Evaluate $sumlimits_{k=1}^{infty}frac{B(k,k)}{k}$ where $B$ denotes the beta function












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How can I evaluate $displaystylesum_{k=1}^{infty}frac{B(k,k)}{k}$ Here B is the beta function






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    – Did
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How can I evaluate $displaystylesum_{k=1}^{infty}frac{B(k,k)}{k}$ Here B is the beta function






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$begingroup$


How can I evaluate $displaystylesum_{k=1}^{infty}frac{B(k,k)}{k}$ Here B is the beta function






sequences-and-series summation gamma-function beta-function







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How can I evaluate $displaystylesum_{k=1}^{infty}frac{B(k,k)}{k}$ Here B is the beta function






sequences-and-series summation gamma-function beta-function




sequences-and-series summation gamma-function beta-function






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edited Jan 27 at 11:26









Did

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asked Jan 27 at 11:03









ben tenysonben tenyson

414




414












  • $begingroup$
    No displaystyle in titles please.
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    Jan 27 at 11:26


















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    – Did
    Jan 27 at 11:26
















$begingroup$
No displaystyle in titles please.
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Jan 27 at 11:26




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3 Answers
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$begingroup$

Here is a solution:



begin{align*}
sum_{k=1}^{infty} frac{B(k,k)}{k}
&= sum_{k=1}^{infty} frac{1}{k} int_{0}^{1} x^{k-1}(1-x)^{k-1} , mathrm{d} x \
&= - int_{0}^{1} frac{log(1 - x(1-x))}{x(1-x)} , mathrm{d}x \
&= -2 int_{0}^{1} frac{log(1 - x + x^2)}{x} , mathrm{d}x \
&= 2 int_{0}^{1} frac{log(1 + x) - log(1 + x^3)}{x} , mathrm{d}x \
&= frac{4}{3} int_{0}^{1} frac{log(1 + x)}{x} , mathrm{d}x \
&= frac{4}{3} sum_{n=1}^{infty} frac{(-1)^{n-1}}{n^2} \
&= frac{4}{3} left(1 - frac{2}{2^2}right)zeta(2) \
&= frac{pi^2}{9}.
end{align*}



Here are explanations for some non-trivial steps:




  • In the third step, we exploited $frac{log(1 - x(1-x))}{x(1-x)} = frac{log(1 - x(1-x))}{x} + frac{log(1 - x(1-x))}{1-x}$. By symmetry, integrals of the two terms in the right-hand side coincide.


  • $int_{0}^{1} frac{log(1+x^3)}{x} , mathrm{d}x = frac{1}{3} int_{0}^{1} frac{log(1+u)}{u} , mathrm{d}u$ by the substitution $u = x^3$.







share|cite|improve this answer











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    3












    $begingroup$

    You can also use the power series of the squared arcsine:
    $$ sum limits_{k=1}^infty frac{operatorname{B}(k,k)}{k} = sum limits_{k=1}^infty frac{k!^2}{k^3 (2k-1)!} = 2sum limits_{k=1}^infty frac{1}{k^2 {2k choose k}} = 4 arcsin^2 left(frac{1}{2}right) = frac{pi^2}{9} , .$$






    share|cite|improve this answer









    $endgroup$





















      -1












      $begingroup$

      The Beta function with two identical arguments is nothing but



      $$frac{Gamma(k)Gamma(k)}{Gamma(2k)} = frac{sqrt{pi } 2^{1-2 k} Gamma (k)}{Gamma left(k+frac{1}{2}right)} = frac{sqrt{pi}2^{1-2k}(k-1)!}{frac{(2k)!}{4^kk!}sqrt{pi}} = frac{2^{1-2k}(k-1)!k! 4^k}{(2k)!} = frac{2k!(k-1)!}{(2k)!}$$



      And since your sum runs only over the natural numbers, the first terms of the series can be easily computed (let me avoid for the moment the division by $k$):



      $$1,frac{1}{6},frac{1}{30},frac{1}{140},frac{1}{630},frac{1}{2772} ldots$$



      Now let's divide by $k$ as it is in your sum, to get:



      $$frac{1}{k}frac{2k!(k-1)!}{(2k)!} = frac{2(k-1)!^2}{(2k)!}$$



      Which we can easily sum to $$frac{pi^2}{9}$$






      share|cite|improve this answer











      $endgroup$













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        3 Answers
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        3 Answers
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        5












