How to prove convergence to $1-frac{1}{log(n)}$












0












$begingroup$


I am trying to prove that:



$$left(frac{1}{2}+frac{1}{2}left(frac{1}{3}+frac{2}{3}left(frac{1}{5}+frac{4}{5}left(...+frac{p_{m-2}-1}{p_{m-2}}left(frac{1}{p_{m-1}}+frac{p_{m-1}-1}{p_{m-1}}left(frac{1}{p_{m}}right)right)...right)right)right)right)approx1-frac{1}{log(n)}$$



Where $m=pi(sqrt{n})$



Some help would be really welcomed! Thanks in advance.






sequences-and-series convergence prime-numbers







share|cite|improve this question











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  • 2




    $begingroup$
    Use $log$ for $log$ instead of $log$ for $log$.
    $endgroup$
    – Shaun
    Dec 21 '18 at 19:23










  • $begingroup$
    $m=pi(sqrt{n})$? $m$ should be an integer?
    $endgroup$
    – Martin Rosenau
    Dec 21 '18 at 19:23










  • $begingroup$
    @MartinRosenau I think $pi(cdot)$ is the prime counting function.
    $endgroup$
    – Alex Vong
    Dec 21 '18 at 19:31








  • 1




    $begingroup$
    I don't understand what is the LHS. Are you trying to estimate the probability that $n$ is not prime ? Then assuming a probability distribution such that $Pr[n not equiv 0 bmod p_j, j in 1 ldots J] approx prod_{j=1}^J Pr[ n not equiv 0 bmod p_j] approx prod_{j=1}^J (1-frac{1}{p_j})$ then $Pr[n$ is not prime $] = 1-Pr[n$ is prime $]=1-prod_{ p le sqrt{n}} Pr[ n not equiv 0 bmod p] approx 1- prod_{ p le sqrt{n}} (1-frac{1}{p})$. That the last expression is $approx 1-frac{1}{log n}$ is quite equivalent to the PNT
    $endgroup$
    – reuns
    Dec 21 '18 at 22:47












  • $begingroup$
    Thanks for your format advice @Shaun; it is already fixed.
    $endgroup$
    – Juan Moreno
    Dec 22 '18 at 10:20
















0












$begingroup$


I am trying to prove that:



$$left(frac{1}{2}+frac{1}{2}left(frac{1}{3}+frac{2}{3}left(frac{1}{5}+frac{4}{5}left(...+frac{p_{m-2}-1}{p_{m-2}}left(frac{1}{p_{m-1}}+frac{p_{m-1}-1}{p_{m-1}}left(frac{1}{p_{m}}right)right)...right)right)right)right)approx1-frac{1}{log(n)}$$



Where $m=pi(sqrt{n})$



Some help would be really welcomed! Thanks in advance.






sequences-and-series convergence prime-numbers







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Use $log$ for $log$ instead of $log$ for $log$.
    $endgroup$
    – Shaun
    Dec 21 '18 at 19:23










  • $begingroup$
    $m=pi(sqrt{n})$? $m$ should be an integer?
    $endgroup$
    – Martin Rosenau
    Dec 21 '18 at 19:23










  • $begingroup$
    @MartinRosenau I think $pi(cdot)$ is the prime counting function.
    $endgroup$
    – Alex Vong
    Dec 21 '18 at 19:31








  • 1




    $begingroup$
    I don't understand what is the LHS. Are you trying to estimate the probability that $n$ is not prime ? Then assuming a probability distribution such that $Pr[n not equiv 0 bmod p_j, j in 1 ldots J] approx prod_{j=1}^J Pr[ n not equiv 0 bmod p_j] approx prod_{j=1}^J (1-frac{1}{p_j})$ then $Pr[n$ is not prime $] = 1-Pr[n$ is prime $]=1-prod_{ p le sqrt{n}} Pr[ n not equiv 0 bmod p] approx 1- prod_{ p le sqrt{n}} (1-frac{1}{p})$. That the last expression is $approx 1-frac{1}{log n}$ is quite equivalent to the PNT
    $endgroup$
    – reuns
    Dec 21 '18 at 22:47












  • $begingroup$
    Thanks for your format advice @Shaun; it is already fixed.
    $endgroup$
    – Juan Moreno
    Dec 22 '18 at 10:20














