conjecture $sum_{k=1}^{infty}{2k choose k}^2frac{k}{2^{4k}(2k-1)^2}=frac{1}{pi}$












2












$begingroup$


A conjecture:
$$sum_{k=1}^{infty}{2k choose k}^2frac{k}{2^{4k}(2k-1)^2}=frac{1}{pi}$$



Can anybody numerically verify if this is anywhere near to the the claim answer?



update:



wolfram's calculator shows that it is true. How can we proves it?






sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Sure, YOU can. Use any mathematical program (sage, maple, ...) or wolfram (which is online) to add the first 100 terms, and you can check if this makes sense.
    $endgroup$
    – A. Pongrácz
    Dec 29 '18 at 7:49










  • $begingroup$
    show me the program, i cant find it
    $endgroup$
    – Endgame
    Dec 29 '18 at 7:53






  • 1




    $begingroup$
    wolframalpha.com
    $endgroup$
    – A. Pongrácz
    Dec 29 '18 at 8:01










  • $begingroup$
    Wolfram says it's true, and not just numerically.
    $endgroup$
    – Ivan Neretin
    Dec 29 '18 at 8:12
















2












$begingroup$


A conjecture:
$$sum_{k=1}^{infty}{2k choose k}^2frac{k}{2^{4k}(2k-1)^2}=frac{1}{pi}$$



Can anybody numerically verify if this is anywhere near to the the claim answer?



update:



wolfram's calculator shows that it is true. How can we proves it?






sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Sure, YOU can. Use any mathematical program (sage, maple, ...) or wolfram (which is online) to add the first 100 terms, and you can check if this makes sense.
    $endgroup$
    – A. Pongrácz
    Dec 29 '18 at 7:49










  • $begingroup$
    show me the program, i cant find it
    $endgroup$
    – Endgame
    Dec 29 '18 at 7:53






  • 1




    $begingroup$
    wolframalpha.com
    $endgroup$
    – A. Pongrácz
    Dec 29 '18 at 8:01










  • $begingroup$
    Wolfram says it's true, and not just numerically.
    $endgroup$
    – Ivan Neretin
    Dec 29 '18 at 8:12














2












2








2


1



$begingroup$


A conjecture:
$$sum_{k=1}^{infty}{2k choose k}^2frac{k}{2^{4k}(2k-1)^2}=frac{1}{pi}$$



Can anybody numerically verify if this is anywhere near to the the claim answer?



update:



wolfram's calculator shows that it is true. How can we proves it?






sequences-and-series







share|cite|improve this question











$endgroup$




A conjecture:
$$sum_{k=1}^{infty}{2k choose k}^2frac{k}{2^{4k}(2k-1)^2}=frac{1}{pi}$$



Can anybody numerically verify if this is anywhere near to the the claim answer?



update:



wolfram's calculator shows that it is true. How can we proves it?






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 Dec 29 '18 at 8:40







Endgame

















asked Dec 29 '18 at 7:46









EndgameEndgame

1




1








  • 1




    $begingroup$
    Sure, YOU can. Use any mathematical program (sage, maple, ...) or wolfram (which is online) to add the first 100 terms, and you can check if this makes sense.
    $endgroup$
    – A. Pongrácz
    Dec 29 '18 at 7:49










  • $begingroup$
    show me the program, i cant find it
    $endgroup$
    – Endgame
    Dec 29 '18 at 7:53






  • 1




    $begingroup$
    wolframalpha.com
    $endgroup$
    – A. Pongrácz
    Dec 29 '18 at 8:01










  • $begingroup$
    Wolfram says it's true, and not just numerically.
    $endgroup$
    – Ivan Neretin
    Dec 29 '18 at 8:12














  • 1




    $begingroup$
    Sure, YOU can. Use any mathematical program (sage, maple, ...) or wolfram (which is online) to add the first 100 terms, and you can check if this makes sense.
    $endgroup$
    – A. Pongrácz
    Dec 29 '18 at 7:49










  • $begingroup$
    show me the program, i cant find it
    $endgroup$
    – Endgame
    Dec 29 '18 at 7:53






  • 1




    $begingroup$
    wolframalpha.com
    $endgroup$
    – A. Pongrácz
    Dec 29 '18 at 8:01










  • $begingroup$
    Wolfram says it's true, and not just numerically.
    $endgroup$
    – Ivan Neretin
    Dec 29 '18 at 8:12








1




1




$begingroup$
Sure, YOU can. Use any mathematical program (sage, maple, ...) or wolfram (which is online) to add the first 100 terms, and you can check if this makes sense.
$endgroup$
– A. Pongrácz
Dec 29 '18 at 7:49




