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












1














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
























  • Is it ${2n choose n}^2 4^{-2n}$ or ${2k choose k}^2 4^{-2k}$ in the headline?
    – Andreas
    22 hours ago












  • Why don't you try the methods of the paper? ;)
    – Jack D'Aurizio
    15 hours ago










  • 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".
    – Jack D'Aurizio
    14 hours ago










  • $(3)$ 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.
    – Jack D'Aurizio
    14 hours ago










  • $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^4$$ is much more obscure: see math.stackexchange.com/questions/2506266/…
    – Jack D'Aurizio
    14 hours ago
















1














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
























  • Is it ${2n choose n}^2 4^{-2n}$ or ${2k choose k}^2 4^{-2k}$ in the headline?
    – Andreas
    22 hours ago












  • Why don't you try the methods of the paper? ;)
    – Jack D'Aurizio
    15 hours ago










  • 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".
    – Jack D'Aurizio
    14 hours ago










  • $(3)$ 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.
    – Jack D'Aurizio
    14 hours ago










  • $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^4$$ is much more obscure: see math.stackexchange.com/questions/2506266/…
    – Jack D'Aurizio
    14 hours ago














1












1








1


3





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















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 22 hours ago

























asked 22 hours ago









user583851

4037




4037












  • Is it ${2n choose n}^2 4^{-2n}$ or ${2k choose k}^2 4^{-2k}$ in the headline?
    – Andreas
    22 hours ago












  • Why don't you try the methods of the paper? ;)
    – Jack D'Aurizio
    15 hours ago










  • 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".
    – Jack D'Aurizio
    14 hours ago










  • $(3)$ 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.
    – Jack D'Aurizio
    14 hours ago










  • $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^4$$ is much more obscure: see math.stackexchange.com/questions/2506266/…
    – Jack D'Aurizio
    14 hours ago


















  • Is it ${2n choose n}^2 4^{-2n}$ or ${2k choose k}^2 4^{-2k}$ in the headline?
    – Andreas
    22 hours ago












  • Why don't you try the methods of the paper? ;)
    – Jack D'Aurizio
    15 hours ago










  • 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".
    – Jack D'Aurizio
    14 hours ago










  • $(3)$ 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.
    – Jack D'Aurizio
    14 hours ago










  • $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^4$$ is much more obscure: see math.stackexchange.com/questions/2506266/…
    – Jack D'Aurizio
    14 hours ago
















Is it ${2n choose n}^2 4^{-2n}$ or ${2k choose k}^2 4^{-2k}$ in the headline?
– Andreas
22 hours ago






Is it ${2n choose n}^2 4^{-2n}$ or ${2k choose k}^2 4^{-2k}$ in the headline?
– Andreas
22 hours ago














Why don't you try the methods of the paper? ;)
– Jack D'Aurizio
15 hours ago




Why don't you try the methods of the paper? ;)
– Jack D'Aurizio
15 hours ago












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".
– Jack D'Aurizio
14 hours ago




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".
– Jack D'Aurizio
14 hours ago












$(3)$ 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.
– Jack D'Aurizio
14 hours ago




$(3)$ 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.
– Jack D'Aurizio
14 hours ago












$$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^4$$ is much more obscure: see math.stackexchange.com/questions/2506266/…
– Jack D'Aurizio
14 hours ago




$$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^4$$ is much more obscure: see math.stackexchange.com/questions/2506266/…
– Jack D'Aurizio
14 hours ago










1 Answer
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3














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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    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




























      3














      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


























        3












        3








        3






        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














        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 14 hours ago

























        answered 14 hours ago









        Jack D'Aurizio

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