Calculate limit of Gamma function












2














Show that



$$lim _{x to infty} log left( frac{ sqrt{x} Gammaleft(frac{x}{2}right) } {Gamma left( frac{x+1}{2}right)} right) = frac{1}{2} log(2),$$ where $Gamma$ is the Gamma function.



I reduced this problem to calculate the limit:
$$lim_{x to infty} frac{Bleft(frac{x}{2},frac{x}{2}right)}{Bleft(frac{x+1}{2},frac{x+1}{2}right)} = 2, $$ where $B$ is the Beta function, but I don't know if this is useful or if there is other way to calculate this limit.



Any help will be very appreciated, thanks!










share|cite|improve this question





























    2














    Show that



    $$lim _{x to infty} log left( frac{ sqrt{x} Gammaleft(frac{x}{2}right) } {Gamma left( frac{x+1}{2}right)} right) = frac{1}{2} log(2),$$ where $Gamma$ is the Gamma function.



    I reduced this problem to calculate the limit:
    $$lim_{x to infty} frac{Bleft(frac{x}{2},frac{x}{2}right)}{Bleft(frac{x+1}{2},frac{x+1}{2}right)} = 2, $$ where $B$ is the Beta function, but I don't know if this is useful or if there is other way to calculate this limit.



    Any help will be very appreciated, thanks!










    share|cite|improve this question



























      2












      2








      2


      0





      Show that



      $$lim _{x to infty} log left( frac{ sqrt{x} Gammaleft(frac{x}{2}right) } {Gamma left( frac{x+1}{2}right)} right) = frac{1}{2} log(2),$$ where $Gamma$ is the Gamma function.



      I reduced this problem to calculate the limit:
      $$lim_{x to infty} frac{Bleft(frac{x}{2},frac{x}{2}right)}{Bleft(frac{x+1}{2},frac{x+1}{2}right)} = 2, $$ where $B$ is the Beta function, but I don't know if this is useful or if there is other way to calculate this limit.



      Any help will be very appreciated, thanks!










      share|cite|improve this question















      Show that



      $$lim _{x to infty} log left( frac{ sqrt{x} Gammaleft(frac{x}{2}right) } {Gamma left( frac{x+1}{2}right)} right) = frac{1}{2} log(2),$$ where $Gamma$ is the Gamma function.



      I reduced this problem to calculate the limit:
      $$lim_{x to infty} frac{Bleft(frac{x}{2},frac{x}{2}right)}{Bleft(frac{x+1}{2},frac{x+1}{2}right)} = 2, $$ where $B$ is the Beta function, but I don't know if this is useful or if there is other way to calculate this limit.



      Any help will be very appreciated, thanks!







      limits statistical-inference gamma-function probability-limit-theorems beta-function






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      share|cite|improve this question













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      edited Nov 27 at 0:54









      Larry

      1,6122722




      1,6122722










      asked Nov 27 at 0:37









      user392559

      37618




      37618






















          3 Answers
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          1














          $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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          begin{align}
          &bbox[10px,#ffd]{lim _{x to infty}
          lnpars{root{x}Gammapars{x/2} over
          Gammapars{bracks{x + 1}/2}}} =
          {1 over 2},lnpars{2} + lim _{x to infty}
          lnpars{root{x}Gammapars{x} over Gammapars{x + 1/2}}
          \[5mm] = &
          {1 over 2},lnpars{2} + lim _{x to infty}
          lnpars{root{x}bracks{x - 1}! over bracks{x - 1/2}!}
          \[5mm] = &
          {1 over 2},lnpars{2} + lim _{x to infty}
          lnpars{root{x},{root{2pi}bracks{x - 1}^{x - 1/2}
          expo{-pars{x - 1}} over
          root{2pi}bracks{x - 1/2}^{x}expo{-pars{x - 1/2}}}}
          \[5mm] = &
          {1 over 2},lnpars{2} + lim _{x to infty}
          lnpars{root{x},{x^{x - 1/2}bracks{1 - 1/x}^{x - 1/2}
          root{expo{}} over
          x^{x}bracks{1 - pars{1/2}/x}^{x}}}
          \[5mm] = &
          {1 over 2},lnpars{2} +
          underbrace{lnpars{{expo{-1}root{expo{}} overexpo{-1/2}}}}
          _{ds{= 0}} = bbx{{1 over 2},lnpars{2}}approx 0.3466
          end{align}






          share|cite|improve this answer





























            1














            Hint

            Use $$logGamma(x)=xlog x-x+frac12logfrac{2pi}x+o(1)$$ as $xtoinfty$.