        $begingroup$

        Here is a solution:



        begin{align*}
        sum_{k=1}^{infty} frac{B(k,k)}{k}
        &= sum_{k=1}^{infty} frac{1}{k} int_{0}^{1} x^{k-1}(1-x)^{k-1} , mathrm{d} x \
        &= - int_{0}^{1} frac{log(1 - x(1-x))}{x(1-x)} , mathrm{d}x \
        &= -2 int_{0}^{1} frac{log(1 - x + x^2)}{x} , mathrm{d}x \
        &= 2 int_{0}^{1} frac{log(1 + x) - log(1 + x^3)}{x} , mathrm{d}x \
        &= frac{4}{3} int_{0}^{1} frac{log(1 + x)}{x} , mathrm{d}x \
        &= frac{4}{3} sum_{n=1}^{infty} frac{(-1)^{n-1}}{n^2} \
        &= frac{4}{3} left(1 - frac{2}{2^2}right)zeta(2) \
        &= frac{pi^2}{9}.
        end{align*}



        Here are explanations for some non-trivial steps:




        • In the third step, we exploited $frac{log(1 - x(1-x))}{x(1-x)} = frac{log(1 - x(1-x))}{x} + frac{log(1 - x(1-x))}{1-x}$. By symmetry, integrals of the two terms in the right-hand side coincide.


        • $int_{0}^{1} frac{log(1+x^3)}{x} , mathrm{d}x = frac{1}{3} int_{0}^{1} frac{log(1+u)}{u} , mathrm{d}u$ by the substitution $u = x^3$.







        share|cite|improve this answer











        $endgroup$


















          5












          $begingroup$

          Here is a solution:



          begin{align*}
          sum_{k=1}^{infty} frac{B(k,k)}{k}
          &= sum_{k=1}^{infty} frac{1}{k} int_{0}^{1} x^{k-1}(1-x)^{k-1} , mathrm{d} x \
          &= - int_{0}^{1} frac{log(1 - x(1-x))}{x(1-x)} , mathrm{d}x \
          &= -2 int_{0}^{1} frac{log(1 - x + x^2)}{x} , mathrm{d}x \
          &= 2 int_{0}^{1} frac{log(1 + x) - log(1 + x^3)}{x} , mathrm{d}x \
          &= frac{4}{3} int_{0}^{1} frac{log(1 + x)}{x} , mathrm{d}x \
          &= frac{4}{3} sum_{n=1}^{infty} frac{(-1)^{n-1}}{n^2} \
          &= frac{4}{3} left(1 - frac{2}{2^2}right)zeta(2) \
          &= frac{pi^2}{9}.
          end{align*}



          Here are explanations for some non-trivial steps:




          • In the third step, we exploited $frac{log(1 - x(1-x))}{x(1-x)} = frac{log(1 - x(1-x))}{x} + frac{log(1 - x(1-x))}{1-x}$. By symmetry, integrals of the two terms in the right-hand side coincide.


          • $int_{0}^{1} frac{log(1+x^3)}{x} , mathrm{d}x = frac{1}{3} int_{0}^{1} frac{log(1+u)}{u} , mathrm{d}u$ by the substitution $u = x^3$.







          share|cite|improve this answer











          $endgroup$
















            5












            5








            5





            $begingroup$

            Here is a solution:



            begin{align*}
            sum_{k=1}^{infty} frac{B(k,k)}{k}
            &= sum_{k=1}^{infty} frac{1}{k} int_{0}^{1} x^{k-1}(1-x)^{k-1} , mathrm{d} x \
            &= - int_{0}^{1} frac{log(1 - x(1-x))}{x(1-x)} , mathrm{d}x \
            &= -2 int_{0}^{1} frac{log(1 - x + x^2)}{x} , mathrm{d}x \
            &= 2 int_{0}^{1} frac{log(1 + x) - log(1 + x^3)}{x} , mathrm{d}x \
            &= frac{4}{3} int_{0}^{1} frac{log(1 + x)}{x} , mathrm{d}x \
            &= frac{4}{3} sum_{n=1}^{infty} frac{(-1)^{n-1}}{n^2} \
            &= frac{4}{3} left(1 - frac{2}{2^2}right)zeta(2) \
            &= frac{pi^2}{9}.
            end{align*}



            Here are explanations for some non-trivial steps:




            • In the third step, we exploited $frac{log(1 - x(1-x))}{x(1-x)} = frac{log(1 - x(1-x))}{x} + frac{log(1 - x(1-x))}{1-x}$. By symmetry, integrals of the two terms in the right-hand side coincide.


