Why shouldn't this prove the Prime Number Theorem?












3












$begingroup$


Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ can be interpreted as the probability that a randomly chosen integer is $k$-free.



Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form



$$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$



since the probability that an integer is ``$1$-free'' is zero ?






nt.number-theory prime-numbers







share|cite|improve this question







New contributor




Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 11




    $begingroup$
    It is true that the PNT is equivalent to $sum_{n leq x} frac{mu(n)}{n} = o(1)$. It is also relatively easy to prove that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = 0$. The hard part is proving that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = lim_{x to infty} sum_{n leq x} frac{mu(n)}{n}$. This is highly nontrivial!
    $endgroup$
    – Peter Humphries
    5 hours ago






  • 11




    $begingroup$
    In general, limit of sums of series $neq$ sum of limits of series. In this particular case, the equality does hold, but it requires intricate arguments to prove, which you see in any proof of PNT.
    $endgroup$
    – Wojowu
    5 hours ago






  • 6




    $begingroup$
    I think this question should be reopened, and the comments made by Peter Humphries and Wojowu posted as an answer. The question might be borderline too elementary for MO but it is natural and I'm sure I'm not the only one to have been confused by this at some (embarrassingly recent) point, it's a bit silly to close when, in effect, the answer is there.
    $endgroup$
    – Gro-Tsen
    4 hours ago






  • 1




    $begingroup$
    I agree with Fourton and have voted accordingly
    $endgroup$
    – Yemon Choi
    4 hours ago
















3












$begingroup$


Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ can be interpreted as the probability that a randomly chosen integer is $k$-free.



Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form



$$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$



since the probability that an integer is ``$1$-free'' is zero ?






nt.number-theory prime-numbers







share|cite|improve this question







New contributor




Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 11




    $begingroup$
    It is true that the PNT is equivalent to $sum_{n leq x} frac{mu(n)}{n} = o(1)$. It is also relatively easy to prove that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = 0$. The hard part is proving that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = lim_{x to infty} sum_{n leq x} frac{mu(n)}{n}$. This is highly nontrivial!
    $endgroup$
    – Peter Humphries
    5 hours ago






  • 11




    $begingroup$
    In general, limit of sums of series $neq$ sum of limits of series. In this particular case, the equality does hold, but it requires intricate arguments to prove, which you see in any proof of PNT.
    $endgroup$
    – Wojowu
    5 hours ago






  • 6




    $begingroup$
    I think this question should be reopened, and the comments made by Peter Humphries and Wojowu posted as an answer. The question might be borderline too elementary for MO but it is natural and I'm sure I'm not the only one to have been confused by this at some (embarrassingly recent) point, it's a bit silly to close when, in effect, the answer is there.
    $endgroup$
    – Gro-Tsen
    4 hours ago






  • 1




    $begingroup$
    I agree with Fourton and have voted accordingly
    $endgroup$
    – Yemon Choi
    4 hours ago














3












3








3





$begingroup$


Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ can be interpreted as the probability that a randomly chosen integer is $k$-free.



Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form



$$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$



since the probability that an integer is ``$1$-free'' is zero ?






nt.number-theory prime-numbers







share|cite|improve this question







New contributor




Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ can be interpreted as the probability that a randomly chosen integer is $k$-free.



Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form



$$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$



since the probability that an integer is ``$1$-free'' is zero ?






nt.number-theory prime-numbers




nt.number-theory prime-numbers






share|cite|improve this question







New contributor




Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 5 hours ago









Fourton.Fourton.

