Riemann Zeta - Euler product convergence?












0












$begingroup$


I'm fascinated by the Riemann Zeta hypothesis - I haven't yet taken a course in complex analysis but I'm curious what this sentence means on Wikipedia:




The convergence of the Euler product shows that ζ(s) has no zeros in
this region, as none of the factors have zeros.




I'm wondering if a) the convergence of the Euler product comes purely from the fact that the p-series diverges when the exponent is greater than 1?
b) What the sentence from Wikipedia is essentially saying - I'm not understanding what factors are being referred to or how the convergence shows that the Zeta function has no zeroes for $Re(s) > 1$.



Edit: I was looking over this answer and it seems to be similar, but how do we know that, as this author wrote, $(1-p^{-s})^{-1}neq 0$ for all primes $p$? Can't it converge to 0, meaning that $(1-p^{-s})^{-1} = 0$ for at least one prime $p$?






sequences-and-series complex-analysis riemann-zeta euler-product







share|cite|improve this question











$endgroup$












  • $begingroup$
    The idea is that $zeta(s) =prod_p frac{1}{1-p^{-s}} = exp(F(s))$ where $F(s) = -sum_p log(1-p^{-s}) = sum_{p^k} frac{p^{-sk}}{k}$ is analytic on $Re(s) > 1$, thus $zeta(s)$ has no zeros there.
    $endgroup$
    – reuns
    Jul 6 '17 at 6:16










  • $begingroup$
    $(1-p^{-s})^{-1}$ is between $1$ and $2$ for every real positive $p$.
    $endgroup$
    – N74
    Jul 6 '17 at 6:18












  • $begingroup$
    Thanks for the answers. So like here: math.stackexchange.com/a/1097980/268374, why do we know $(1-p^{-s})^{-1}neq 0$ for all primes $p$?
    $endgroup$
    – rb612
    Jul 6 '17 at 6:20






  • 1




    $begingroup$
    Also note that when $s < 0$ : $prod_p frac{1}{1-p^{-s}} = 0$ (the product is said to diverge to $0$). That's why it is important to restrict to $Re(s) > 1$, where we have the equality $sum_{n=1}^infty n^{-s} = prod_p frac{1}{1-p^{-s}}$.
    $endgroup$
    – reuns
    Jul 6 '17 at 6:24








  • 1




    $begingroup$
    That's what I said : here the product converges and its factor are non-zero for $Re(s) > 1$ really means that $-sum_p log(1-p^{-s}) = log(prod_p frac{1}{1-p^{-s}})=log zeta(s)$ is analytic for $Re(s) > 1$.
    $endgroup$
    – reuns
    Jul 6 '17 at 6:29


















0












$begingroup$


I'm fascinated by the Riemann Zeta hypothesis - I haven't yet taken a course in complex analysis but I'm curious what this sentence means on Wikipedia:




The convergence of the Euler product shows that ζ(s) has no zeros in
this region, as none of the factors have zeros.




I'm wondering if a) the convergence of the Euler product comes purely from the fact that the p-series diverges when the exponent is greater than 1?
b) What the sentence from Wikipedia is essentially saying - I'm not understanding what factors are being referred to or how the convergence shows that the Zeta function has no zeroes for $Re(s) > 1$.



Edit: I was looking over this answer and it seems to be similar, but how do we know that, as this author wrote, $(1-p^{-s})^{-1}neq 0$ for all primes $p$? Can't it converge to 0, meaning that $(1-p^{-s})^{-1} = 0$ for at least one prime $p$?






sequences-and-series complex-analysis riemann-zeta euler-product







share|cite|improve this question











$endgroup$












  • $begingroup$
    The idea is that $zeta(s) =prod_p frac{1}{1-p^{-s}} = exp(F(s))$ where $F(s) = -sum_p log(1-p^{-s}) = sum_{p^k} frac{p^{-sk}}{k}$ is analytic on $Re(s) > 1$, thus $zeta(s)$ has no zeros there.
    $endgroup$
    – reuns
    Jul 6 '17 at 6:16










