Concerning this sum $sum_{k=1}^{infty}{2k choose k}^2 4^{-2k}cdot frac{1}{(ak)^3-ak}$












1












$begingroup$


I was looking at this Ramanujan phi-Function



Let:
$$R(a)=sum_{k=1}^{infty}frac{1}{(ak)^3-ak}tag1$$
and this paper of the form,



$$sum_{k=1}^{infty}frac{{2k choose k}}{4^k}tag2$$ and
$$sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2tag 3$$



By combining them together we have



$$S(a)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(ak)^3-ak}tag4$$ and



$$T(a)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(ak)^3-ak}tag5$$



The conjectured closed form for $(4)$ and $(5)$, where $a=2$ are



$$S(2)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(2k)^3-2k}=frac{pi}{4}-ln(2)tag6$$ and



$$T(2)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(2k)^3-2k}=frac{6G-piln(4)-1}{pi}tag7$$



Where G is the Catalan's constant.



How can we prove these conjectures?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Is it ${2n choose n}^2 4^{-2n}$ or ${2k choose k}^2 4^{-2k}$ in the headline?
    $endgroup$
    – Andreas
    Jan 3 at 9:43












  • $begingroup$
    Why don't you try the methods of the paper? ;)
    $endgroup$
    – Jack D'Aurizio
    Jan 3 at 17:04










  • $begingroup$
    Please notice that $(2)$ is actually a divergent series, since $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pi n}}$, so "this paper of the form divergent series" does not make much sense. Better to say "this paper about the weight $frac{1}{4^n}binom{2n}{n}$ and its square".
    $endgroup$
    – Jack D'Aurizio
    Jan 3 at 17:20










  • $begingroup$
    $(3)$ is also a divergent series (same reason), while $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^3$$ is related to the lemniscate constant (hence to $Gammaleft(frac{1}{4}right)$) by many good reasons; among them, Clausen's formula for the square of particular $phantom{}_2 F_1$s.
    $endgroup$
    – Jack D'Aurizio
    Jan 3 at 17:32










  • $begingroup$
    $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^4$$ is much more obscure: see math.stackexchange.com/questions/2506266/…
    $endgroup$
    – Jack D'Aurizio
    Jan 3 at 17:34
















1












$begingroup$


I was looking at this Ramanujan phi-Function



Let:
$$R(a)=sum_{k=1}^{infty}frac{1}{(ak)^3-ak}tag1$$
and this paper of the form,



$$sum_{k=1}^{infty}frac{{2k choose k}}{4^k}tag2$$ and
$$sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2tag 3$$



By combining them together we have



$$S(a)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(ak)^3-ak}tag4$$ and



$$T(a)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(ak)^3-ak}tag5$$



The conjectured closed form for $(4)$ and $(5)$, where $a=2$ are



$$S(2)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(2k)^3-2k}=frac{pi}{4}-ln(2)tag6$$ and



$$T(2)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(2k)^3-2k}=frac{6G-piln(4)-1}{pi}tag7$$



Where G is the Catalan's constant.



How can we prove these conjectures?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Is it ${2n choose n}^2 4^{-2n}$ or ${2k choose k}^2 4^{-2k}$ in the headline?
    $endgroup$
    – Andreas
    Jan 3 at 9:43












  • $begingroup$
    Why don't you try the methods of the paper? ;)
    $endgroup$
    – Jack D'Aurizio
    Jan 3 at 17:04










  • $begingroup$
    Please notice that $(2)$ is actually a divergent series, since $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pi n}}$, so "this paper of the form divergent series" does not make much sense. Better to say "this paper about the weight $frac{1}{4^n}binom{2n}{n}$ and its square".
    $endgroup$
    – Jack D'Aurizio
    Jan 3 at 17:20










  • $begingroup$
    $(3)$ is also a divergent series (same reason), while $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^3$$ is related to the lemniscate constant (hence to $Gammaleft(frac{1}{4}right)$) by many good reasons; among them, Clausen's formula for the square of particular $phantom{}_2 F_1$s.
    $endgroup$
    – Jack D'Aurizio
    Jan 3 at 17:32