            share|cite|improve this answer

















            • 1




              I can see how to use this to calculate the limit I want, but why is this true?
              – user392559
              Nov 27 at 1:08






            • 1




              It is called Stirling's formula, proof can be found here.
              – Kemono Chen
              Nov 27 at 1:11



















            1














            $$log left( frac{ sqrt{x}, Gammaleft(frac{x}{2}right) } {Gamma left( frac{x+1}{2}right)} right)=frac 12 log(x)+log left(Gamma left(frac{x}{2}right)right)-log left(Gamma left(frac{x+1}{2}right)right)$$



            Now, using Stirling approximation, as Kemono Chen commented,
            $$logleft(Gamma left(tright)right)=t (log (t)-1)+frac{1}{2} left(-log left({t}right)+log (2 pi
            )right)+frac{1}{12 t}+Oleft(frac{1}{t^3}right)$$
            Use this formula, simplify and continue with Taylor expansion to get
            $$log left( frac{ sqrt{x}, Gammaleft(frac{x}{2}right) } {Gamma left( frac{x+1}{2}right)} right)=frac{log (2)}{2}+frac{1}{4 x}+Oleft(frac{1}{x^3}right)$$ which shows the limit and how it is approached.






            share|cite|improve this answer





















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              3 Answers
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              3 Answers
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              active

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              1














              $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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              begin{align}
              &bbox[10px,#ffd]{lim _{x to infty}
              lnpars{root{x}Gammapars{x/2} over
              Gammapars{bracks{x + 1}/2}}} =
              {1 over 2},lnpars{2} + lim _{x to infty}
              lnpars{root{x}Gammapars{x} over Gammapars{x + 1/2}}
              \[5mm] = &
              {1 over 2},lnpars{2} + lim _{x to infty}
              lnpars{root{x}bracks{x - 1}! over bracks{x - 1/2}!}
              \[5mm] = &
              {1 over 2},lnpars{2} + lim _{x to infty}
              lnpars{root{x},{root{2pi}bracks{x - 1}^{x - 1/2}
              expo{-pars{x - 1}} over
              root{2pi}bracks{x - 1/2}^{x}expo{-pars{x - 1/2}}}}
              \[5mm] = &
              {1 over 2},lnpars{2} + lim _{x to infty}
              lnpars{root{x},{x^{x - 1/2}bracks{1 - 1/x}^{x - 1/2}
              root{expo{}} over
              x^{x}bracks{1 - pars{1/2}/x}^{x}}}
              \[5mm] = &
              {1 over 2},lnpars{2} +
              underbrace{lnpars{{expo{-1}root{expo{}} overexpo{-1/2}}}}
              _{ds{= 0}} = bbx{{1 over 2},lnpars{2}}approx 0.3466
              end{align}






              share|cite|improve this answer


























                1














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                begin{align}
                &bbox[10px,#ffd]{lim _{x to infty}
                lnpars{root{x}Gammapars{x/2} over
                Gammapars{bracks{x + 1}/2}}} =
                {1 over 2},lnpars{2} + lim _{x to infty}
                lnpars{root{x}Gammapars{x} over Gammapars{x + 1/2}}
                \[5mm] = &
                {1 over 2},lnpars{2} + lim _{x to infty}
                lnpars{root{x}bracks{x - 1}! over bracks{x - 1/2}!}
                \[5mm] = &
                {1 over 2},lnpars{2} + lim _{x to infty}
                lnpars{root{x},{root{2pi}bracks{x - 1}^{x - 1/2}
                expo{-pars{x - 1}} over
                root{2pi}bracks{x - 1/2}^{x}expo{-pars{x - 1/2}}}}
                \[5mm] = &
                {1 over 2},lnpars{2} + lim _{x to infty}
                lnpars{root{x},{x^{x - 1/2}bracks{1 - 1/x}^{x - 1/2}
                root{expo{}} over
                x^{x}bracks{1 - pars{1/2}/x}^{x}}}
                \[5mm] = &
                {1 over 2},lnpars{2} +
                underbrace{lnpars{{expo{-1}root{expo{}} overexpo{-1/2}}}}
                _{ds{= 0}} = bbx{{1 over 2},lnpars{2}}approx 0.3466
                end{align}