  • $begingroup$
    $(1-p^{-s})^{-1}$ is between $1$ and $2$ for every real positive $p$.
    $endgroup$
    – N74
    Jul 6 '17 at 6:18












  • $begingroup$
    Thanks for the answers. So like here: math.stackexchange.com/a/1097980/268374, why do we know $(1-p^{-s})^{-1}neq 0$ for all primes $p$?
    $endgroup$
    – rb612
    Jul 6 '17 at 6:20






  • 1




    $begingroup$
    Also note that when $s < 0$ : $prod_p frac{1}{1-p^{-s}} = 0$ (the product is said to diverge to $0$). That's why it is important to restrict to $Re(s) > 1$, where we have the equality $sum_{n=1}^infty n^{-s} = prod_p frac{1}{1-p^{-s}}$.
    $endgroup$
    – reuns
    Jul 6 '17 at 6:24








  • 1




    $begingroup$
    That's what I said : here the product converges and its factor are non-zero for $Re(s) > 1$ really means that $-sum_p log(1-p^{-s}) = log(prod_p frac{1}{1-p^{-s}})=log zeta(s)$ is analytic for $Re(s) > 1$.
    $endgroup$
    – reuns
    Jul 6 '17 at 6:29
















0












0








0





$begingroup$


I'm fascinated by the Riemann Zeta hypothesis - I haven't yet taken a course in complex analysis but I'm curious what this sentence means on Wikipedia:




The convergence of the Euler product shows that ζ(s) has no zeros in
this region, as none of the factors have zeros.




I'm wondering if a) the convergence of the Euler product comes purely from the fact that the p-series diverges when the exponent is greater than 1?
b) What the sentence from Wikipedia is essentially saying - I'm not understanding what factors are being referred to or how the convergence shows that the Zeta function has no zeroes for $Re(s) > 1$.



Edit: I was looking over this answer and it seems to be similar, but how do we know that, as this author wrote, $(1-p^{-s})^{-1}neq 0$ for all primes $p$? Can't it converge to 0, meaning that $(1-p^{-s})^{-1} = 0$ for at least one prime $p$?






sequences-and-series complex-analysis riemann-zeta euler-product







share|cite|improve this question











$endgroup$




I'm fascinated by the Riemann Zeta hypothesis - I haven't yet taken a course in complex analysis but I'm curious what this sentence means on Wikipedia:




The convergence of the Euler product shows that ζ(s) has no zeros in
this region, as none of the factors have zeros.




I'm wondering if a) the convergence of the Euler product comes purely from the fact that the p-series diverges when the exponent is greater than 1?
b) What the sentence from Wikipedia is essentially saying - I'm not understanding what factors are being referred to or how the convergence shows that the Zeta function has no zeroes for $Re(s) > 1$.



Edit: I was looking over this answer and it seems to be similar, but how do we know that, as this author wrote, $(1-p^{-s})^{-1}neq 0$ for all primes $p$? Can't it converge to 0, meaning that $(1-p^{-s})^{-1} = 0$ for at least one prime $p$?






sequences-and-series complex-analysis riemann-zeta euler-product




sequences-and-series complex-analysis riemann-zeta euler-product






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 6 '17 at 6:15









reuns

21.4k21352




21.4k21352










asked Jul 6 '17 at 6:04









rb612rb612

1,8531924




1,8531924












  • $begingroup$
    The idea is that $zeta(s) =prod_p frac{1}{1-p^{-s}} = exp(F(s))$ where $F(s) = -sum_p log(1-p^{-s}) = sum_{p^k} frac{p^{-sk}}{k}$ is analytic on $Re(s) > 1$, thus $zeta(s)$ has no zeros there.
    $endgroup$
    – reuns
    Jul 6 '17 at 6:16










  • $begingroup$
    $(1-p^{-s})^{-1}$ is between $1$ and $2$ for every real positive $p$.
    $endgroup$
    – N74
    Jul 6 '17 at 6:18












  • $begingroup$
    Thanks for the answers. So like here: math.stackexchange.com/a/1097980/268374, why do we know $(1-p^{-s})^{-1}neq 0$ for all primes $p$?
    $endgroup$
    – rb612
    Jul 6 '17 at 6:20