$begingroup$
Sure, YOU can. Use any mathematical program (sage, maple, ...) or wolfram (which is online) to add the first 100 terms, and you can check if this makes sense.
$endgroup$
– A. Pongrácz
Dec 29 '18 at 7:49












$begingroup$
show me the program, i cant find it
$endgroup$
– Endgame
Dec 29 '18 at 7:53




$begingroup$
show me the program, i cant find it
$endgroup$
– Endgame
Dec 29 '18 at 7:53




1




1




$begingroup$
wolframalpha.com
$endgroup$
– A. Pongrácz
Dec 29 '18 at 8:01




$begingroup$
wolframalpha.com
$endgroup$
– A. Pongrácz
Dec 29 '18 at 8:01












$begingroup$
Wolfram says it's true, and not just numerically.
$endgroup$
– Ivan Neretin
Dec 29 '18 at 8:12




$begingroup$
Wolfram says it's true, and not just numerically.
$endgroup$
– Ivan Neretin
Dec 29 '18 at 8:12










2 Answers
2






active

oldest

votes


















4












$begingroup$

Using $(7)$ and $(9)$ from this answer, we get
$$
begin{align}
sum_{k=1}^inftyfrac{binom{2k}{k}^2}{2^{4k}}frac{k}{(2k-1)^2}
&=sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}left(color{#00F}{frac{1/2}{(2k-1)^2}}color{#090}{+frac{1/2}{2k-1}}right)\
&=left(color{#00F}{frac2pi}color{#090}{-frac1pi}right)\[6pt]
&=frac1pi
end{align}
$$






share|cite|improve this answer









$endgroup$





















    7












    $begingroup$

    Not only it is true but we can also get the partial sum
    $$S_p=sum_{k=1}^{p}{2k choose k}^2frac{k}{2^{4k}(2k-1)^2}=frac{ p, (p+1)^2 }{2^{4 p+2},(2 p+1)^2}binom{2p+2}{p+1}^2$$
    Taking logarithms and using Stirling approximation
    $$log(S_p)=-log (pi )-frac{1}{4 p}+frac{1}{96 p^3}+Oleft(frac{1}{p^5}right)$$ Continuing with Taylor
    $$S_p=e^{log(S_p)}=frac 1 pi left(1-frac{1}{4 p}+frac{1}{32 p^2}+frac{1}{128 p^3}-frac{5}{2048
    p^4}+Oleft(frac{1}{p^5}right)right)$$



    For example
    $$S_{10}=frac{10667118605}{34359738368}approx 0.3104540113$$ while the above expansion would give
    $$S_{10} approx frac 1 pi frac{3994911}{4096000}approx 0.3104540200$$






    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      Using $(7)$ and $(9)$ from this answer, we get
      $$
      begin{align}
      sum_{k=1}^inftyfrac{binom{2k}{k}^2}{2^{4k}}frac{k}{(2k-1)^2}
      &=sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}left(color{#00F}{frac{1/2}{(2k-1)^2}}color{#090}{+frac{1/2}{2k-1}}right)\
      &=left(color{#00F}{frac2pi}color{#090}{-frac1pi}right)\[6pt]
      &=frac1pi
      end{align}
      $$






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Using $(7)$ and $(9)$ from this answer, we get
        $$
        begin{align}
        sum_{k=1}^inftyfrac{binom{2k}{k}^2}{2^{4k}}frac{k}{(2k-1)^2}
        &=sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}left(color{#00F}{frac{1/2}{(2k-1)^2}}color{#090}{+frac{1/2}{2k-1}}right)\
        &=left(color{#00F}{frac2pi}color{#090}{-frac1pi}right)\[6pt]
        &=frac1pi
        end{align}
        $$






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Using $(7)$ and $(9)$ from this answer, we get
          $$
          begin{align}
          sum_{k=1}^inftyfrac{binom{2k}{k}^2}{2^{4k}}frac{k}{(2k-1)^2}
          &=sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}left(color{#00F}{frac{1/2}{(2k-1)^2}}color{#090}{+frac{1/2}{2k-1}}right)\
          &=left(color{#00F}{frac2pi}color{#090}{-frac1pi}right)\[6pt]
          &=frac1pi
          end{align}
          $$






          share|cite|improve this answer









          $endgroup$



          Using $(7)$ and $(9)$ from this answer, we get
          $$
          begin{align}
          sum_{k=1}^inftyfrac{binom{2k}{k}^2}{2^{4k}}frac{k}{(2k-1)^2}
          &=sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}left(color{#00F}{frac{1/2}{(2k-1)^2}}color{#090}{+frac{1/2}{2k-1}}right)\
          &=left(color{#00F}{frac2pi}color{#090}{-frac1pi}right)\[6pt]
          &=frac1pi
          end{align}
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 30 '18 at 22:49