                share|cite|improve this answer
























                  1












                  1








                  1






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                  begin{align}
                  &bbox[10px,#ffd]{lim _{x to infty}
                  lnpars{root{x}Gammapars{x/2} over
                  Gammapars{bracks{x + 1}/2}}} =
                  {1 over 2},lnpars{2} + lim _{x to infty}
                  lnpars{root{x}Gammapars{x} over Gammapars{x + 1/2}}
                  \[5mm] = &
                  {1 over 2},lnpars{2} + lim _{x to infty}
                  lnpars{root{x}bracks{x - 1}! over bracks{x - 1/2}!}
                  \[5mm] = &
                  {1 over 2},lnpars{2} + lim _{x to infty}
                  lnpars{root{x},{root{2pi}bracks{x - 1}^{x - 1/2}
                  expo{-pars{x - 1}} over
                  root{2pi}bracks{x - 1/2}^{x}expo{-pars{x - 1/2}}}}
                  \[5mm] = &
                  {1 over 2},lnpars{2} + lim _{x to infty}
                  lnpars{root{x},{x^{x - 1/2}bracks{1 - 1/x}^{x - 1/2}
                  root{expo{}} over
                  x^{x}bracks{1 - pars{1/2}/x}^{x}}}
                  \[5mm] = &
                  {1 over 2},lnpars{2} +
                  underbrace{lnpars{{expo{-1}root{expo{}} overexpo{-1/2}}}}
                  _{ds{= 0}} = bbx{{1 over 2},lnpars{2}}approx 0.3466
                  end{align}






                  share|cite|improve this answer












                  $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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                  newcommand{verts}[1]{leftvert,{#1},rightvert}$

                  begin{align}
                  &bbox[10px,#ffd]{lim _{x to infty}
                  lnpars{root{x}Gammapars{x/2} over
                  Gammapars{bracks{x + 1}/2}}} =
                  {1 over 2},lnpars{2} + lim _{x to infty}
                  lnpars{root{x}Gammapars{x} over Gammapars{x + 1/2}}
                  \[5mm] = &
                  {1 over 2},lnpars{2} + lim _{x to infty}
                  lnpars{root{x}bracks{x - 1}! over bracks{x - 1/2}!}
                  \[5mm] = &
                  {1 over 2},lnpars{2} + lim _{x to infty}
                  lnpars{root{x},{root{2pi}bracks{x - 1}^{x - 1/2}
                  expo{-pars{x - 1}} over
                  root{2pi}bracks{x - 1/2}^{x}expo{-pars{x - 1/2}}}}
                  \[5mm] = &
                  {1 over 2},lnpars{2} + lim _{x to infty}
                  lnpars{root{x},{x^{x - 1/2}bracks{1 - 1/x}^{x - 1/2}
                  root{expo{}} over
                  x^{x}bracks{1 - pars{1/2}/x}^{x}}}
                  \[5mm] = &
                  {1 over 2},lnpars{2} +
                  underbrace{lnpars{{expo{-1}root{expo{}} overexpo{-1/2}}}}
                  _{ds{= 0}} = bbx{{1 over 2},lnpars{2}}approx 0.3466
                  end{align}







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                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 27 at 22:08









                  Felix Marin

                  67k7107140




                  67k7107140























                      1














                      Hint

                      Use $$logGamma(x)=xlog x-x+frac12logfrac{2pi}x+o(1)$$ as $xtoinfty$.






                      share|cite|improve this answer

















                      • 1




                        I can see how to use this to calculate the limit I want, but why is this true?
                        – user392559
                        Nov 27 at 1:08






                      • 1




                        It is called Stirling's formula, proof can be found here.
                        – Kemono Chen
                        Nov 27 at 1:11
















                      1














                      Hint

                      Use $$logGamma(x)=xlog x-x+frac12logfrac{2pi}x+o(1)$$ as $xtoinfty$.






                      share|cite|improve this answer

















                      • 1




                        I can see how to use this to calculate the limit I want, but why is this true?
                        – user392559
                        Nov 27 at 1:08






                      • 1




                        It is called Stirling's formula, proof can be found here.
                        – Kemono Chen
                        Nov 27 at 1:11














                      1












                      1








                      1






                      Hint

                      Use $$logGamma(x)=xlog x-x+frac12logfrac{2pi}x+o(1)$$ as $xtoinfty$.






                      share|cite|improve this answer












                      Hint

                      Use $$logGamma(x)=xlog x-x+frac12logfrac{2pi}x+o(1)$$ as $xtoinfty$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 27 at 0:53









                      Kemono Chen

                      2,527436




                      2,527436








                      • 1




                        I can see how to use this to calculate the limit I want, but why is this true?
                        – user392559
                        Nov 27 at 1:08






                      • 1




                        It is called Stirling's formula, proof can be found here.
                        – Kemono Chen
                        Nov 27 at 1:11














                      • 1




                        I can see how to use this to calculate the limit I want, but why is this true?
                        – user392559
                        Nov 27 at 1:08






                      • 1




                        It is called Stirling's formula, proof can be found here.
                        – Kemono Chen
                        Nov 27 at 1:11