  • 1




    $begingroup$
    Also note that when $s < 0$ : $prod_p frac{1}{1-p^{-s}} = 0$ (the product is said to diverge to $0$). That's why it is important to restrict to $Re(s) > 1$, where we have the equality $sum_{n=1}^infty n^{-s} = prod_p frac{1}{1-p^{-s}}$.
    $endgroup$
    – reuns
    Jul 6 '17 at 6:24








  • 1




    $begingroup$
    That's what I said : here the product converges and its factor are non-zero for $Re(s) > 1$ really means that $-sum_p log(1-p^{-s}) = log(prod_p frac{1}{1-p^{-s}})=log zeta(s)$ is analytic for $Re(s) > 1$.
    $endgroup$
    – reuns
    Jul 6 '17 at 6:29




















  • $begingroup$
    The idea is that $zeta(s) =prod_p frac{1}{1-p^{-s}} = exp(F(s))$ where $F(s) = -sum_p log(1-p^{-s}) = sum_{p^k} frac{p^{-sk}}{k}$ is analytic on $Re(s) > 1$, thus $zeta(s)$ has no zeros there.
    $endgroup$
    – reuns
    Jul 6 '17 at 6:16










  • $begingroup$
    $(1-p^{-s})^{-1}$ is between $1$ and $2$ for every real positive $p$.
    $endgroup$
    – N74
    Jul 6 '17 at 6:18












  • $begingroup$
    Thanks for the answers. So like here: math.stackexchange.com/a/1097980/268374, why do we know $(1-p^{-s})^{-1}neq 0$ for all primes $p$?
    $endgroup$
    – rb612
    Jul 6 '17 at 6:20






  • 1




    $begingroup$
    Also note that when $s < 0$ : $prod_p frac{1}{1-p^{-s}} = 0$ (the product is said to diverge to $0$). That's why it is important to restrict to $Re(s) > 1$, where we have the equality $sum_{n=1}^infty n^{-s} = prod_p frac{1}{1-p^{-s}}$.
    $endgroup$
    – reuns
    Jul 6 '17 at 6:24








  • 1




    $begingroup$
    That's what I said : here the product converges and its factor are non-zero for $Re(s) > 1$ really means that $-sum_p log(1-p^{-s}) = log(prod_p frac{1}{1-p^{-s}})=log zeta(s)$ is analytic for $Re(s) > 1$.
    $endgroup$
    – reuns
    Jul 6 '17 at 6:29


















$begingroup$
The idea is that $zeta(s) =prod_p frac{1}{1-p^{-s}} = exp(F(s))$ where $F(s) = -sum_p log(1-p^{-s}) = sum_{p^k} frac{p^{-sk}}{k}$ is analytic on $Re(s) > 1$, thus $zeta(s)$ has no zeros there.
$endgroup$
– reuns
Jul 6 '17 at 6:16




$begingroup$
The idea is that $zeta(s) =prod_p frac{1}{1-p^{-s}} = exp(F(s))$ where $F(s) = -sum_p log(1-p^{-s}) = sum_{p^k} frac{p^{-sk}}{k}$ is analytic on $Re(s) > 1$, thus $zeta(s)$ has no zeros there.
$endgroup$
– reuns
Jul 6 '17 at 6:16












$begingroup$
$(1-p^{-s})^{-1}$ is between $1$ and $2$ for every real positive $p$.
$endgroup$
– N74
Jul 6 '17 at 6:18






$begingroup$
$(1-p^{-s})^{-1}$ is between $1$ and $2$ for every real positive $p$.
$endgroup$
– N74
Jul 6 '17 at 6:18














$begingroup$
Thanks for the answers. So like here: math.stackexchange.com/a/1097980/268374, why do we know $(1-p^{-s})^{-1}neq 0$ for all primes $p$?
$endgroup$
– rb612
Jul 6 '17 at 6:20




$begingroup$
Thanks for the answers. So like here: math.stackexchange.com/a/1097980/268374, why do we know $(1-p^{-s})^{-1}neq 0$ for all primes $p$?
$endgroup$
– rb612
Jul 6 '17 at 6:20




