Does this series converge to a rational multiple of $pi^2$?












6












$begingroup$


If we define the sum



$$S_2=sum_{n_2=1}^inftyfrac{1}{(n_2)^2} $$



in which $n_2$ are products of an even number of prime factors, together with 1, so $n_2=1,4,6,...,15,16,21,22,...,24,$ etc., the sum seems to be a rational multiple of $pi^2$. Of course it converges by comparison to $zeta(2).$



The first $k=9,997,745$ of these (including 1) sum to approximately $frac{7pi^2}{60}$ and the ratio $S_k/(7pi^2/60)=0.999999978...$ (there are 7 nines in this decimal).



My unscientific test is that if Mathematica rounds to one it bears a second look. In this case I looked at various proofs of the Basel problem and connections with 2-primes $pq$ to see if there was a way of showing this.




Is this already known to be true? If not, can someone show it? Also any comments on the likelihood of the relation given the approximation are welcome. How much numerical similarity is enough to guess in this way?




I am still working on this, and if I find a proof or a reference I will post it.






sequences-and-series







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$endgroup$








  • 1




    $begingroup$
    A side note: $1 + zeta(2) P(2) = sum_{n=1}^{infty} frac{omega(n)}{n^2}$, where $omega(n)$ is the number of distinct prime divisors of $n$ and $P$ is the prime zeta function. Although you are interested in $sum_{n = 2|omega(n)}^{infty}frac{1}{n^2}$ I wonder if you can utilise some kind of mobius inversion or if you would just go in circles?
    $endgroup$
    – DanielOnMSE
    Dec 6 '18 at 14:59


















6












$begingroup$


If we define the sum



$$S_2=sum_{n_2=1}^inftyfrac{1}{(n_2)^2} $$



in which $n_2$ are products of an even number of prime factors, together with 1, so $n_2=1,4,6,...,15,16,21,22,...,24,$ etc., the sum seems to be a rational multiple of $pi^2$. Of course it converges by comparison to $zeta(2).$



The first $k=9,997,745$ of these (including 1) sum to approximately $frac{7pi^2}{60}$ and the ratio $S_k/(7pi^2/60)=0.999999978...$ (there are 7 nines in this decimal).



My unscientific test is that if Mathematica rounds to one it bears a second look. In this case I looked at various proofs of the Basel problem and connections with 2-primes $pq$ to see if there was a way of showing this.




Is this already known to be true? If not, can someone show it? Also any comments on the likelihood of the relation given the approximation are welcome. How much numerical similarity is enough to guess in this way?




I am still working on this, and if I find a proof or a reference I will post it.






sequences-and-series







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    A side note: $1 + zeta(2) P(2) = sum_{n=1}^{infty} frac{omega(n)}{n^2}$, where $omega(n)$ is the number of distinct prime divisors of $n$ and $P$ is the prime zeta function. Although you are interested in $sum_{n = 2|omega(n)}^{infty}frac{1}{n^2}$ I wonder if you can utilise some kind of mobius inversion or if you would just go in circles?
    $endgroup$
    – DanielOnMSE
    Dec 6 '18 at 14:59
















6












6








6


1



$begingroup$


If we define the sum



$$S_2=sum_{n_2=1}^inftyfrac{1}{(n_2)^2} $$



in which $n_2$ are products of an even number of prime factors, together with 1, so $n_2=1,4,6,...,15,16,21,22,...,24,$ etc., the sum seems to be a rational multiple of $pi^2$. Of course it converges by comparison to $zeta(2).$



The first $k=9,997,745$ of these (including 1) sum to approximately $frac{7pi^2}{60}$ and the ratio $S_k/(7pi^2/60)=0.999999978...$ (there are 7 nines in this decimal).



My unscientific test is that if Mathematica rounds to one it bears a second look. In this case I looked at various proofs of the Basel problem and connections with 2-primes $pq$ to see if there was a way of showing this.