0












0








0





$begingroup$


I am trying to prove that:



$$left(frac{1}{2}+frac{1}{2}left(frac{1}{3}+frac{2}{3}left(frac{1}{5}+frac{4}{5}left(...+frac{p_{m-2}-1}{p_{m-2}}left(frac{1}{p_{m-1}}+frac{p_{m-1}-1}{p_{m-1}}left(frac{1}{p_{m}}right)right)...right)right)right)right)approx1-frac{1}{log(n)}$$



Where $m=pi(sqrt{n})$



Some help would be really welcomed! Thanks in advance.






sequences-and-series convergence prime-numbers







share|cite|improve this question











$endgroup$




I am trying to prove that:



$$left(frac{1}{2}+frac{1}{2}left(frac{1}{3}+frac{2}{3}left(frac{1}{5}+frac{4}{5}left(...+frac{p_{m-2}-1}{p_{m-2}}left(frac{1}{p_{m-1}}+frac{p_{m-1}-1}{p_{m-1}}left(frac{1}{p_{m}}right)right)...right)right)right)right)approx1-frac{1}{log(n)}$$



Where $m=pi(sqrt{n})$



Some help would be really welcomed! Thanks in advance.






sequences-and-series convergence prime-numbers




sequences-and-series convergence prime-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 14:11









Larry

2,41331129




2,41331129










asked Dec 21 '18 at 19:17









Juan MorenoJuan Moreno

163




163








  • 2




    $begingroup$
    Use $log$ for $log$ instead of $log$ for $log$.
    $endgroup$
    – Shaun
    Dec 21 '18 at 19:23










  • $begingroup$
    $m=pi(sqrt{n})$? $m$ should be an integer?
    $endgroup$
    – Martin Rosenau
    Dec 21 '18 at 19:23










  • $begingroup$
    @MartinRosenau I think $pi(cdot)$ is the prime counting function.
    $endgroup$
    – Alex Vong
    Dec 21 '18 at 19:31








  • 1




    $begingroup$
    I don't understand what is the LHS. Are you trying to estimate the probability that $n$ is not prime ? Then assuming a probability distribution such that $Pr[n not equiv 0 bmod p_j, j in 1 ldots J] approx prod_{j=1}^J Pr[ n not equiv 0 bmod p_j] approx prod_{j=1}^J (1-frac{1}{p_j})$ then $Pr[n$ is not prime $] = 1-Pr[n$ is prime $]=1-prod_{ p le sqrt{n}} Pr[ n not equiv 0 bmod p] approx 1- prod_{ p le sqrt{n}} (1-frac{1}{p})$. That the last expression is $approx 1-frac{1}{log n}$ is quite equivalent to the PNT
    $endgroup$
    – reuns
    Dec 21 '18 at 22:47












  • $begingroup$
    Thanks for your format advice @Shaun; it is already fixed.
    $endgroup$
    – Juan Moreno
    Dec 22 '18 at 10:20














  • 2




    $begingroup$
    Use $log$ for $log$ instead of $log$ for $log$.
    $endgroup$
    – Shaun
    Dec 21 '18 at 19:23










  • $begingroup$
    $m=pi(sqrt{n})$? $m$ should be an integer?
    $endgroup$
    – Martin Rosenau
    Dec 21 '18 at 19:23










  • $begingroup$
    @MartinRosenau I think $pi(cdot)$ is the prime counting function.
    $endgroup$
    – Alex Vong
    Dec 21 '18 at 19:31








  • 1




    $begingroup$
    I don't understand what is the LHS. Are you trying to estimate the probability that $n$ is not prime ? Then assuming a probability distribution such that $Pr[n not equiv 0 bmod p_j, j in 1 ldots J] approx prod_{j=1}^J Pr[ n not equiv 0 bmod p_j] approx prod_{j=1}^J (1-frac{1}{p_j})$ then $Pr[n$ is not prime $] = 1-Pr[n$ is prime $]=1-prod_{ p le sqrt{n}} Pr[ n not equiv 0 bmod p] approx 1- prod_{ p le sqrt{n}} (1-frac{1}{p})$. That the last expression is $approx 1-frac{1}{log n}$ is quite equivalent to the PNT
    $endgroup$
    – reuns
    Dec 21 '18 at 22:47












  • $begingroup$
    Thanks for your format advice @Shaun; it is already fixed.
    $endgroup$
    – Juan Moreno
    Dec 22 '18 at 10:20








2




2




$begingroup$
Use $log$ for $log$ instead of $log$ for $log$.
$endgroup$
– Shaun
Dec 21 '18 at 19:23