            • $int_{0}^{1} frac{log(1+x^3)}{x} , mathrm{d}x = frac{1}{3} int_{0}^{1} frac{log(1+u)}{u} , mathrm{d}u$ by the substitution $u = x^3$.







            share|cite|improve this answer











            $endgroup$



            Here is a solution:



            begin{align*}
            sum_{k=1}^{infty} frac{B(k,k)}{k}
            &= sum_{k=1}^{infty} frac{1}{k} int_{0}^{1} x^{k-1}(1-x)^{k-1} , mathrm{d} x \
            &= - int_{0}^{1} frac{log(1 - x(1-x))}{x(1-x)} , mathrm{d}x \
            &= -2 int_{0}^{1} frac{log(1 - x + x^2)}{x} , mathrm{d}x \
            &= 2 int_{0}^{1} frac{log(1 + x) - log(1 + x^3)}{x} , mathrm{d}x \
            &= frac{4}{3} int_{0}^{1} frac{log(1 + x)}{x} , mathrm{d}x \
            &= frac{4}{3} sum_{n=1}^{infty} frac{(-1)^{n-1}}{n^2} \
            &= frac{4}{3} left(1 - frac{2}{2^2}right)zeta(2) \
            &= frac{pi^2}{9}.
            end{align*}



            Here are explanations for some non-trivial steps:




            • In the third step, we exploited $frac{log(1 - x(1-x))}{x(1-x)} = frac{log(1 - x(1-x))}{x} + frac{log(1 - x(1-x))}{1-x}$. By symmetry, integrals of the two terms in the right-hand side coincide.


            • $int_{0}^{1} frac{log(1+x^3)}{x} , mathrm{d}x = frac{1}{3} int_{0}^{1} frac{log(1+u)}{u} , mathrm{d}u$ by the substitution $u = x^3$.








            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 27 at 11:28

























            answered Jan 27 at 11:20









            Sangchul LeeSangchul Lee

            96.2k12171281




            96.2k12171281























                3












                $begingroup$

                You can also use the power series of the squared arcsine:
                $$ sum limits_{k=1}^infty frac{operatorname{B}(k,k)}{k} = sum limits_{k=1}^infty frac{k!^2}{k^3 (2k-1)!} = 2sum limits_{k=1}^infty frac{1}{k^2 {2k choose k}} = 4 arcsin^2 left(frac{1}{2}right) = frac{pi^2}{9} , .$$






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  You can also use the power series of the squared arcsine:
                  $$ sum limits_{k=1}^infty frac{operatorname{B}(k,k)}{k} = sum limits_{k=1}^infty frac{k!^2}{k^3 (2k-1)!} = 2sum limits_{k=1}^infty frac{1}{k^2 {2k choose k}} = 4 arcsin^2 left(frac{1}{2}right) = frac{pi^2}{9} , .$$






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    You can also use the power series of the squared arcsine:
                    $$ sum limits_{k=1}^infty frac{operatorname{B}(k,k)}{k} = sum limits_{k=1}^infty frac{k!^2}{k^3 (2k-1)!} = 2sum limits_{k=1}^infty frac{1}{k^2 {2k choose k}} = 4 arcsin^2 left(frac{1}{2}right) = frac{pi^2}{9} , .$$






                    share|cite|improve this answer









                    $endgroup$



                    You can also use the power series of the squared arcsine:
                    $$ sum limits_{k=1}^infty frac{operatorname{B}(k,k)}{k} = sum limits_{k=1}^infty frac{k!^2}{k^3 (2k-1)!} = 2sum limits_{k=1}^infty frac{1}{k^2 {2k choose k}} = 4 arcsin^2 left(frac{1}{2}right) = frac{pi^2}{9} , .$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 27 at 11:24









                    ComplexYetTrivialComplexYetTrivial

                    4,9682631




                    4,9682631























                        -1












                        $begingroup$

                        The Beta function with two identical arguments is nothing but



                        $$frac{Gamma(k)Gamma(k)}{Gamma(2k)} = frac{sqrt{pi } 2^{1-2 k} Gamma (k)}{Gamma left(k+frac{1}{2}right)} = frac{sqrt{pi}2^{1-2k}(k-1)!}{frac{(2k)!}{4^kk!}sqrt{pi}} = frac{2^{1-2k}(k-1)!k! 4^k}{(2k)!} = frac{2k!(k-1)!}{(2k)!}$$