Is this already known to be true? If not, can someone show it? Also any comments on the likelihood of the relation given the approximation are welcome. How much numerical similarity is enough to guess in this way?




I am still working on this, and if I find a proof or a reference I will post it.






sequences-and-series







share|cite|improve this question









$endgroup$




If we define the sum



$$S_2=sum_{n_2=1}^inftyfrac{1}{(n_2)^2} $$



in which $n_2$ are products of an even number of prime factors, together with 1, so $n_2=1,4,6,...,15,16,21,22,...,24,$ etc., the sum seems to be a rational multiple of $pi^2$. Of course it converges by comparison to $zeta(2).$



The first $k=9,997,745$ of these (including 1) sum to approximately $frac{7pi^2}{60}$ and the ratio $S_k/(7pi^2/60)=0.999999978...$ (there are 7 nines in this decimal).



My unscientific test is that if Mathematica rounds to one it bears a second look. In this case I looked at various proofs of the Basel problem and connections with 2-primes $pq$ to see if there was a way of showing this.




Is this already known to be true? If not, can someone show it? Also any comments on the likelihood of the relation given the approximation are welcome. How much numerical similarity is enough to guess in this way?




I am still working on this, and if I find a proof or a reference I will post it.






sequences-and-series




sequences-and-series






share|cite|improve this question













share|cite|improve this question











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asked Dec 6 '18 at 14:32









danieldaniel

6,25722259




6,25722259








  • 1




    $begingroup$
    A side note: $1 + zeta(2) P(2) = sum_{n=1}^{infty} frac{omega(n)}{n^2}$, where $omega(n)$ is the number of distinct prime divisors of $n$ and $P$ is the prime zeta function. Although you are interested in $sum_{n = 2|omega(n)}^{infty}frac{1}{n^2}$ I wonder if you can utilise some kind of mobius inversion or if you would just go in circles?
    $endgroup$
    – DanielOnMSE
    Dec 6 '18 at 14:59
















  • 1




    $begingroup$
    A side note: $1 + zeta(2) P(2) = sum_{n=1}^{infty} frac{omega(n)}{n^2}$, where $omega(n)$ is the number of distinct prime divisors of $n$ and $P$ is the prime zeta function. Although you are interested in $sum_{n = 2|omega(n)}^{infty}frac{1}{n^2}$ I wonder if you can utilise some kind of mobius inversion or if you would just go in circles?
    $endgroup$
    – DanielOnMSE
    Dec 6 '18 at 14:59










1




1




$begingroup$
A side note: $1 + zeta(2) P(2) = sum_{n=1}^{infty} frac{omega(n)}{n^2}$, where $omega(n)$ is the number of distinct prime divisors of $n$ and $P$ is the prime zeta function. Although you are interested in $sum_{n = 2|omega(n)}^{infty}frac{1}{n^2}$ I wonder if you can utilise some kind of mobius inversion or if you would just go in circles?
$endgroup$
– DanielOnMSE
Dec 6 '18 at 14:59






$begingroup$
A side note: $1 + zeta(2) P(2) = sum_{n=1}^{infty} frac{omega(n)}{n^2}$, where $omega(n)$ is the number of distinct prime divisors of $n$ and $P$ is the prime zeta function. Although you are interested in $sum_{n = 2|omega(n)}^{infty}frac{1}{n^2}$ I wonder if you can utilise some kind of mobius inversion or if you would just go in circles?
$endgroup$
– DanielOnMSE
Dec 6 '18 at 14:59












2 Answers
2






active

oldest

votes


















11












$begingroup$

Consider the product over all prime numbers
$$P=prod_pleft(1+frac{1}{p^2}right)^{-1}=prod_pleft(1-frac{1}{p^2}+frac{1}{(p^2)^2}+dotsright).$$
If you think about how a term of this series will look like when you multiply it out, you will find that there is a term $1/n^2$ for every $n$ with an even number of prime factors, and a term $-1/n^2$ for every $n$ with an odd number of prime factors. Therefore, we find that $P=2S_2-zeta(2)$. Since we know the value of $zeta(2)$ and want to find the value of $S_2$, it's enough to find the value of $P$.