  • $begingroup$
    $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^4$$ is much more obscure: see math.stackexchange.com/questions/2506266/…
    $endgroup$
    – Jack D'Aurizio
    Jan 3 at 17:34














1












1








1


3



$begingroup$


I was looking at this Ramanujan phi-Function



Let:
$$R(a)=sum_{k=1}^{infty}frac{1}{(ak)^3-ak}tag1$$
and this paper of the form,



$$sum_{k=1}^{infty}frac{{2k choose k}}{4^k}tag2$$ and
$$sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2tag 3$$



By combining them together we have



$$S(a)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(ak)^3-ak}tag4$$ and



$$T(a)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(ak)^3-ak}tag5$$



The conjectured closed form for $(4)$ and $(5)$, where $a=2$ are



$$S(2)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(2k)^3-2k}=frac{pi}{4}-ln(2)tag6$$ and



$$T(2)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(2k)^3-2k}=frac{6G-piln(4)-1}{pi}tag7$$



Where G is the Catalan's constant.



How can we prove these conjectures?










share|cite|improve this question











$endgroup$




I was looking at this Ramanujan phi-Function



Let:
$$R(a)=sum_{k=1}^{infty}frac{1}{(ak)^3-ak}tag1$$
and this paper of the form,



$$sum_{k=1}^{infty}frac{{2k choose k}}{4^k}tag2$$ and
$$sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2tag 3$$



By combining them together we have



$$S(a)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(ak)^3-ak}tag4$$ and



$$T(a)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(ak)^3-ak}tag5$$



The conjectured closed form for $(4)$ and $(5)$, where $a=2$ are



$$S(2)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(2k)^3-2k}=frac{pi}{4}-ln(2)tag6$$ and



$$T(2)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(2k)^3-2k}=frac{6G-piln(4)-1}{pi}tag7$$



Where G is the Catalan's constant.



How can we prove these conjectures?







sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 10:10







user583851

















asked Jan 3 at 9:21









user583851user583851

518110




518110












  • $begingroup$
    Is it ${2n choose n}^2 4^{-2n}$ or ${2k choose k}^2 4^{-2k}$ in the headline?
    $endgroup$
    – Andreas
    Jan 3 at 9:43












  • $begingroup$
    Why don't you try the methods of the paper? ;)
    $endgroup$
    – Jack D'Aurizio
    Jan 3 at 17:04










  • $begingroup$
    Please notice that $(2)$ is actually a divergent series, since $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pi n}}$, so "this paper of the form divergent series" does not make much sense. Better to say "this paper about the weight $frac{1}{4^n}binom{2n}{n}$ and its square".
    $endgroup$
    – Jack D'Aurizio
    Jan 3 at 17:20










  • $begingroup$
    $(3)$ is also a divergent series (same reason), while $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^3$$ is related to the lemniscate constant (hence to $Gammaleft(frac{1}{4}right)$) by many good reasons; among them, Clausen's formula for the square of particular $phantom{}_2 F_1$s.
    $endgroup$
    – Jack D'Aurizio
    Jan 3 at 17:32










  • $begingroup$
    $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^4$$ is much more obscure: see math.stackexchange.com/questions/2506266/…
    $endgroup$
    – Jack D'Aurizio
    Jan 3 at 17:34


















  • $begingroup$
    Is it ${2n choose n}^2 4^{-2n}$ or ${2k choose k}^2 4^{-2k}$ in the headline?
    $endgroup$
    – Andreas
    Jan 3 at 9:43












  • $begingroup$
    Why don't you try the methods of the paper? ;)
    $endgroup$
    – Jack D'Aurizio
    Jan 3 at 17:04










  • $begingroup$
    Please notice that $(2)$ is actually a divergent series, since $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pi n}}$, so "this paper of the form divergent series" does not make much sense. Better to say "this paper about the weight $frac{1}{4^n}binom{2n}{n}$ and its square".
    $endgroup$
    – Jack D'Aurizio
    Jan 3 at 17:20










