Is Apéry's constant a rational multiple of $ pi ^ 3$?












2












$begingroup$


It is well known that the values of the Riemann zeta function for even positive numbers are of the form:



$$zeta(2k) = rm rational * pi ^{2k},$$



and more specifically $zeta (2k)=(-1)^{{k+1}}{frac {B_{{2k}}(2pi )^{{2k}}}{2(2k)!}}!$. It is not that far-fetched to consider that



$$zeta(2k + 1) = rm rational * pi ^{2k + 1}.$$



Specifically for Apéry's constant (which is $zeta(3)$), did someone prove something like that? The proof should be something like:




$frac{zeta(3)}{pi^3}$ is rational / irrational / transcendental.




EDIT: Even if the question is still open (which I can see it is from the comments), is there any new development on this matter lately? Just curious.










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    No one has proved anything like this.
    $endgroup$
    – Lord Shark the Unknown
    Sep 16 '17 at 18:02






  • 4




    $begingroup$
    This is a well-known question, see this MO-question.
    $endgroup$
    – Dietrich Burde
    Sep 16 '17 at 18:04






  • 2




    $begingroup$
    A significant amount of effort has gone into this, but the nature of Apery's constant is still largely mysterious.
    $endgroup$
    – George Coote
    Sep 16 '17 at 18:07






  • 4




    $begingroup$
    I am not sure whether the problem is open. But I would be rather surprised if $large frac{zeta(3)}{pi^3}$ turned out to be rational. My guess is that it is even transcendental (of course, only a guess). The continued fraction I calculated with PARI/GP with $20 000$ digits accuracy, contains $19501$ entries not exceeding $134656$. So, if the constant IS rational, numerator and denominator must be very large.
    $endgroup$
    – Peter
    Sep 16 '17 at 18:10








  • 2




    $begingroup$
    Cross-site duplicate
    $endgroup$
    – Wojowu
    Sep 16 '17 at 18:21


















2












$begingroup$


It is well known that the values of the Riemann zeta function for even positive numbers are of the form:



$$zeta(2k) = rm rational * pi ^{2k},$$



and more specifically $zeta (2k)=(-1)^{{k+1}}{frac {B_{{2k}}(2pi )^{{2k}}}{2(2k)!}}!$. It is not that far-fetched to consider that



$$zeta(2k + 1) = rm rational * pi ^{2k + 1}.$$



Specifically for Apéry's constant (which is $zeta(3)$), did someone prove something like that? The proof should be something like:




$frac{zeta(3)}{pi^3}$ is rational / irrational / transcendental.




EDIT: Even if the question is still open (which I can see it is from the comments), is there any new development on this matter lately? Just curious.










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    No one has proved anything like this.
    $endgroup$
    – Lord Shark the Unknown
    Sep 16 '17 at 18:02






  • 4




    $begingroup$
    This is a well-known question, see this MO-question.
    $endgroup$
    – Dietrich Burde
    Sep 16 '17 at 18:04






  • 2




    $begingroup$
    A significant amount of effort has gone into this, but the nature of Apery's constant is still largely mysterious.
    $endgroup$
    – George Coote
    Sep 16 '17 at 18:07






  • 4




    $begingroup$
    I am not sure whether the problem is open. But I would be rather surprised if $large frac{zeta(3)}{pi^3}$ turned out to be rational. My guess is that it is even transcendental (of course, only a guess). The continued fraction I calculated with PARI/GP with $20 000$ digits accuracy, contains $19501$ entries not exceeding $134656$. So, if the constant IS rational, numerator and denominator must be very large.
    $endgroup$
    – Peter
    Sep 16 '17 at 18:10








  • 2




    $begingroup$
    Cross-site duplicate
    $endgroup$
    – Wojowu
    Sep 16 '17 at 18:21
















2












2








2





$begingroup$


It is well known that the values of the Riemann zeta function for even positive numbers are of the form:



$$zeta(2k) = rm rational * pi ^{2k},$$



and more specifically $zeta (2k)=(-1)^{{k+1}}{frac {B_{{2k}}(2pi )^{{2k}}}{2(2k)!}}!$. It is not that far-fetched to consider that



$$zeta(2k + 1) = rm rational * pi ^{2k + 1}.$$



Specifically for Apéry's constant (which is $zeta(3)$), did someone prove something like that? The proof should be something like:




$frac{zeta(3)}{pi^3}$ is rational / irrational / transcendental.