                        And since your sum runs only over the natural numbers, the first terms of the series can be easily computed (let me avoid for the moment the division by $k$):



                        $$1,frac{1}{6},frac{1}{30},frac{1}{140},frac{1}{630},frac{1}{2772} ldots$$



                        Now let's divide by $k$ as it is in your sum, to get:



                        $$frac{1}{k}frac{2k!(k-1)!}{(2k)!} = frac{2(k-1)!^2}{(2k)!}$$



                        Which we can easily sum to $$frac{pi^2}{9}$$






                        share|cite|improve this answer











                        $endgroup$


















                          -1












                          $begingroup$

                          The Beta function with two identical arguments is nothing but



                          $$frac{Gamma(k)Gamma(k)}{Gamma(2k)} = frac{sqrt{pi } 2^{1-2 k} Gamma (k)}{Gamma left(k+frac{1}{2}right)} = frac{sqrt{pi}2^{1-2k}(k-1)!}{frac{(2k)!}{4^kk!}sqrt{pi}} = frac{2^{1-2k}(k-1)!k! 4^k}{(2k)!} = frac{2k!(k-1)!}{(2k)!}$$



                          And since your sum runs only over the natural numbers, the first terms of the series can be easily computed (let me avoid for the moment the division by $k$):



                          $$1,frac{1}{6},frac{1}{30},frac{1}{140},frac{1}{630},frac{1}{2772} ldots$$



                          Now let's divide by $k$ as it is in your sum, to get:



                          $$frac{1}{k}frac{2k!(k-1)!}{(2k)!} = frac{2(k-1)!^2}{(2k)!}$$



                          Which we can easily sum to $$frac{pi^2}{9}$$






                          share|cite|improve this answer











                          $endgroup$
















                            -1












                            -1








                            -1





                            $begingroup$

                            The Beta function with two identical arguments is nothing but



                            $$frac{Gamma(k)Gamma(k)}{Gamma(2k)} = frac{sqrt{pi } 2^{1-2 k} Gamma (k)}{Gamma left(k+frac{1}{2}right)} = frac{sqrt{pi}2^{1-2k}(k-1)!}{frac{(2k)!}{4^kk!}sqrt{pi}} = frac{2^{1-2k}(k-1)!k! 4^k}{(2k)!} = frac{2k!(k-1)!}{(2k)!}$$



                            And since your sum runs only over the natural numbers, the first terms of the series can be easily computed (let me avoid for the moment the division by $k$):



                            $$1,frac{1}{6},frac{1}{30},frac{1}{140},frac{1}{630},frac{1}{2772} ldots$$



                            Now let's divide by $k$ as it is in your sum, to get:



                            $$frac{1}{k}frac{2k!(k-1)!}{(2k)!} = frac{2(k-1)!^2}{(2k)!}$$



                            Which we can easily sum to $$frac{pi^2}{9}$$






                            share|cite|improve this answer











                            $endgroup$



                            The Beta function with two identical arguments is nothing but



                            $$frac{Gamma(k)Gamma(k)}{Gamma(2k)} = frac{sqrt{pi } 2^{1-2 k} Gamma (k)}{Gamma left(k+frac{1}{2}right)} = frac{sqrt{pi}2^{1-2k}(k-1)!}{frac{(2k)!}{4^kk!}sqrt{pi}} = frac{2^{1-2k}(k-1)!k! 4^k}{(2k)!} = frac{2k!(k-1)!}{(2k)!}$$



                            And since your sum runs only over the natural numbers, the first terms of the series can be easily computed (let me avoid for the moment the division by $k$):



                            $$1,frac{1}{6},frac{1}{30},frac{1}{140},frac{1}{630},frac{1}{2772} ldots$$



                            Now let's divide by $k$ as it is in your sum, to get:



                            $$frac{1}{k}frac{2k!(k-1)!}{(2k)!} = frac{2(k-1)!^2}{(2k)!}$$



                            Which we can easily sum to $$frac{pi^2}{9}$$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jan 29 at 21:24

























                            answered Jan 27 at 11:21









                            Von NeumannVon Neumann

                            16.5k72545




                            16.5k72545






























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