1




1




$begingroup$
Also note that when $s < 0$ : $prod_p frac{1}{1-p^{-s}} = 0$ (the product is said to diverge to $0$). That's why it is important to restrict to $Re(s) > 1$, where we have the equality $sum_{n=1}^infty n^{-s} = prod_p frac{1}{1-p^{-s}}$.
$endgroup$
– reuns
Jul 6 '17 at 6:24






$begingroup$
Also note that when $s < 0$ : $prod_p frac{1}{1-p^{-s}} = 0$ (the product is said to diverge to $0$). That's why it is important to restrict to $Re(s) > 1$, where we have the equality $sum_{n=1}^infty n^{-s} = prod_p frac{1}{1-p^{-s}}$.
$endgroup$
– reuns
Jul 6 '17 at 6:24






1




1




$begingroup$
That's what I said : here the product converges and its factor are non-zero for $Re(s) > 1$ really means that $-sum_p log(1-p^{-s}) = log(prod_p frac{1}{1-p^{-s}})=log zeta(s)$ is analytic for $Re(s) > 1$.
$endgroup$
– reuns
Jul 6 '17 at 6:29






$begingroup$
That's what I said : here the product converges and its factor are non-zero for $Re(s) > 1$ really means that $-sum_p log(1-p^{-s}) = log(prod_p frac{1}{1-p^{-s}})=log zeta(s)$ is analytic for $Re(s) > 1$.
$endgroup$
– reuns
Jul 6 '17 at 6:29












1 Answer
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0












$begingroup$

The Riemann Zeta function defined as



$$
zeta(s) = sum_{n=1}^{infty} n^{-s}
$$



for $Re(s)>1$ is convergent and admits the Euler product representation



$$
zeta(s) = prod_p (1-p^{-s})^{-1}.
$$



Since none of the factors in the product equal zero (i.e., $p^{-s}$ is never equal to $1$), $zeta(s)$ is never equal to zero in the half-plane $Re(s)>1$.






share|cite|improve this answer









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    1 Answer
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    1 Answer
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    0












    $begingroup$

    The Riemann Zeta function defined as



    $$
    zeta(s) = sum_{n=1}^{infty} n^{-s}
    $$



    for $Re(s)>1$ is convergent and admits the Euler product representation



    $$
    zeta(s) = prod_p (1-p^{-s})^{-1}.
    $$



    Since none of the factors in the product equal zero (i.e., $p^{-s}$ is never equal to $1$), $zeta(s)$ is never equal to zero in the half-plane $Re(s)>1$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The Riemann Zeta function defined as



      $$
      zeta(s) = sum_{n=1}^{infty} n^{-s}
      $$



      for $Re(s)>1$ is convergent and admits the Euler product representation



      $$
      zeta(s) = prod_p (1-p^{-s})^{-1}.
      $$



      Since none of the factors in the product equal zero (i.e., $p^{-s}$ is never equal to $1$), $zeta(s)$ is never equal to zero in the half-plane $Re(s)>1$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The Riemann Zeta function defined as



        $$
        zeta(s) = sum_{n=1}^{infty} n^{-s}
        $$



        for $Re(s)>1$ is convergent and admits the Euler product representation



        $$
        zeta(s) = prod_p (1-p^{-s})^{-1}.
        $$



        Since none of the factors in the product equal zero (i.e., $p^{-s}$ is never equal to $1$), $zeta(s)$ is never equal to zero in the half-plane $Re(s)>1$.






        share|cite|improve this answer









        $endgroup$



        The Riemann Zeta function defined as



        $$
        zeta(s) = sum_{n=1}^{infty} n^{-s}
        $$



        for $Re(s)>1$ is convergent and admits the Euler product representation



        $$
        zeta(s) = prod_p (1-p^{-s})^{-1}.
        $$



        Since none of the factors in the product equal zero (i.e., $p^{-s}$ is never equal to $1$), $zeta(s)$ is never equal to zero in the half-plane $Re(s)>1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 17 '18 at 10:22









        KlangenKlangen

        1,74411334




        1,74411334






























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