EDIT: Even if the question is still open (which I can see it is from the comments), is there any new development on this matter lately? Just curious.










share|cite|improve this question











$endgroup$




It is well known that the values of the Riemann zeta function for even positive numbers are of the form:



$$zeta(2k) = rm rational * pi ^{2k},$$



and more specifically $zeta (2k)=(-1)^{{k+1}}{frac {B_{{2k}}(2pi )^{{2k}}}{2(2k)!}}!$. It is not that far-fetched to consider that



$$zeta(2k + 1) = rm rational * pi ^{2k + 1}.$$



Specifically for Apéry's constant (which is $zeta(3)$), did someone prove something like that? The proof should be something like:




$frac{zeta(3)}{pi^3}$ is rational / irrational / transcendental.




EDIT: Even if the question is still open (which I can see it is from the comments), is there any new development on this matter lately? Just curious.







complex-analysis number-theory riemann-zeta






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 11:39









Klangen

1,72811334




1,72811334










asked Sep 16 '17 at 18:01









Sagi ShadurSagi Shadur

340116




340116








  • 3




    $begingroup$
    No one has proved anything like this.
    $endgroup$
    – Lord Shark the Unknown
    Sep 16 '17 at 18:02






  • 4




    $begingroup$
    This is a well-known question, see this MO-question.
    $endgroup$
    – Dietrich Burde
    Sep 16 '17 at 18:04






  • 2




    $begingroup$
    A significant amount of effort has gone into this, but the nature of Apery's constant is still largely mysterious.
    $endgroup$
    – George Coote
    Sep 16 '17 at 18:07






  • 4




    $begingroup$
    I am not sure whether the problem is open. But I would be rather surprised if $large frac{zeta(3)}{pi^3}$ turned out to be rational. My guess is that it is even transcendental (of course, only a guess). The continued fraction I calculated with PARI/GP with $20 000$ digits accuracy, contains $19501$ entries not exceeding $134656$. So, if the constant IS rational, numerator and denominator must be very large.
    $endgroup$
    – Peter
    Sep 16 '17 at 18:10








  • 2




    $begingroup$
    Cross-site duplicate
    $endgroup$
    – Wojowu
    Sep 16 '17 at 18:21
















  • 3




    $begingroup$
    No one has proved anything like this.
    $endgroup$
    – Lord Shark the Unknown
    Sep 16 '17 at 18:02






  • 4




    $begingroup$
    This is a well-known question, see this MO-question.
    $endgroup$
    – Dietrich Burde
    Sep 16 '17 at 18:04






  • 2




    $begingroup$
    A significant amount of effort has gone into this, but the nature of Apery's constant is still largely mysterious.
    $endgroup$
    – George Coote
    Sep 16 '17 at 18:07






  • 4




    $begingroup$
    I am not sure whether the problem is open. But I would be rather surprised if $large frac{zeta(3)}{pi^3}$ turned out to be rational. My guess is that it is even transcendental (of course, only a guess). The continued fraction I calculated with PARI/GP with $20 000$ digits accuracy, contains $19501$ entries not exceeding $134656$. So, if the constant IS rational, numerator and denominator must be very large.
    $endgroup$
    – Peter
    Sep 16 '17 at 18:10








  • 2




    $begingroup$
    Cross-site duplicate
    $endgroup$
    – Wojowu
    Sep 16 '17 at 18:21










3




3




$begingroup$
No one has proved anything like this.
$endgroup$
– Lord Shark the Unknown
Sep 16 '17 at 18:02




$begingroup$
No one has proved anything like this.
$endgroup$
– Lord Shark the Unknown
Sep 16 '17 at 18:02




4




4




$begingroup$
This is a well-known question, see this MO-question.
$endgroup$
– Dietrich Burde
Sep 16 '17 at 18:04




$begingroup$
This is a well-known question, see this MO-question.
$endgroup$
– Dietrich Burde
Sep 16 '17 at 18:04




