primes,
>
> The comparable sum to compare it to is the reciprocal of the squares.
>
> Consider to the denominators compared to the squares.
>
> 3 5 11 17 31 41 59 67 83 10 127 157 179 191 211 241 277 283 331 ...
>
> 1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 384 361 400 ...
>
> After 191<196 the squares remain larger so 1/196 < 1/191
>
> Any series like this, where you select from an infinite number of
> will diverge.
1/3+1/5+1/11+1/17+1/31+1/41+1/59+1/67+1/83+1/109+1/127+1/157+1/179+1/191+1/2
>
> Milton L. Brown
> > [Original Message]
> > From: crgreathouse <crgreathouse@g...>
> > To: <primenumbers@yahoogroups.com>
> > Date: 11/25/2005 4:55:44 AM
> > Subject: [PrimeNumbers] Re: primeth primes
> >
> > Certainly the denominators need not decrease at a continuing rate for
> > convergence. Take
> >
> > 1/3 + 1/4 + 1/9 + 1/10 + 1/27 + 1/28 + 1/81 + 1/82 + ...
> >
> > which is about 0.90406.
> >
> > I believe even the Sophie Germain prime reciprocals converge  I'm
> > not sure of this, of course.
> >
> > 1/2 + 1/3 + 1/5 + 1/11 + 1/23 + ...
> >
> > Asymptotically, p_k ~ k(ln k + ln ln k + 1), so p_{p_k} ~ k(ln k + ln
> > ln k + 1)(ln (k(ln k + ln ln k + 1)) + ln ln (k(ln k + ln ln k + 1)) +
> > 1). In fact, this is a strict lower bound on p_k (Dusart 1999), so it
> > seems this could be used to prove the sum converges.
> >
> > Lacking a formal proof, I'll take the most significant term k(ln k)^2.
> > Integrate 2...infty dk/(k(ln k)^2): with x=ln k, this is integral ln
> > 2...infty dx/x^2 ~= 0.7213475.
> >
> > Does someone want to tighten this up or check it over?
> >
> > Charles Greathouse
> >
> >  In primenumbers@yahoogroups.com, "Milton Brown" <miltbrown@e...>
> > wrote:
> > > I believe that your sum does not converge.
> > >
> > > Yours is s =
> > >
> >
>
> > > 11+1/241+1/277+1/283+1/331+...
which
> > >
> > > Notice the difference in the denominators
> > >
> > > 2 6 14 10 18 8 16 26 16 30 22 12 20 30 36 6 50 ...
> > >
> > > You will see that we are not guaranteed that denominators are
> > decreasing at
> > > a continuing rate as we are for the reciprocals of the sqauares,
> > > converges:
> > >
> > > 1/2^2+1/2^3+1/2^4+...
> > >
> > > whose differences are 5 7 9 11 13 15 ... where we are guaranteed
> > > increasingly decreasing denominators.
> > >
> > > Milton L. Brown
> > >
> > > > [Original Message]
> > > > From: Werner D. Sand <Theo.3.1415@w...>
> > > > To: <primenumbers@yahoogroups.com>
> > > > Date: 11/23/2005 10:26:45 AM
> > > > Subject: [PrimeNumbers] primeth primes
> > > >
> > > > The sum of the reciprocal primeth primes p(p(n))
> > > > s = 1/3 + 1/5 + 1/11 + 1/17 + 1/31 + seems to
> > > > converge to 1.
> >
> >
> >
> >
> >
> >
> > Unsubscribe by an email to: primenumbersunsubscribe@yahoogroups.com
> > The Prime Pages : http://www.primepages.org/
> >
> >
> > Yahoo! Groups Links
> >
> >
> >
> >
> >
> >
> 0.7213475 is my (rough, heuristic) upper bound for the value of the
sum from 2 on; the corresponding bound on the entire sum is 1.0546808.
The purpose of my post was to show that the sum is convergent, as well
as to give a method that could be used (without, in my opinion, too
much effort) to create concrete bounds for the sum.
Charles Greathouse
 In primenumbers@yahoogroups.com, "Milton Brown" <miltbrown@e...>
wrote:> Your number 0.7213475 is certainly not correct.
