Why shouldn't this prove the Prime Number Theorem? Planned maintenance scheduled April 23,...



Why shouldn't this prove the Prime Number Theorem?



Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Heuristic argument for the prime number theorem?Why is the Chebyshev function relevant to the Prime Number TheoremWhy could Mertens not prove the prime number theorem?Probability that randomly chosen integers from a restricted set of natural numbers are coprimeCan the following quantitative version of Chen's theorem be obtained?Any way to prove Prime Number Theorem using Hyperbolic Geometry?Any ways to Simplify Daboussi's Argument for Prime Number Theorem?Effective prime number theoremIs the number $sum_{ptext{ prime}}p^{-2}$ known to be irrational?Landau's theorem using nth roots












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 ?










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












$begingroup$


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 ?










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 ?










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






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




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.





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











$endgroup$














    Your Answer








    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "504"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });






    Fourton. is a new contributor. Be nice, and check out our Code of Conduct.










    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f328552%2fwhy-shouldnt-this-prove-the-prime-number-theorem%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    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











    $endgroup$


















      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








        answered 12 mins ago


























        community wiki





        kodlu























            Fourton. is a new contributor. Be nice, and check out our Code of Conduct.










            draft saved

            draft discarded


















            Fourton. is a new contributor. Be nice, and check out our Code of Conduct.













            Fourton. is a new contributor. Be nice, and check out our Code of Conduct.












            Fourton. is a new contributor. Be nice, and check out our Code of Conduct.
















            Thanks for contributing an answer to MathOverflow!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f328552%2fwhy-shouldnt-this-prove-the-prime-number-theorem%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            “%fieldName is a required field.”, in Magento2 REST API Call for GET Method Type The Next...

            How to change City field to a dropdown in Checkout step Magento 2Magento 2 : How to change UI field(s)...

            變成蝙蝠會怎樣? 參考資料 外部連結 导航菜单Thomas Nagel, "What is it like to be a...