          robjohn♦robjohn

          271k27316643




          271k27316643























              7












              $begingroup$

              Not only it is true but we can also get the partial sum
              $$S_p=sum_{k=1}^{p}{2k choose k}^2frac{k}{2^{4k}(2k-1)^2}=frac{ p, (p+1)^2 }{2^{4 p+2},(2 p+1)^2}binom{2p+2}{p+1}^2$$
              Taking logarithms and using Stirling approximation
              $$log(S_p)=-log (pi )-frac{1}{4 p}+frac{1}{96 p^3}+Oleft(frac{1}{p^5}right)$$ Continuing with Taylor
              $$S_p=e^{log(S_p)}=frac 1 pi left(1-frac{1}{4 p}+frac{1}{32 p^2}+frac{1}{128 p^3}-frac{5}{2048
              p^4}+Oleft(frac{1}{p^5}right)right)$$



              For example
              $$S_{10}=frac{10667118605}{34359738368}approx 0.3104540113$$ while the above expansion would give
              $$S_{10} approx frac 1 pi frac{3994911}{4096000}approx 0.3104540200$$






              share|cite|improve this answer









              $endgroup$


















                7












                $begingroup$

                Not only it is true but we can also get the partial sum
                $$S_p=sum_{k=1}^{p}{2k choose k}^2frac{k}{2^{4k}(2k-1)^2}=frac{ p, (p+1)^2 }{2^{4 p+2},(2 p+1)^2}binom{2p+2}{p+1}^2$$
                Taking logarithms and using Stirling approximation
                $$log(S_p)=-log (pi )-frac{1}{4 p}+frac{1}{96 p^3}+Oleft(frac{1}{p^5}right)$$ Continuing with Taylor
                $$S_p=e^{log(S_p)}=frac 1 pi left(1-frac{1}{4 p}+frac{1}{32 p^2}+frac{1}{128 p^3}-frac{5}{2048
                p^4}+Oleft(frac{1}{p^5}right)right)$$



                For example
                $$S_{10}=frac{10667118605}{34359738368}approx 0.3104540113$$ while the above expansion would give
                $$S_{10} approx frac 1 pi frac{3994911}{4096000}approx 0.3104540200$$






                share|cite|improve this answer









                $endgroup$
















                  7












                  7








                  7





                  $begingroup$

                  Not only it is true but we can also get the partial sum
                  $$S_p=sum_{k=1}^{p}{2k choose k}^2frac{k}{2^{4k}(2k-1)^2}=frac{ p, (p+1)^2 }{2^{4 p+2},(2 p+1)^2}binom{2p+2}{p+1}^2$$
                  Taking logarithms and using Stirling approximation
                  $$log(S_p)=-log (pi )-frac{1}{4 p}+frac{1}{96 p^3}+Oleft(frac{1}{p^5}right)$$ Continuing with Taylor
                  $$S_p=e^{log(S_p)}=frac 1 pi left(1-frac{1}{4 p}+frac{1}{32 p^2}+frac{1}{128 p^3}-frac{5}{2048
                  p^4}+Oleft(frac{1}{p^5}right)right)$$



                  For example
                  $$S_{10}=frac{10667118605}{34359738368}approx 0.3104540113$$ while the above expansion would give
                  $$S_{10} approx frac 1 pi frac{3994911}{4096000}approx 0.3104540200$$






                  share|cite|improve this answer









                  $endgroup$



                  Not only it is true but we can also get the partial sum
                  $$S_p=sum_{k=1}^{p}{2k choose k}^2frac{k}{2^{4k}(2k-1)^2}=frac{ p, (p+1)^2 }{2^{4 p+2},(2 p+1)^2}binom{2p+2}{p+1}^2$$
                  Taking logarithms and using Stirling approximation
                  $$log(S_p)=-log (pi )-frac{1}{4 p}+frac{1}{96 p^3}+Oleft(frac{1}{p^5}right)$$ Continuing with Taylor
                  $$S_p=e^{log(S_p)}=frac 1 pi left(1-frac{1}{4 p}+frac{1}{32 p^2}+frac{1}{128 p^3}-frac{5}{2048
                  p^4}+Oleft(frac{1}{p^5}right)right)$$



                  For example
                  $$S_{10}=frac{10667118605}{34359738368}approx 0.3104540113$$ while the above expansion would give
                  $$S_{10} approx frac 1 pi frac{3994911}{4096000}approx 0.3104540200$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 29 '18 at 9:22









                  Claude LeiboviciClaude Leibovici

                  126k1158135




                  126k1158135






























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