1/3+1/5+1/11+1/17+1/31+1/41+1/59+1/67+1/83+1/109+1/127+1/157+1/179+1/191+1/2
>
> The sum of the first 11 terms
>
> 1/3+1/5+1/11+1/17+1/31+1/41+1/59+1/67+1/83+1/109+1/127
>
> is 0.80071164
>
> Milton L. Brown
>
> > [Original Message]
> > From: crgreathouse <crgreathouse@g...>
> > To: <primenumbers@yahoogroups.com>
> > Date: 11/25/2005 4:55:44 AM
> > Subject: [PrimeNumbers] Re: primeth primes
> >
> > Certainly the denominators need not decrease at a continuing rate for
> > convergence. Take
> >
> > 1/3 + 1/4 + 1/9 + 1/10 + 1/27 + 1/28 + 1/81 + 1/82 + ...
> >
> > which is about 0.90406.
> >
> > I believe even the Sophie Germain prime reciprocals converge  I'm
> > not sure of this, of course.
> >
> > 1/2 + 1/3 + 1/5 + 1/11 + 1/23 + ...
> >
> > Asymptotically, p_k ~ k(ln k + ln ln k + 1), so p_{p_k} ~ k(ln k + ln
> > ln k + 1)(ln (k(ln k + ln ln k + 1)) + ln ln (k(ln k + ln ln k + 1)) +
> > 1). In fact, this is a strict lower bound on p_k (Dusart 1999), so it
> > seems this could be used to prove the sum converges.
> >
> > Lacking a formal proof, I'll take the most significant term k(ln k)^2.
> > Integrate 2...infty dk/(k(ln k)^2): with x=ln k, this is integral ln
> > 2...infty dx/x^2 ~= 0.7213475.
> >
> > Does someone want to tighten this up or check it over?
> >
> > Charles Greathouse
> >
> >  In primenumbers@yahoogroups.com, "Milton Brown" <miltbrown@e...>
> > wrote:
> > > I believe that your sum does not converge.
> > >
> > > Yours is s =
> > >
> >
>
> > > 11+1/241+1/277+1/283+1/331+...
which
> > >
> > > Notice the difference in the denominators
> > >
> > > 2 6 14 10 18 8 16 26 16 30 22 12 20 30 36 6 50 ...
> > >
> > > You will see that we are not guaranteed that denominators are
> > decreasing at
> > > a continuing rate as we are for the reciprocals of the sqauares,
> > > converges:
> > >
> > > 1/2^2+1/2^3+1/2^4+...
> > >
> > > whose differences are 5 7 9 11 13 15 ... where we are guaranteed
> > > increasingly decreasing denominators.
> > >
> > > Milton L. Brown
> > >
> > > > [Original Message]
> > > > From: Werner D. Sand <Theo.3.1415@w...>
> > > > To: <primenumbers@yahoogroups.com>
> > > > Date: 11/23/2005 10:26:45 AM
> > > > Subject: [PrimeNumbers] primeth primes
> > > >
> > > > The sum of the reciprocal primeth primes p(p(n))
> > > > s = 1/3 + 1/5 + 1/11 + 1/17 + 1/31 + seems to
> > > > converge to 1.
> >
> >
> >
> >
> >
> >
> > Unsubscribe by an email to: primenumbersunsubscribe@yahoogroups.com
> > The Prime Pages : http://www.primepages.org/
> >
> >
> > Yahoo! Groups Links
> >
> >
> >
> >
> >
> >
>  Andrey Kulsha wrote:
> reaches 1 for p(p(n)) near exp(25) (so n is about 145e6),
...
> and converges to the value of 1.04...
>
> n p(n) p(p(n)) sum(1/p(p(n)))
> 125791 1666589 26717767 0.978334479698888415
> 6037968 105097561 2147483563 0.992403845704372124
My program agrees with Andrey's table.
An extension shows his estimate for sum>1 is good (it's n=148189304):
n p(n) p(p(n)) sum(1/p(p(n)))
10000000 179424673 3767321791 0.993782079779048
20000000 373587883 8132396761 0.995553008182270
30000000 573259391 12736743289 0.996528886890023
40000000 776531401 17500350067 0.997196394520154
50000000 982451653 22383430879 0.997700609931158
60000000 1190494759 27363009697 0.998104122267683
70000000 1400305337 32423573141 0.998439516689021
80000000 1611623773 37553793989 0.998725881592459
90000000 1824261409 42745397623 0.998975327892169
100000000 2038074743 47991893393 0.999196010613537
110000000 2252945251 53287844179 0.999393678625278
120000000 2468776129 58629206579 0.999572530777542
130000000 2685457421 64011376427 0.999735723538573
140000000 2902958801 69432251137 0.999885688879310
148189304 3081648379 73898684653 1.000000000008883
150000000 3121238909 74889755137 1.000024339564021
160000000 3340200037 80380156427 1.000153205905392
170000000 3559788179 85901366807 1.000273531253509
180000000 3780008329 91452739153 1.000386339616163
190000000 4000846301 97033105289 1.000492482024935
200000000 4222234741 102639953363 1.000592674128579

Jens Kruse Andersen  Retraction:
Whoops, somebody pointed out my integration mistake.