$begingroup$
Use $log$ for $log$ instead of $log$ for $log$.
$endgroup$
– Shaun
Dec 21 '18 at 19:23












$begingroup$
$m=pi(sqrt{n})$? $m$ should be an integer?
$endgroup$
– Martin Rosenau
Dec 21 '18 at 19:23




$begingroup$
$m=pi(sqrt{n})$? $m$ should be an integer?
$endgroup$
– Martin Rosenau
Dec 21 '18 at 19:23












$begingroup$
@MartinRosenau I think $pi(cdot)$ is the prime counting function.
$endgroup$
– Alex Vong
Dec 21 '18 at 19:31






$begingroup$
@MartinRosenau I think $pi(cdot)$ is the prime counting function.
$endgroup$
– Alex Vong
Dec 21 '18 at 19:31






1




1




$begingroup$
I don't understand what is the LHS. Are you trying to estimate the probability that $n$ is not prime ? Then assuming a probability distribution such that $Pr[n not equiv 0 bmod p_j, j in 1 ldots J] approx prod_{j=1}^J Pr[ n not equiv 0 bmod p_j] approx prod_{j=1}^J (1-frac{1}{p_j})$ then $Pr[n$ is not prime $] = 1-Pr[n$ is prime $]=1-prod_{ p le sqrt{n}} Pr[ n not equiv 0 bmod p] approx 1- prod_{ p le sqrt{n}} (1-frac{1}{p})$. That the last expression is $approx 1-frac{1}{log n}$ is quite equivalent to the PNT
$endgroup$
– reuns
Dec 21 '18 at 22:47






$begingroup$
I don't understand what is the LHS. Are you trying to estimate the probability that $n$ is not prime ? Then assuming a probability distribution such that $Pr[n not equiv 0 bmod p_j, j in 1 ldots J] approx prod_{j=1}^J Pr[ n not equiv 0 bmod p_j] approx prod_{j=1}^J (1-frac{1}{p_j})$ then $Pr[n$ is not prime $] = 1-Pr[n$ is prime $]=1-prod_{ p le sqrt{n}} Pr[ n not equiv 0 bmod p] approx 1- prod_{ p le sqrt{n}} (1-frac{1}{p})$. That the last expression is $approx 1-frac{1}{log n}$ is quite equivalent to the PNT
$endgroup$
– reuns
Dec 21 '18 at 22:47














$begingroup$
Thanks for your format advice @Shaun; it is already fixed.
$endgroup$
– Juan Moreno
Dec 22 '18 at 10:20




$begingroup$
Thanks for your format advice @Shaun; it is already fixed.
$endgroup$
– Juan Moreno
Dec 22 '18 at 10:20










1 Answer
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I have found that Merten's third theorem, which remarkably can be proved without the assistance of the Prime Number Theorem, establishes that:



$lim_{xto infty}log{(x)}prod_{ple{x}}left(1-frac{1}{p}right)=e^{-gamma}$



Using this theorem, the proof I was looking for follows easily.






share|cite|improve this answer









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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    I have found that Merten's third theorem, which remarkably can be proved without the assistance of the Prime Number Theorem, establishes that:



    $lim_{xto infty}log{(x)}prod_{ple{x}}left(1-frac{1}{p}right)=e^{-gamma}$



    Using this theorem, the proof I was looking for follows easily.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      I have found that Merten's third theorem, which remarkably can be proved without the assistance of the Prime Number Theorem, establishes that:



      $lim_{xto infty}log{(x)}prod_{ple{x}}left(1-frac{1}{p}right)=e^{-gamma}$



      Using this theorem, the proof I was looking for follows easily.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        I have found that Merten's third theorem, which remarkably can be proved without the assistance of the Prime Number Theorem, establishes that:



        $lim_{xto infty}log{(x)}prod_{ple{x}}left(1-frac{1}{p}right)=e^{-gamma}$



        Using this theorem, the proof I was looking for follows easily.






        share|cite|improve this answer









        $endgroup$



        I have found that Merten's third theorem, which remarkably can be proved without the assistance of the Prime Number Theorem, establishes that:



        $lim_{xto infty}log{(x)}prod_{ple{x}}left(1-frac{1}{p}right)=e^{-gamma}$



        Using this theorem, the proof I was looking for follows easily.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 22 '18 at 18:30









        Juan MorenoJuan Moreno

        163




        163






























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