But observe
$$P=prod_pleft(1+frac{1}{p^2}right)^{-1}=prod_pfrac{(1-1/p^4)^{-1}}{(1-1/p^2)^{-1}}=frac{prod_p(1-1/p^4)^{-1}}{prod_p(1-1/p^2)^{-1}}=frac{zeta(4)}{zeta(2)}=frac{pi^4/90}{pi^2/6}=frac{pi^2}{15}.$$



Comparing this with the above,
$$S_2=frac{P+zeta(2)}{2}=frac{pi^2/15+pi^2/6}{2}=frac{7pi^2}{60}.$$






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Very nice indeed :)
    $endgroup$
    – gandalf61
    Dec 6 '18 at 14:50



















1












$begingroup$

About the alternative version,



$$begin{eqnarray*} sum_{substack{ngeq 1\ omega(n)text{ is even}}}!!!!!!frac{1}{n^2}&=&frac{1}{2}left[zeta(2)+sum_{ngeq 1}frac{(-1)^{omega(n)}}{n^2}right]=frac{pi^2}{12}+frac{1}{2}prod_{p}left(1-frac{1}{p^2}-frac{1}{p^4}-frac{1}{p^6}-ldotsright)\&=&frac{pi^2}{12}+frac{1}{2}prod_{p}frac{p^2-2}{p^2-1}=frac{pi^2}{12}left[1+prod_{p}left(1-frac{2}{p^2}right)right]=zeta(2)C_{text{Feller-Tornier}}end{eqnarray*} $$
where the Feller-Tornier constant (close to $frac{2}{3}$) is related to the probability for two consecutive integers of being both squarefree.






share|cite|improve this answer









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    2 Answers
    2






    active

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    2 Answers
    2






    active

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    active

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    active

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    11












    $begingroup$

    Consider the product over all prime numbers
    $$P=prod_pleft(1+frac{1}{p^2}right)^{-1}=prod_pleft(1-frac{1}{p^2}+frac{1}{(p^2)^2}+dotsright).$$
    If you think about how a term of this series will look like when you multiply it out, you will find that there is a term $1/n^2$ for every $n$ with an even number of prime factors, and a term $-1/n^2$ for every $n$ with an odd number of prime factors. Therefore, we find that $P=2S_2-zeta(2)$. Since we know the value of $zeta(2)$ and want to find the value of $S_2$, it's enough to find the value of $P$.



    But observe
    $$P=prod_pleft(1+frac{1}{p^2}right)^{-1}=prod_pfrac{(1-1/p^4)^{-1}}{(1-1/p^2)^{-1}}=frac{prod_p(1-1/p^4)^{-1}}{prod_p(1-1/p^2)^{-1}}=frac{zeta(4)}{zeta(2)}=frac{pi^4/90}{pi^2/6}=frac{pi^2}{15}.$$



    Comparing this with the above,
    $$S_2=frac{P+zeta(2)}{2}=frac{pi^2/15+pi^2/6}{2}=frac{7pi^2}{60}.$$






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      Very nice indeed :)
      $endgroup$
      – gandalf61
      Dec 6 '18 at 14:50
















    11












    $begingroup$

    Consider the product over all prime numbers
    $$P=prod_pleft(1+frac{1}{p^2}right)^{-1}=prod_pleft(1-frac{1}{p^2}+frac{1}{(p^2)^2}+dotsright).$$
    If you think about how a term of this series will look like when you multiply it out, you will find that there is a term $1/n^2$ for every $n$ with an even number of prime factors, and a term $-1/n^2$ for every $n$ with an odd number of prime factors. Therefore, we find that $P=2S_2-zeta(2)$. Since we know the value of $zeta(2)$ and want to find the value of $S_2$, it's enough to find the value of $P$.