My wife suggests I should say "I was too busy talking to my wife"
or "I just put that there to see if anyone was paying attention."
All I can say in my own defense is don't integrate in your head
while typing email and whoops classic intro calc mistake.
Of course int(1/[x*log(x)^2]) should be 1/log(x) and not 1/(3*log(x)
^3). That would change the error term dramatically.
Sorry for the mistake.
Adam
 In primenumbers@yahoogroups.com, "Adam" <a_math_guy@y...> wrote:
>
> Message 17230 of 17232 indicates that pn>=n(log(n)+log(log(n))+1)
so
> we conclude that pn>n*log(n) and then "compose this with itself"
to
> get p(pn)>pn*log(pn)>n*log(n)*log[n*log(n)]>n*log(n)^2 so 1/p(pn)
<1/
> [n*log(n)^2].
>
> The tail end of the sum sum(1/p(pn),n=k+1..infinity) is bounded
above
> by sum(1/[n*log(n)^2],n=k+1..infinity) which is bounded above by
> integral(1/[x*log(x)^2],x=k..infinity) which is 1/[3*log(k)^3].
>
> You add up the first four terms 1/p(p1)+1/p(p2)+1/p(p3)+1/p(p4)=1/p
(2)
> +1/p(3)+1/p(5)+1/p(7)=1/3+1/5+1/11+1/17<.68306596 and then you
have
> the error term of int(1/[x*log(x)^2],x=3..infinity)=1/[3*log(3)^3]
> <.25138849 so the whole sum is <.68306596+.25138849=.93445445.
>
> For n=7 I get an upper bound of .814611570570018210618938079323
>
> Adam
>
>  In primenumbers@yahoogroups.com, "Werner D. Sand"
> <Theo.3.1415@w...> wrote:
> >
> > The sum of the reciprocal primeth primes p(p(n))
> >
> > s = 1/3 + 1/5 + 1/11 + 1/17 + 1/31 + seems to
> >
> > converge to 1.
> >
>   In primenumbers@yahoogroups.com, "crgreathouse"
<crgreathouse@g...> wrote:>
Well, obviously Milton has proven the twin prime conjecture to be false!
> > Any series like this, where you select from an infinite number of
> > primes, will diverge.
> What about the reciprocals of the twin primes
>
> B=(1/3+1/5)+(1/5+1/7)+(1/(11)+1/(13))+(1/(17)+1/(19))+....
>
> which converges to about 1.902160583?
Dick  Thanks for your comments.
However, you are missing part of my message.
Please include the entire message in the future.
And try to think positively.
Thanks.
Here it is:
"As I indicated, this sum does not converge.
The comparable sum to compare it to is the reciprocal of the squares.
Consider to the denominators compared to the squares.
3 5 11 17 31 41 59 67 83 10 127 157 179 191 211 241 277 283 331 ...
1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 384 361 400 ...
After 191<196 the squares remain larger so 1/196 < 1/191
Any series like this, where you select from an infinite number of primes,
will diverge."
Milton L. Brown
> [Original Message]
> From: Dick <richard042@...>
> To: <primenumbers@yahoogroups.com>
> Date: 11/26/2005 9:54:24 PM
> Subject: [PrimeNumbers] Re: primeth primes
>
>  In primenumbers@yahoogroups.com, "crgreathouse"
> <crgreathouse@g...> wrote:
> >
> > > Any series like this, where you select from an infinite number of
> > > primes, will diverge.
>
> > What about the reciprocals of the twin primes
> >
> > B=(1/3+1/5)+(1/5+1/7)+(1/(11)+1/(13))+(1/(17)+1/(19))+....
> >
> > which converges to about 1.902160583?
>
> Well, obviously Milton has proven the twin prime conjecture to be false!
>
> Dick
>
>
>
>
>
>
> Unsubscribe by an email to: primenumbersunsubscribe@yahoogroups.com
> The Prime Pages : http://www.primepages.org/
>
>
> Yahoo! Groups Links
>
>
>
>
>
> > > My program agrees with Andrey's table.
