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?






sequences-and-series







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?






sequences-and-series







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?






sequences-and-series







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




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

508110




508110












  • $begingroup$
    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











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    1 Answer
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    active

    oldest

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    active

    oldest

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    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

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        291k33284666






























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            Cabo Verde

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              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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            Gyllenstierna

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