    But observe
    $$P=prod_pleft(1+frac{1}{p^2}right)^{-1}=prod_pfrac{(1-1/p^4)^{-1}}{(1-1/p^2)^{-1}}=frac{prod_p(1-1/p^4)^{-1}}{prod_p(1-1/p^2)^{-1}}=frac{zeta(4)}{zeta(2)}=frac{pi^4/90}{pi^2/6}=frac{pi^2}{15}.$$



    Comparing this with the above,
    $$S_2=frac{P+zeta(2)}{2}=frac{pi^2/15+pi^2/6}{2}=frac{7pi^2}{60}.$$






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      Very nice indeed :)
      $endgroup$
      – gandalf61
      Dec 6 '18 at 14:50














    11












    11








    11





    $begingroup$

    Consider the product over all prime numbers
    $$P=prod_pleft(1+frac{1}{p^2}right)^{-1}=prod_pleft(1-frac{1}{p^2}+frac{1}{(p^2)^2}+dotsright).$$
    If you think about how a term of this series will look like when you multiply it out, you will find that there is a term $1/n^2$ for every $n$ with an even number of prime factors, and a term $-1/n^2$ for every $n$ with an odd number of prime factors. Therefore, we find that $P=2S_2-zeta(2)$. Since we know the value of $zeta(2)$ and want to find the value of $S_2$, it's enough to find the value of $P$.



    But observe
    $$P=prod_pleft(1+frac{1}{p^2}right)^{-1}=prod_pfrac{(1-1/p^4)^{-1}}{(1-1/p^2)^{-1}}=frac{prod_p(1-1/p^4)^{-1}}{prod_p(1-1/p^2)^{-1}}=frac{zeta(4)}{zeta(2)}=frac{pi^4/90}{pi^2/6}=frac{pi^2}{15}.$$



    Comparing this with the above,
    $$S_2=frac{P+zeta(2)}{2}=frac{pi^2/15+pi^2/6}{2}=frac{7pi^2}{60}.$$






    share|cite|improve this answer









    $endgroup$



    Consider the product over all prime numbers
    $$P=prod_pleft(1+frac{1}{p^2}right)^{-1}=prod_pleft(1-frac{1}{p^2}+frac{1}{(p^2)^2}+dotsright).$$
    If you think about how a term of this series will look like when you multiply it out, you will find that there is a term $1/n^2$ for every $n$ with an even number of prime factors, and a term $-1/n^2$ for every $n$ with an odd number of prime factors. Therefore, we find that $P=2S_2-zeta(2)$. Since we know the value of $zeta(2)$ and want to find the value of $S_2$, it's enough to find the value of $P$.



    But observe
    $$P=prod_pleft(1+frac{1}{p^2}right)^{-1}=prod_pfrac{(1-1/p^4)^{-1}}{(1-1/p^2)^{-1}}=frac{prod_p(1-1/p^4)^{-1}}{prod_p(1-1/p^2)^{-1}}=frac{zeta(4)}{zeta(2)}=frac{pi^4/90}{pi^2/6}=frac{pi^2}{15}.$$



    Comparing this with the above,
    $$S_2=frac{P+zeta(2)}{2}=frac{pi^2/15+pi^2/6}{2}=frac{7pi^2}{60}.$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 6 '18 at 14:46









    WojowuWojowu

    18.8k23172




    18.8k23172








    • 2




      $begingroup$
      Very nice indeed :)
      $endgroup$
      – gandalf61
      Dec 6 '18 at 14:50














    • 2




      $begingroup$
      Very nice indeed :)
      $endgroup$
      – gandalf61
      Dec 6 '18 at 14:50








    2




    2




    $begingroup$
    Very nice indeed :)
    $endgroup$
    – gandalf61
    Dec 6 '18 at 14:50