Jose Ramón Brox brox wrote:
> > An extension shows his estimate for sum>1 is good (it's n=148189304):
> I was trying to compute "Werner's constant" up to 10^10.
I used my own C program with the sieve of Eratosthenes.
> It was running in PARIGP for near three days and then
> Windows crashed (blame it on Billy :P). How did you
> do the computation so quickly? A fast processor,
> a lot of memory, a nice computation trick?
It took below a GHz hour with 3 MB ram.
I have not analyzed whether the exact n is reliable after rounding errors.
PARI/GP does not have functions suited to compute more small primes
than fit in memory after increasing primelimit. nextprime(x) is too slow.

Jens Kruse Andersen  Milton Brown <miltbrown@...> wrote:
> Thanks for your comments.
Hi Milton,
> However, you are missing part of my message.
> Please include the entire message in the future.
> And try to think positively.
I responded to crgreathouse's message which was a response to
yoursyour blather had already been excised. Are you suggesting
I am responsible to go back and reassemble a thread into a single
historical document everytime I post inside that thread?
> "As I indicated, this sum does not converge.
Three things:
> The comparable sum to compare it to is the reciprocal of the
> squares.
> Consider to the denominators compared to the squares.
> 3 5 11 17 31 41 59 67 83 10 127 157 179 191 211 241 277 283
> 331 ...
> 1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 384 361
> 400 ...
> After 191<196 the squares remain larger so 1/196 < 1/191
> Any series like this, where you select from an infinite number
> of primes, will diverge."
1) Other posters prior to you provided sufficient proof to
demonstrate that the sequence does indeed converge, and in fact
the limit was calculated and posted.
2) Zeta[1] diverges and Zeta[1+c] converges for any positive real
value of c. As c ranges over 0+>1, there are an infinite
number of convergent series that can have denominators
consistently larger than the squares on a 1 to 1 basis. You seem
to be thinking that the pivot point between convergence and
divergence is Zeta[2], corresponding to c=1, but it is not, the
pivot point is Zeta[1], corresponding to c=0. There are infinite
possibilities for convergent series between c=0 and c=1.
3) What on God's green Earth makes you think that primeth primes
and the squares progress more or less with the same magnitude
when followed on a one to one basis?
Everyone knows your village idiot routine is just an act
Miltonwhy do you bother to keep it up?
Regards
Dick Boland
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http://mail.yahoo.com  From: Milton Brown
Date: 11/25/05 17:57:22
To: crgreathouse; primenumbers@yahoogroups.com
Subject: RE: [PrimeNumbers] Re: primeth primes
As I indicated, this sum does not converge.
The comparable sum to compare it to is the reciprocal of the squares.
Consider to the denominators compared to the squares.
3 5 11 17 31 41 59 67 83 10 127 157 179 191 211 241 277 283 331 ...
1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 384 361 400 ...
After 191<196 the squares remain larger so 1/196 < 1/191
Any series like this, where you select from an infinite number of primes,
will diverge.
Milton L. Brown
Milton, I'm not sure what you are saying here.
Suppose I select primes by the following rule.
Pick not more than 10 primes between consecutive squares to get the
sequence p1, p2, p3,...
Then would you expect the series
1 + 1/P1 + 1/P2 + 1/P3 + .... to converge or diverge?
It's easy to prove that it will converge.
[Nontext portions of this message have been removed]  Milton wrote:
> As I indicated, this sum does not converge.
Based on the above (the entire message quoted to please Milton :),
>
> The comparable sum to compare it to is the reciprocal of the squares.
>
> Consider to the denominators compared to the squares.
>
> 3 5 11 17 31 41 59 67 83 10 127 157 179 191 211 241 277 283 331 ...
>
> 1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 384 361 400 ...
>
> After 191<196 the squares remain larger so 1/196 < 1/191
>
> Any series like this, where you select from an infinite number of primes,
> will diverge.
>
> Milton L. Brown
I guess he assumes the sum of the reciprocals of squares diverges,
and then he thinks the discussed primeth sum must also diverge
because the terms soon become larger.
One minor problem with that argument:
The sum of reciprocal of squares converges to pi^2/6.
I guess Milton was thinking of the harmonic series
1 + 1/2 + 1/3 + 1/4 + ... which is famous for diverging.
(The pi^2/6 sum is also a little famous.)
I wonder whether I'm better at mind reading than Milton is at maths ;)

Jens Kruse Andersen
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