2




2




$begingroup$
A significant amount of effort has gone into this, but the nature of Apery's constant is still largely mysterious.
$endgroup$
– George Coote
Sep 16 '17 at 18:07




$begingroup$
A significant amount of effort has gone into this, but the nature of Apery's constant is still largely mysterious.
$endgroup$
– George Coote
Sep 16 '17 at 18:07




4




4




$begingroup$
I am not sure whether the problem is open. But I would be rather surprised if $large frac{zeta(3)}{pi^3}$ turned out to be rational. My guess is that it is even transcendental (of course, only a guess). The continued fraction I calculated with PARI/GP with $20 000$ digits accuracy, contains $19501$ entries not exceeding $134656$. So, if the constant IS rational, numerator and denominator must be very large.
$endgroup$
– Peter
Sep 16 '17 at 18:10






$begingroup$
I am not sure whether the problem is open. But I would be rather surprised if $large frac{zeta(3)}{pi^3}$ turned out to be rational. My guess is that it is even transcendental (of course, only a guess). The continued fraction I calculated with PARI/GP with $20 000$ digits accuracy, contains $19501$ entries not exceeding $134656$. So, if the constant IS rational, numerator and denominator must be very large.
$endgroup$
– Peter
Sep 16 '17 at 18:10






2




2




$begingroup$
Cross-site duplicate
$endgroup$
– Wojowu
Sep 16 '17 at 18:21






$begingroup$
Cross-site duplicate
$endgroup$
– Wojowu
Sep 16 '17 at 18:21












1 Answer
1






active

oldest

votes


















0












$begingroup$

While Apéry proved in 1978 that $zeta(3)$ is irrational, the irrationality of $frac{zeta(3)}{pi^3}$ is still an open problem. There are some formulas expressing odd zeta values in terms of powers of $pi$, the most known ones are due to Plouffe and Borwein & Bradley. Here are some examples:



$$
begin{aligned}
zeta(3)&=frac{7pi^3}{180}-2sum_{n=1}^infty frac{1}{n^3(e^{2pi n}-1)},\
sum_{n=1}^infty frac{1}{n^3,binom {2n}n} &= -frac{4}{3},zeta(3)+frac{pisqrt{3}}{2cdot 3^2},left(zeta(2, tfrac{1}{3})-zeta(2,tfrac{2}{3}) right).
end{aligned}
$$



Moreover, in this Math.SE post we have:



$$
frac{3}{2},zeta(3) = frac{pi^3}{24}sqrt{2}-2sum_{k=1}^infty frac{1}{k^3(e^{pi ksqrt{2}}-1)}-sum_{k=1}^inftyfrac{1}{k^3(e^{2pi ksqrt{2}}-1)}.
$$



You can also check out this paper by Vepstas, which provides a nice generalization to some of these identities.






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    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
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2432055%2fis-ap%25c3%25a9rys-constant-a-rational-multiple-of-pi-3%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









    0












    $begingroup$

    While Apéry proved in 1978 that $zeta(3)$ is irrational, the irrationality of $frac{zeta(3)}{pi^3}$ is still an open problem. There are some formulas expressing odd zeta values in terms of powers of $pi$, the most known ones are due to Plouffe and Borwein & Bradley. Here are some examples:



    $$
    begin{aligned}
    zeta(3)&=frac{7pi^3}{180}-2sum_{n=1}^infty frac{1}{n^3(e^{2pi n}-1)},\
    sum_{n=1}^infty frac{1}{n^3,binom {2n}n} &= -frac{4}{3},zeta(3)+frac{pisqrt{3}}{2cdot 3^2},left(zeta(2, tfrac{1}{3})-zeta(2,tfrac{2}{3}) right).
    end{aligned}
    $$



    Moreover, in this Math.SE post we have:



    $$
    frac{3}{2},zeta(3) = frac{pi^3}{24}sqrt{2}-2sum_{k=1}^infty frac{1}{k^3(e^{pi ksqrt{2}}-1)}-sum_{k=1}^inftyfrac{1}{k^3(e^{2pi ksqrt{2}}-1)}.
    $$