                      1




                      1




                      I can see how to use this to calculate the limit I want, but why is this true?
                      – user392559
                      Nov 27 at 1:08




                      I can see how to use this to calculate the limit I want, but why is this true?
                      – user392559
                      Nov 27 at 1:08




                      1




                      1




                      It is called Stirling's formula, proof can be found here.
                      – Kemono Chen
                      Nov 27 at 1:11




                      It is called Stirling's formula, proof can be found here.
                      – Kemono Chen
                      Nov 27 at 1:11











                      1














                      $$log left( frac{ sqrt{x}, Gammaleft(frac{x}{2}right) } {Gamma left( frac{x+1}{2}right)} right)=frac 12 log(x)+log left(Gamma left(frac{x}{2}right)right)-log left(Gamma left(frac{x+1}{2}right)right)$$



                      Now, using Stirling approximation, as Kemono Chen commented,
                      $$logleft(Gamma left(tright)right)=t (log (t)-1)+frac{1}{2} left(-log left({t}right)+log (2 pi
                      )right)+frac{1}{12 t}+Oleft(frac{1}{t^3}right)$$
                      Use this formula, simplify and continue with Taylor expansion to get
                      $$log left( frac{ sqrt{x}, Gammaleft(frac{x}{2}right) } {Gamma left( frac{x+1}{2}right)} right)=frac{log (2)}{2}+frac{1}{4 x}+Oleft(frac{1}{x^3}right)$$ which shows the limit and how it is approached.






                      share|cite|improve this answer


























                        1














                        $$log left( frac{ sqrt{x}, Gammaleft(frac{x}{2}right) } {Gamma left( frac{x+1}{2}right)} right)=frac 12 log(x)+log left(Gamma left(frac{x}{2}right)right)-log left(Gamma left(frac{x+1}{2}right)right)$$



                        Now, using Stirling approximation, as Kemono Chen commented,
                        $$logleft(Gamma left(tright)right)=t (log (t)-1)+frac{1}{2} left(-log left({t}right)+log (2 pi
                        )right)+frac{1}{12 t}+Oleft(frac{1}{t^3}right)$$
                        Use this formula, simplify and continue with Taylor expansion to get
                        $$log left( frac{ sqrt{x}, Gammaleft(frac{x}{2}right) } {Gamma left( frac{x+1}{2}right)} right)=frac{log (2)}{2}+frac{1}{4 x}+Oleft(frac{1}{x^3}right)$$ which shows the limit and how it is approached.






                        share|cite|improve this answer
























                          1












                          1








                          1






                          $$log left( frac{ sqrt{x}, Gammaleft(frac{x}{2}right) } {Gamma left( frac{x+1}{2}right)} right)=frac 12 log(x)+log left(Gamma left(frac{x}{2}right)right)-log left(Gamma left(frac{x+1}{2}right)right)$$



                          Now, using Stirling approximation, as Kemono Chen commented,
                          $$logleft(Gamma left(tright)right)=t (log (t)-1)+frac{1}{2} left(-log left({t}right)+log (2 pi
                          )right)+frac{1}{12 t}+Oleft(frac{1}{t^3}right)$$
                          Use this formula, simplify and continue with Taylor expansion to get
                          $$log left( frac{ sqrt{x}, Gammaleft(frac{x}{2}right) } {Gamma left( frac{x+1}{2}right)} right)=frac{log (2)}{2}+frac{1}{4 x}+Oleft(frac{1}{x^3}right)$$ which shows the limit and how it is approached.






                          share|cite|improve this answer












                          $$log left( frac{ sqrt{x}, Gammaleft(frac{x}{2}right) } {Gamma left( frac{x+1}{2}right)} right)=frac 12 log(x)+log left(Gamma left(frac{x}{2}right)right)-log left(Gamma left(frac{x+1}{2}right)right)$$



                          Now, using Stirling approximation, as Kemono Chen commented,
                          $$logleft(Gamma left(tright)right)=t (log (t)-1)+frac{1}{2} left(-log left({t}right)+log (2 pi
                          )right)+frac{1}{12 t}+Oleft(frac{1}{t^3}right)$$
                          Use this formula, simplify and continue with Taylor expansion to get
                          $$log left( frac{ sqrt{x}, Gammaleft(frac{x}{2}right) } {Gamma left( frac{x+1}{2}right)} right)=frac{log (2)}{2}+frac{1}{4 x}+Oleft(frac{1}{x^3}right)$$ which shows the limit and how it is approached.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 27 at 4:29









                          Claude Leibovici

                          119k1157132




                          119k1157132






























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