  • $begingroup$
    $(3)$ is also a divergent series (same reason), while $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^3$$ is related to the lemniscate constant (hence to $Gammaleft(frac{1}{4}right)$) by many good reasons; among them, Clausen's formula for the square of particular $phantom{}_2 F_1$s.
    $endgroup$
    – Jack D'Aurizio
    Jan 3 at 17:32










  • $begingroup$
    $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^4$$ is much more obscure: see math.stackexchange.com/questions/2506266/…
    $endgroup$
    – Jack D'Aurizio
    Jan 3 at 17:34
















$begingroup$
Is it ${2n choose n}^2 4^{-2n}$ or ${2k choose k}^2 4^{-2k}$ in the headline?
$endgroup$
– Andreas
Jan 3 at 9:43






$begingroup$
Is it ${2n choose n}^2 4^{-2n}$ or ${2k choose k}^2 4^{-2k}$ in the headline?
$endgroup$
– Andreas
Jan 3 at 9:43














$begingroup$
Why don't you try the methods of the paper? ;)
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:04




$begingroup$
Why don't you try the methods of the paper? ;)
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:04












$begingroup$
Please notice that $(2)$ is actually a divergent series, since $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pi n}}$, so "this paper of the form divergent series" does not make much sense. Better to say "this paper about the weight $frac{1}{4^n}binom{2n}{n}$ and its square".
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:20




$begingroup$
Please notice that $(2)$ is actually a divergent series, since $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pi n}}$, so "this paper of the form divergent series" does not make much sense. Better to say "this paper about the weight $frac{1}{4^n}binom{2n}{n}$ and its square".
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:20












$begingroup$
$(3)$ is also a divergent series (same reason), while $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^3$$ is related to the lemniscate constant (hence to $Gammaleft(frac{1}{4}right)$) by many good reasons; among them, Clausen's formula for the square of particular $phantom{}_2 F_1$s.
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:32




$begingroup$
$(3)$ is also a divergent series (same reason), while $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^3$$ is related to the lemniscate constant (hence to $Gammaleft(frac{1}{4}right)$) by many good reasons; among them, Clausen's formula for the square of particular $phantom{}_2 F_1$s.
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:32












$begingroup$
$$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^4$$ is much more obscure: see math.stackexchange.com/questions/2506266/…
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:34




$begingroup$
$$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^4$$ is much more obscure: see math.stackexchange.com/questions/2506266/…
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:34










1 Answer
1






active

oldest

votes


















3












$begingroup$

Let us tackle $(4)$ first. The ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]$ is $frac{1}{sqrt{1-x}}$ and
$$ frac{1}{(an)^3-an}=frac{1}{(an-1)an(an+1)} = int_{0}^{1} x^{an}frac{(1-x)^2}{2x^2},dx $$
hence
$$ S(a) = int_{0}^{1}left(frac{1}{sqrt{1-x^a}}-1right)cdotfrac{(1-x)^2}{2x^2},dx $$
and similarly, since the ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]^2$ is $frac{2}{pi}K(x)$,
$$ T(a) = frac{2}{pi}int_{0}^{1}left(K(x^a)-frac{pi}{2}right)frac{(1-x)^2}{2x^2},dx. $$
For specific values of $a$ (like $a=2$ or $a=4$) integration by parts, suitable substitutions and known FL-expansions greatly simplify the underlying hypergeometric structure of these integrals. I won't expect a simple closed form in the general case.



(6) and (7) are proved in the mentioned paper, not just conjectured.






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%2f3060392%2fconcerning-this-sum-sum-k-1-infty2k-choose-k2-4-2k-cdot-frac1%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









    3












    $begingroup$

    Let us tackle $(4)$ first. The ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]$ is $frac{1}{sqrt{1-x}}$ and
    $$ frac{1}{(an)^3-an}=frac{1}{(an-1)an(an+1)} = int_{0}^{1} x^{an}frac{(1-x)^2}{2x^2},dx $$
    hence
    $$ S(a) = int_{0}^{1}left(frac{1}{sqrt{1-x^a}}-1right)cdotfrac{(1-x)^2}{2x^2},dx $$
    and similarly, since the ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]^2$ is $frac{2}{pi}K(x)$,
    $$ T(a) = frac{2}{pi}int_{0}^{1}left(K(x^a)-frac{pi}{2}right)frac{(1-x)^2}{2x^2},dx. $$
    For specific values of $a$ (like $a=2$ or $a=4$) integration by parts, suitable substitutions and known FL-expansions greatly simplify the underlying hypergeometric structure of these integrals. I won't expect a simple closed form in the general case.



