Finding Bounds in proof of Stirling's Formula












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In a proof of Stirling's Formula, my professor claims that



$frac 12 $($ln$k + $ln$(k+1)) $le$ $int_k^{k+1}$$ln$x $dx$ $le$ $frac 12$($ln$k + $ln$(k+1)) + $frac 1{k^2}$.



I can see that the first inequality is true by concavity of the natural log function (hence the area of the trapezoid beneath natural log is less than the integral of the function itself), but can someone provide insight into the second inequality??



I know $frac 1{k^2}$ must be a bound for the area beneath $ln$x but above the trapezoid described by the leftmost expression, but it is not clear how to check that this is true.






logarithms factorial







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    2












    $begingroup$


    In a proof of Stirling's Formula, my professor claims that



    $frac 12 $($ln$k + $ln$(k+1)) $le$ $int_k^{k+1}$$ln$x $dx$ $le$ $frac 12$($ln$k + $ln$(k+1)) + $frac 1{k^2}$.



    I can see that the first inequality is true by concavity of the natural log function (hence the area of the trapezoid beneath natural log is less than the integral of the function itself), but can someone provide insight into the second inequality??



    I know $frac 1{k^2}$ must be a bound for the area beneath $ln$x but above the trapezoid described by the leftmost expression, but it is not clear how to check that this is true.






    logarithms factorial







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      In a proof of Stirling's Formula, my professor claims that



      $frac 12 $($ln$k + $ln$(k+1)) $le$ $int_k^{k+1}$$ln$x $dx$ $le$ $frac 12$($ln$k + $ln$(k+1)) + $frac 1{k^2}$.



      I can see that the first inequality is true by concavity of the natural log function (hence the area of the trapezoid beneath natural log is less than the integral of the function itself), but can someone provide insight into the second inequality??



      I know $frac 1{k^2}$ must be a bound for the area beneath $ln$x but above the trapezoid described by the leftmost expression, but it is not clear how to check that this is true.






      logarithms factorial







      share|cite|improve this question











      $endgroup$




      In a proof of Stirling's Formula, my professor claims that



      $frac 12 $($ln$k + $ln$(k+1)) $le$ $int_k^{k+1}$$ln$x $dx$ $le$ $frac 12$($ln$k + $ln$(k+1)) + $frac 1{k^2}$.



      I can see that the first inequality is true by concavity of the natural log function (hence the area of the trapezoid beneath natural log is less than the integral of the function itself), but can someone provide insight into the second inequality??



      I know $frac 1{k^2}$ must be a bound for the area beneath $ln$x but above the trapezoid described by the leftmost expression, but it is not clear how to check that this is true.






      logarithms factorial




      logarithms factorial






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 11 '18 at 10:20









      Shaun

      9,083113683




      9,083113683










      asked Dec 11 '18 at 8:47









      Greg HudgensGreg Hudgens

      111




      111






















          1 Answer
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          $begingroup$

          My attempt:



          begin{align}
          int_{ln(k)}^{ln(k+1)} &= int_0^1 ln(k+x) dx = int_0^{1/2} ln(k+x) + int_{1/2}^{1} ln(k+1 - x) dx \
          &= int_0^{1/2} ln(k) + frac{x}{k} - frac{x^2}{2k^2} + o(frac{1}{k^2}) dx \
          &quad + int_{1/2}^1 ln(k+1) - frac{x}{k+1} - frac{x^2}{(k+1)^2} + o(frac{1}{(k+1)^2}) dx \
          &= frac{1}{2} (ln(k) + ln(k+1)) + frac{1}{8k} + frac{1}{8(k+1)} - frac{1}{2(k+1)} \
          &quad - frac{1}{48k^2} + frac{1}{48(k+1)^2} - frac{1}{6k^2} + o(frac{1}{k^2})
          end{align}



          Clearly the terms of order $1/k$ and $1/k^2$ have negative sums, hence the bound is actually quite brutal, but probably sufficient for what your professor wants to do.






          share|cite|improve this answer











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            1 Answer
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            active

