Show that $ lim_{jtoinfty}prod_{n=j}^{mj}frac{pi}{2tan^{-1}(kn)}=m^{frac{2}{kpi}}$












1












$begingroup$


MathWorld states that



$displaystyle lim_{jtoinfty}prod_{n=j}^{2j}frac{pi}{2tan^{-1}n}=4^{frac{1}{pi}}$
(equation 130)



With a calculator, I find a general formula:



$displaystyle lim_{jtoinfty}prod_{n=j}^{mj}frac{pi}{2tan^{-1}(kn)}=m^{frac{2}{kpi}}$



But I have not found a proof for this. Any proof would be appreciated.






sequences-and-series







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    MathWorld states that



    $displaystyle lim_{jtoinfty}prod_{n=j}^{2j}frac{pi}{2tan^{-1}n}=4^{frac{1}{pi}}$
    (equation 130)



    With a calculator, I find a general formula:



    $displaystyle lim_{jtoinfty}prod_{n=j}^{mj}frac{pi}{2tan^{-1}(kn)}=m^{frac{2}{kpi}}$



    But I have not found a proof for this. Any proof would be appreciated.






    sequences-and-series







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      MathWorld states that



      $displaystyle lim_{jtoinfty}prod_{n=j}^{2j}frac{pi}{2tan^{-1}n}=4^{frac{1}{pi}}$
      (equation 130)



      With a calculator, I find a general formula:



      $displaystyle lim_{jtoinfty}prod_{n=j}^{mj}frac{pi}{2tan^{-1}(kn)}=m^{frac{2}{kpi}}$



      But I have not found a proof for this. Any proof would be appreciated.






      sequences-and-series







      share|cite|improve this question











      $endgroup$




      MathWorld states that



      $displaystyle lim_{jtoinfty}prod_{n=j}^{2j}frac{pi}{2tan^{-1}n}=4^{frac{1}{pi}}$
      (equation 130)



      With a calculator, I find a general formula:



      $displaystyle lim_{jtoinfty}prod_{n=j}^{mj}frac{pi}{2tan^{-1}(kn)}=m^{frac{2}{kpi}}$



      But I have not found a proof for this. Any proof would be appreciated.






      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 21 '18 at 22:54







      Larry

















      asked Aug 26 '18 at 23:15









      LarryLarry

      2,55031131




      2,55031131






















          1 Answer
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          $begingroup$

          It is easier to analyze the behavior of the limit if you take logarithm. Indeed, write



          $$ P_j = P_j(m,k) = prod_{n=j}^{mj} frac{pi}{2arctan(kn)} $$



          for the expression inside the limit. Then using the relation $arctan(x) = frac{pi}{2} - arctan(frac{1}{x})$ which holds for $x > 0$, we find that



          $$ logleft(frac{pi}{2arctan(x)}right)
          = -logleft(1 - tfrac{2}{pi}arctanleft(tfrac{1}{x}right) right)
          = tfrac{2}{pi x} + mathcal{O}left(x^{-2}right)
          quad text{as} quad x to infty.$$



          So it follows that



          $$
          log P_j
          = sum_{n=j}^{mj} left( logfrac{pi}{2} - logarctan(kn)right)
          = sum_{n=j}^{mj} left( frac{2}{pi kn} + mathcal{O}(n^{-2}) right).
          $$



          By noting that this is a Riemann sum with a vanishing error term, we realize that $log P_n$ converges to $int_{1}^{m} frac{2}{pi kx} , dx
          = frac{2}{pi k}log m$ as $ntoinfty$. Exponentiation then yields $ P_n to m^{frac{2}{pi k}} $ as expected.






          share|cite|improve this answer









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            4












            $begingroup$

            It is easier to analyze the behavior of the limit if you take logarithm. Indeed, write



            $$ P_j = P_j(m,k) = prod_{n=j}^{mj} frac{pi}{2arctan(kn)} $$



            for the expression inside the limit. Then using the relation $arctan(x) = frac{pi}{2} - arctan(frac{1}{x})$ which holds for $x > 0$, we find that



            $$ logleft(frac{pi}{2arctan(x)}right)
            = -logleft(1 - tfrac{2}{pi}arctanleft(tfrac{1}{x}right) right)
            = tfrac{2}{pi x} + mathcal{O}left(x^{-2}right)
            quad text{as} quad x to infty.$$



            So it follows that



            $$
            log P_j
            = sum_{n=j}^{mj} left( logfrac{pi}{2} - logarctan(kn)right)
            = sum_{n=j}^{mj} left( frac{2}{pi kn} + mathcal{O}(n^{-2}) right).
            $$



            By noting that this is a Riemann sum with a vanishing error term, we realize that $log P_n$ converges to $int_{1}^{m} frac{2}{pi kx} , dx
            = frac{2}{pi k}log m$ as $ntoinfty$. Exponentiation then yields $ P_n to m^{frac{2}{pi k}} $ as expected.