    (6) and (7) are proved in the mentioned paper, not just conjectured.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      Let us tackle $(4)$ first. The ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]$ is $frac{1}{sqrt{1-x}}$ and
      $$ frac{1}{(an)^3-an}=frac{1}{(an-1)an(an+1)} = int_{0}^{1} x^{an}frac{(1-x)^2}{2x^2},dx $$
      hence
      $$ S(a) = int_{0}^{1}left(frac{1}{sqrt{1-x^a}}-1right)cdotfrac{(1-x)^2}{2x^2},dx $$
      and similarly, since the ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]^2$ is $frac{2}{pi}K(x)$,
      $$ T(a) = frac{2}{pi}int_{0}^{1}left(K(x^a)-frac{pi}{2}right)frac{(1-x)^2}{2x^2},dx. $$
      For specific values of $a$ (like $a=2$ or $a=4$) integration by parts, suitable substitutions and known FL-expansions greatly simplify the underlying hypergeometric structure of these integrals. I won't expect a simple closed form in the general case.



      (6) and (7) are proved in the mentioned paper, not just conjectured.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        Let us tackle $(4)$ first. The ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]$ is $frac{1}{sqrt{1-x}}$ and
        $$ frac{1}{(an)^3-an}=frac{1}{(an-1)an(an+1)} = int_{0}^{1} x^{an}frac{(1-x)^2}{2x^2},dx $$
        hence
        $$ S(a) = int_{0}^{1}left(frac{1}{sqrt{1-x^a}}-1right)cdotfrac{(1-x)^2}{2x^2},dx $$
        and similarly, since the ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]^2$ is $frac{2}{pi}K(x)$,
        $$ T(a) = frac{2}{pi}int_{0}^{1}left(K(x^a)-frac{pi}{2}right)frac{(1-x)^2}{2x^2},dx. $$
        For specific values of $a$ (like $a=2$ or $a=4$) integration by parts, suitable substitutions and known FL-expansions greatly simplify the underlying hypergeometric structure of these integrals. I won't expect a simple closed form in the general case.



        (6) and (7) are proved in the mentioned paper, not just conjectured.






        share|cite|improve this answer











        $endgroup$



        Let us tackle $(4)$ first. The ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]$ is $frac{1}{sqrt{1-x}}$ and
        $$ frac{1}{(an)^3-an}=frac{1}{(an-1)an(an+1)} = int_{0}^{1} x^{an}frac{(1-x)^2}{2x^2},dx $$
        hence
        $$ S(a) = int_{0}^{1}left(frac{1}{sqrt{1-x^a}}-1right)cdotfrac{(1-x)^2}{2x^2},dx $$
        and similarly, since the ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]^2$ is $frac{2}{pi}K(x)$,
        $$ T(a) = frac{2}{pi}int_{0}^{1}left(K(x^a)-frac{pi}{2}right)frac{(1-x)^2}{2x^2},dx. $$
        For specific values of $a$ (like $a=2$ or $a=4$) integration by parts, suitable substitutions and known FL-expansions greatly simplify the underlying hypergeometric structure of these integrals. I won't expect a simple closed form in the general case.



        (6) and (7) are proved in the mentioned paper, not just conjectured.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 3 at 17:22

























        answered Jan 3 at 17:16









        Jack D'AurizioJack D'Aurizio

        292k33284672




        292k33284672






























            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%2f3060392%2fconcerning-this-sum-sum-k-1-infty2k-choose-k2-4-2k-cdot-frac1%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

            Probability when a professor distributes a quiz and homework assignment to a class of n students.

            Aardman Animations

            Are they similar matrix