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            0












            $begingroup$

            My attempt:



            begin{align}
            int_{ln(k)}^{ln(k+1)} &= int_0^1 ln(k+x) dx = int_0^{1/2} ln(k+x) + int_{1/2}^{1} ln(k+1 - x) dx \
            &= int_0^{1/2} ln(k) + frac{x}{k} - frac{x^2}{2k^2} + o(frac{1}{k^2}) dx \
            &quad + int_{1/2}^1 ln(k+1) - frac{x}{k+1} - frac{x^2}{(k+1)^2} + o(frac{1}{(k+1)^2}) dx \
            &= frac{1}{2} (ln(k) + ln(k+1)) + frac{1}{8k} + frac{1}{8(k+1)} - frac{1}{2(k+1)} \
            &quad - frac{1}{48k^2} + frac{1}{48(k+1)^2} - frac{1}{6k^2} + o(frac{1}{k^2})
            end{align}



            Clearly the terms of order $1/k$ and $1/k^2$ have negative sums, hence the bound is actually quite brutal, but probably sufficient for what your professor wants to do.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              My attempt:



              begin{align}
              int_{ln(k)}^{ln(k+1)} &= int_0^1 ln(k+x) dx = int_0^{1/2} ln(k+x) + int_{1/2}^{1} ln(k+1 - x) dx \
              &= int_0^{1/2} ln(k) + frac{x}{k} - frac{x^2}{2k^2} + o(frac{1}{k^2}) dx \
              &quad + int_{1/2}^1 ln(k+1) - frac{x}{k+1} - frac{x^2}{(k+1)^2} + o(frac{1}{(k+1)^2}) dx \
              &= frac{1}{2} (ln(k) + ln(k+1)) + frac{1}{8k} + frac{1}{8(k+1)} - frac{1}{2(k+1)} \
              &quad - frac{1}{48k^2} + frac{1}{48(k+1)^2} - frac{1}{6k^2} + o(frac{1}{k^2})
              end{align}



              Clearly the terms of order $1/k$ and $1/k^2$ have negative sums, hence the bound is actually quite brutal, but probably sufficient for what your professor wants to do.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                My attempt:



                begin{align}
                int_{ln(k)}^{ln(k+1)} &= int_0^1 ln(k+x) dx = int_0^{1/2} ln(k+x) + int_{1/2}^{1} ln(k+1 - x) dx \
                &= int_0^{1/2} ln(k) + frac{x}{k} - frac{x^2}{2k^2} + o(frac{1}{k^2}) dx \
                &quad + int_{1/2}^1 ln(k+1) - frac{x}{k+1} - frac{x^2}{(k+1)^2} + o(frac{1}{(k+1)^2}) dx \
                &= frac{1}{2} (ln(k) + ln(k+1)) + frac{1}{8k} + frac{1}{8(k+1)} - frac{1}{2(k+1)} \
                &quad - frac{1}{48k^2} + frac{1}{48(k+1)^2} - frac{1}{6k^2} + o(frac{1}{k^2})
                end{align}



                Clearly the terms of order $1/k$ and $1/k^2$ have negative sums, hence the bound is actually quite brutal, but probably sufficient for what your professor wants to do.






                share|cite|improve this answer











                $endgroup$



                My attempt:



                begin{align}
                int_{ln(k)}^{ln(k+1)} &= int_0^1 ln(k+x) dx = int_0^{1/2} ln(k+x) + int_{1/2}^{1} ln(k+1 - x) dx \
                &= int_0^{1/2} ln(k) + frac{x}{k} - frac{x^2}{2k^2} + o(frac{1}{k^2}) dx \
                &quad + int_{1/2}^1 ln(k+1) - frac{x}{k+1} - frac{x^2}{(k+1)^2} + o(frac{1}{(k+1)^2}) dx \
                &= frac{1}{2} (ln(k) + ln(k+1)) + frac{1}{8k} + frac{1}{8(k+1)} - frac{1}{2(k+1)} \
                &quad - frac{1}{48k^2} + frac{1}{48(k+1)^2} - frac{1}{6k^2} + o(frac{1}{k^2})
                end{align}



                Clearly the terms of order $1/k$ and $1/k^2$ have negative sums, hence the bound is actually quite brutal, but probably sufficient for what your professor wants to do.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 11 '18 at 12:38

























                answered Dec 11 '18 at 10:34









                Gâteau-GalloisGâteau-Gallois

                362112




                362112






























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