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              It is easier to analyze the behavior of the limit if you take logarithm. Indeed, write



              $$ P_j = P_j(m,k) = prod_{n=j}^{mj} frac{pi}{2arctan(kn)} $$



              for the expression inside the limit. Then using the relation $arctan(x) = frac{pi}{2} - arctan(frac{1}{x})$ which holds for $x > 0$, we find that



              $$ logleft(frac{pi}{2arctan(x)}right)
              = -logleft(1 - tfrac{2}{pi}arctanleft(tfrac{1}{x}right) right)
              = tfrac{2}{pi x} + mathcal{O}left(x^{-2}right)
              quad text{as} quad x to infty.$$



              So it follows that



              $$
              log P_j
              = sum_{n=j}^{mj} left( logfrac{pi}{2} - logarctan(kn)right)
              = sum_{n=j}^{mj} left( frac{2}{pi kn} + mathcal{O}(n^{-2}) right).
              $$



              By noting that this is a Riemann sum with a vanishing error term, we realize that $log P_n$ converges to $int_{1}^{m} frac{2}{pi kx} , dx
              = frac{2}{pi k}log m$ as $ntoinfty$. Exponentiation then yields $ P_n to m^{frac{2}{pi k}} $ as expected.






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                It is easier to analyze the behavior of the limit if you take logarithm. Indeed, write



                $$ P_j = P_j(m,k) = prod_{n=j}^{mj} frac{pi}{2arctan(kn)} $$



                for the expression inside the limit. Then using the relation $arctan(x) = frac{pi}{2} - arctan(frac{1}{x})$ which holds for $x > 0$, we find that



                $$ logleft(frac{pi}{2arctan(x)}right)
                = -logleft(1 - tfrac{2}{pi}arctanleft(tfrac{1}{x}right) right)
                = tfrac{2}{pi x} + mathcal{O}left(x^{-2}right)
                quad text{as} quad x to infty.$$



                So it follows that



                $$
                log P_j
                = sum_{n=j}^{mj} left( logfrac{pi}{2} - logarctan(kn)right)
                = sum_{n=j}^{mj} left( frac{2}{pi kn} + mathcal{O}(n^{-2}) right).
                $$



                By noting that this is a Riemann sum with a vanishing error term, we realize that $log P_n$ converges to $int_{1}^{m} frac{2}{pi kx} , dx
                = frac{2}{pi k}log m$ as $ntoinfty$. Exponentiation then yields $ P_n to m^{frac{2}{pi k}} $ as expected.






                share|cite|improve this answer









                $endgroup$



                It is easier to analyze the behavior of the limit if you take logarithm. Indeed, write



                $$ P_j = P_j(m,k) = prod_{n=j}^{mj} frac{pi}{2arctan(kn)} $$



                for the expression inside the limit. Then using the relation $arctan(x) = frac{pi}{2} - arctan(frac{1}{x})$ which holds for $x > 0$, we find that



                $$ logleft(frac{pi}{2arctan(x)}right)
                = -logleft(1 - tfrac{2}{pi}arctanleft(tfrac{1}{x}right) right)
                = tfrac{2}{pi x} + mathcal{O}left(x^{-2}right)
                quad text{as} quad x to infty.$$



                So it follows that



                $$
                log P_j
                = sum_{n=j}^{mj} left( logfrac{pi}{2} - logarctan(kn)right)
                = sum_{n=j}^{mj} left( frac{2}{pi kn} + mathcal{O}(n^{-2}) right).
                $$



                By noting that this is a Riemann sum with a vanishing error term, we realize that $log P_n$ converges to $int_{1}^{m} frac{2}{pi kx} , dx
                = frac{2}{pi k}log m$ as $ntoinfty$. Exponentiation then yields $ P_n to m^{frac{2}{pi k}} $ as expected.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 26 '18 at 23:37









                Sangchul LeeSangchul Lee

                96.6k12173283




                96.6k12173283






























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