422




422




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New contributor





Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Fourton. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 11




    $begingroup$
    It is true that the PNT is equivalent to $sum_{n leq x} frac{mu(n)}{n} = o(1)$. It is also relatively easy to prove that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = 0$. The hard part is proving that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = lim_{x to infty} sum_{n leq x} frac{mu(n)}{n}$. This is highly nontrivial!
    $endgroup$
    – Peter Humphries
    5 hours ago






  • 11




    $begingroup$
    In general, limit of sums of series $neq$ sum of limits of series. In this particular case, the equality does hold, but it requires intricate arguments to prove, which you see in any proof of PNT.
    $endgroup$
    – Wojowu
    5 hours ago






  • 6




    $begingroup$
    I think this question should be reopened, and the comments made by Peter Humphries and Wojowu posted as an answer. The question might be borderline too elementary for MO but it is natural and I'm sure I'm not the only one to have been confused by this at some (embarrassingly recent) point, it's a bit silly to close when, in effect, the answer is there.
    $endgroup$
    – Gro-Tsen
    4 hours ago






  • 1




    $begingroup$
    I agree with Fourton and have voted accordingly
    $endgroup$
    – Yemon Choi
    4 hours ago














  • 11




    $begingroup$
    It is true that the PNT is equivalent to $sum_{n leq x} frac{mu(n)}{n} = o(1)$. It is also relatively easy to prove that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = 0$. The hard part is proving that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = lim_{x to infty} sum_{n leq x} frac{mu(n)}{n}$. This is highly nontrivial!
    $endgroup$
    – Peter Humphries
    5 hours ago






  • 11




    $begingroup$
    In general, limit of sums of series $neq$ sum of limits of series. In this particular case, the equality does hold, but it requires intricate arguments to prove, which you see in any proof of PNT.
    $endgroup$
    – Wojowu
    5 hours ago






  • 6




    $begingroup$
    I think this question should be reopened, and the comments made by Peter Humphries and Wojowu posted as an answer. The question might be borderline too elementary for MO but it is natural and I'm sure I'm not the only one to have been confused by this at some (embarrassingly recent) point, it's a bit silly to close when, in effect, the answer is there.
    $endgroup$
    – Gro-Tsen
    4 hours ago






  • 1




    $begingroup$
    I agree with Fourton and have voted accordingly
    $endgroup$
    – Yemon Choi
    4 hours ago








11




11




$begingroup$
It is true that the PNT is equivalent to $sum_{n leq x} frac{mu(n)}{n} = o(1)$. It is also relatively easy to prove that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = 0$. The hard part is proving that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = lim_{x to infty} sum_{n leq x} frac{mu(n)}{n}$. This is highly nontrivial!
$endgroup$
– Peter Humphries
5 hours ago




$begingroup$
It is true that the PNT is equivalent to $sum_{n leq x} frac{mu(n)}{n} = o(1)$. It is also relatively easy to prove that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = 0$. The hard part is proving that $lim_{s searrow 1} sum_{n = 1}^{infty} frac{mu(n)}{n^s} = lim_{x to infty} sum_{n leq x} frac{mu(n)}{n}$. This is highly nontrivial!
$endgroup$
– Peter Humphries
5 hours ago




11




11




$begingroup$
In general, limit of sums of series $neq$ sum of limits of series. In this particular case, the equality does hold, but it requires intricate arguments to prove, which you see in any proof of PNT.
$endgroup$
– Wojowu
5 hours ago




$begingroup$
In general, limit of sums of series $neq$ sum of limits of series. In this particular case, the equality does hold, but it requires intricate arguments to prove, which you see in any proof of PNT.
$endgroup$
– Wojowu
5 hours ago




6




6




$begingroup$
I think this question should be reopened, and the comments made by Peter Humphries and Wojowu posted as an answer. The question might be borderline too elementary for MO but it is natural and I'm sure I'm not the only one to have been confused by this at some (embarrassingly recent) point, it's a bit silly to close when, in effect, the answer is there.
$endgroup$
– Gro-Tsen
4 hours ago




$begingroup$
I think this question should be reopened, and the comments made by Peter Humphries and Wojowu posted as an answer. The question might be borderline too elementary for MO but it is natural and I'm sure I'm not the only one to have been confused by this at some (embarrassingly recent) point, it's a bit silly to close when, in effect, the answer is there.
$endgroup$
– Gro-Tsen
4 hours ago




1




1




$begingroup$
I agree with Fourton and have voted accordingly
$endgroup$
– Yemon Choi
4 hours ago




$begingroup$
I agree with Fourton and have voted accordingly
$endgroup$
– Yemon Choi
4 hours ago










1 Answer
1






active

oldest

votes


















2












$begingroup$

You ask:




Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ can be interpreted as the probability that a randomly chosen integer is $k$-free.



Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form



$$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$



since the probability that an integer is ``$1$-free'' is zero ?




As pointed out by the users @wojowu and @PeterHumphries,
it is true that the PNT is equivalent to



$$sum_{n=1}^{infty} frac{mu(n)}{n}=o(1),$$
and it is relatively easy to prove that



$$lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s}=0.$$
The real difficulty lies in proving that



$$lim_{xrightarrow infty} sum_{nleq x} frac{mu(n)}{n}=
lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s},$$
which is highly nontrivial and requires intricate arguments.






share|cite|improve this answer











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    active

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    active

    oldest

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    2












    $begingroup$

    You ask:




    Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ can be interpreted as the probability that a randomly chosen integer is $k$-free.



    Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form



    $$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$



    since the probability that an integer is ``$1$-free'' is zero ?




    As pointed out by the users @wojowu and @PeterHumphries,
    it is true that the PNT is equivalent to



    $$sum_{n=1}^{infty} frac{mu(n)}{n}=o(1),$$
    and it is relatively easy to prove that



    $$lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s}=0.$$
    The real difficulty lies in proving that



    $$lim_{xrightarrow infty} sum_{nleq x} frac{mu(n)}{n}=
    lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s},$$
    which is highly nontrivial and requires intricate arguments.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      You ask:




      Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ can be interpreted as the probability that a randomly chosen integer is $k$-free.



      Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form



      $$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$



      since the probability that an integer is ``$1$-free'' is zero ?




      As pointed out by the users @wojowu and @PeterHumphries,
      it is true that the PNT is equivalent to



      $$sum_{n=1}^{infty} frac{mu(n)}{n}=o(1),$$
      and it is relatively easy to prove that



      $$lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s}=0.$$
      The real difficulty lies in proving that



      $$lim_{xrightarrow infty} sum_{nleq x} frac{mu(n)}{n}=
      lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s},$$
      which is highly nontrivial and requires intricate arguments.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        You ask:




        Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ can be interpreted as the probability that a randomly chosen integer is $k$-free.



        Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form



        $$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$



        since the probability that an integer is ``$1$-free'' is zero ?




        As pointed out by the users @wojowu and @PeterHumphries,
        it is true that the PNT is equivalent to



        $$sum_{n=1}^{infty} frac{mu(n)}{n}=o(1),$$
        and it is relatively easy to prove that



        $$lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s}=0.$$
        The real difficulty lies in proving that



        $$lim_{xrightarrow infty} sum_{nleq x} frac{mu(n)}{n}=
        lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s},$$
        which is highly nontrivial and requires intricate arguments.






        share|cite|improve this answer











        $endgroup$



        You ask:




        Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_{n=1}^{infty} frac{mu(n)}{n^k}$ can be interpreted as the probability that a randomly chosen integer is $k$-free.



        Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form



        $$sum_{n=1}^{infty} frac{mu(n)}{n}=0,$$



        since the probability that an integer is ``$1$-free'' is zero ?




        As pointed out by the users @wojowu and @PeterHumphries,
        it is true that the PNT is equivalent to



        $$sum_{n=1}^{infty} frac{mu(n)}{n}=o(1),$$
        and it is relatively easy to prove that



        $$lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s}=0.$$
        The real difficulty lies in proving that



        $$lim_{xrightarrow infty} sum_{nleq x} frac{mu(n)}{n}=
        lim_{srightarrow 1^+} sum_{n=1}^{infty} frac{mu(n)}{n^s},$$
        which is highly nontrivial and requires intricate arguments.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








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