    $begingroup$
    Very nice indeed :)
    $endgroup$
    – gandalf61
    Dec 6 '18 at 14:50











    1












    $begingroup$

    About the alternative version,



    $$begin{eqnarray*} sum_{substack{ngeq 1\ omega(n)text{ is even}}}!!!!!!frac{1}{n^2}&=&frac{1}{2}left[zeta(2)+sum_{ngeq 1}frac{(-1)^{omega(n)}}{n^2}right]=frac{pi^2}{12}+frac{1}{2}prod_{p}left(1-frac{1}{p^2}-frac{1}{p^4}-frac{1}{p^6}-ldotsright)\&=&frac{pi^2}{12}+frac{1}{2}prod_{p}frac{p^2-2}{p^2-1}=frac{pi^2}{12}left[1+prod_{p}left(1-frac{2}{p^2}right)right]=zeta(2)C_{text{Feller-Tornier}}end{eqnarray*} $$
    where the Feller-Tornier constant (close to $frac{2}{3}$) is related to the probability for two consecutive integers of being both squarefree.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      About the alternative version,



      $$begin{eqnarray*} sum_{substack{ngeq 1\ omega(n)text{ is even}}}!!!!!!frac{1}{n^2}&=&frac{1}{2}left[zeta(2)+sum_{ngeq 1}frac{(-1)^{omega(n)}}{n^2}right]=frac{pi^2}{12}+frac{1}{2}prod_{p}left(1-frac{1}{p^2}-frac{1}{p^4}-frac{1}{p^6}-ldotsright)\&=&frac{pi^2}{12}+frac{1}{2}prod_{p}frac{p^2-2}{p^2-1}=frac{pi^2}{12}left[1+prod_{p}left(1-frac{2}{p^2}right)right]=zeta(2)C_{text{Feller-Tornier}}end{eqnarray*} $$
      where the Feller-Tornier constant (close to $frac{2}{3}$) is related to the probability for two consecutive integers of being both squarefree.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        About the alternative version,



        $$begin{eqnarray*} sum_{substack{ngeq 1\ omega(n)text{ is even}}}!!!!!!frac{1}{n^2}&=&frac{1}{2}left[zeta(2)+sum_{ngeq 1}frac{(-1)^{omega(n)}}{n^2}right]=frac{pi^2}{12}+frac{1}{2}prod_{p}left(1-frac{1}{p^2}-frac{1}{p^4}-frac{1}{p^6}-ldotsright)\&=&frac{pi^2}{12}+frac{1}{2}prod_{p}frac{p^2-2}{p^2-1}=frac{pi^2}{12}left[1+prod_{p}left(1-frac{2}{p^2}right)right]=zeta(2)C_{text{Feller-Tornier}}end{eqnarray*} $$
        where the Feller-Tornier constant (close to $frac{2}{3}$) is related to the probability for two consecutive integers of being both squarefree.






        share|cite|improve this answer









        $endgroup$



        About the alternative version,



        $$begin{eqnarray*} sum_{substack{ngeq 1\ omega(n)text{ is even}}}!!!!!!frac{1}{n^2}&=&frac{1}{2}left[zeta(2)+sum_{ngeq 1}frac{(-1)^{omega(n)}}{n^2}right]=frac{pi^2}{12}+frac{1}{2}prod_{p}left(1-frac{1}{p^2}-frac{1}{p^4}-frac{1}{p^6}-ldotsright)\&=&frac{pi^2}{12}+frac{1}{2}prod_{p}frac{p^2-2}{p^2-1}=frac{pi^2}{12}left[1+prod_{p}left(1-frac{2}{p^2}right)right]=zeta(2)C_{text{Feller-Tornier}}end{eqnarray*} $$
        where the Feller-Tornier constant (close to $frac{2}{3}$) is related to the probability for two consecutive integers of being both squarefree.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 '18 at 16:09









        Jack D'AurizioJack D'Aurizio

        291k33284666




        291k33284666






























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