    You can also check out this paper by Vepstas, which provides a nice generalization to some of these identities.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      While Apéry proved in 1978 that $zeta(3)$ is irrational, the irrationality of $frac{zeta(3)}{pi^3}$ is still an open problem. There are some formulas expressing odd zeta values in terms of powers of $pi$, the most known ones are due to Plouffe and Borwein & Bradley. Here are some examples:



      $$
      begin{aligned}
      zeta(3)&=frac{7pi^3}{180}-2sum_{n=1}^infty frac{1}{n^3(e^{2pi n}-1)},\
      sum_{n=1}^infty frac{1}{n^3,binom {2n}n} &= -frac{4}{3},zeta(3)+frac{pisqrt{3}}{2cdot 3^2},left(zeta(2, tfrac{1}{3})-zeta(2,tfrac{2}{3}) right).
      end{aligned}
      $$



      Moreover, in this Math.SE post we have:



      $$
      frac{3}{2},zeta(3) = frac{pi^3}{24}sqrt{2}-2sum_{k=1}^infty frac{1}{k^3(e^{pi ksqrt{2}}-1)}-sum_{k=1}^inftyfrac{1}{k^3(e^{2pi ksqrt{2}}-1)}.
      $$



      You can also check out this paper by Vepstas, which provides a nice generalization to some of these identities.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        While Apéry proved in 1978 that $zeta(3)$ is irrational, the irrationality of $frac{zeta(3)}{pi^3}$ is still an open problem. There are some formulas expressing odd zeta values in terms of powers of $pi$, the most known ones are due to Plouffe and Borwein & Bradley. Here are some examples:



        $$
        begin{aligned}
        zeta(3)&=frac{7pi^3}{180}-2sum_{n=1}^infty frac{1}{n^3(e^{2pi n}-1)},\
        sum_{n=1}^infty frac{1}{n^3,binom {2n}n} &= -frac{4}{3},zeta(3)+frac{pisqrt{3}}{2cdot 3^2},left(zeta(2, tfrac{1}{3})-zeta(2,tfrac{2}{3}) right).
        end{aligned}
        $$



        Moreover, in this Math.SE post we have:



        $$
        frac{3}{2},zeta(3) = frac{pi^3}{24}sqrt{2}-2sum_{k=1}^infty frac{1}{k^3(e^{pi ksqrt{2}}-1)}-sum_{k=1}^inftyfrac{1}{k^3(e^{2pi ksqrt{2}}-1)}.
        $$



        You can also check out this paper by Vepstas, which provides a nice generalization to some of these identities.






        share|cite|improve this answer









        $endgroup$



        While Apéry proved in 1978 that $zeta(3)$ is irrational, the irrationality of $frac{zeta(3)}{pi^3}$ is still an open problem. There are some formulas expressing odd zeta values in terms of powers of $pi$, the most known ones are due to Plouffe and Borwein & Bradley. Here are some examples:



        $$
        begin{aligned}
        zeta(3)&=frac{7pi^3}{180}-2sum_{n=1}^infty frac{1}{n^3(e^{2pi n}-1)},\
        sum_{n=1}^infty frac{1}{n^3,binom {2n}n} &= -frac{4}{3},zeta(3)+frac{pisqrt{3}}{2cdot 3^2},left(zeta(2, tfrac{1}{3})-zeta(2,tfrac{2}{3}) right).
        end{aligned}
        $$



        Moreover, in this Math.SE post we have:



        $$
        frac{3}{2},zeta(3) = frac{pi^3}{24}sqrt{2}-2sum_{k=1}^infty frac{1}{k^3(e^{pi ksqrt{2}}-1)}-sum_{k=1}^inftyfrac{1}{k^3(e^{2pi ksqrt{2}}-1)}.
        $$



        You can also check out this paper by Vepstas, which provides a nice generalization to some of these identities.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 10:38









        KlangenKlangen

        1,72811334




        1,72811334






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • 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%2fmath.stackexchange.com%2fquestions%2f2432055%2fis-ap%25c3%25a9rys-constant-a-rational-multiple-of-pi-3%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

            Quarter-circle Tiles

            build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

            Mont Emei