Limit as a Riemann sum











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I need to comupte this limit by treating it as a Riemann sum
$$L := lim_{ntoinfty} frac{1}{n}sqrt[n]{(n+1)(n+2)...(n+n)}$$
So first I took the natural log from both sides to get
$$ ln L = lim_{ntoinfty} left(lnfrac{1}{n} + frac{1}{n}left(ln(n+1)+ln(n+2)+...+ln(n+n)right)right)$$
which can then be written as
$$ ln L = lim_{ntoinfty} lnfrac{1}{n} + lim_{ntoinfty}left(frac{ln(n+1)+ln(n+2)+...+ln(n+n)}{n}right)$$
And using the sigma notation
$$ ln L = lim_{ntoinfty} lnfrac{1}{n} + lim_{ntoinfty}sum_{k=1}^{n}left(frac{1}{n}ln(n+k)right)$$
Further analysing this second bit I got that since $ln(n+k) = lnleft(nleft(1+frac{k}{n}right)right) = ln(n) +lnleft(1+frac{k}{n}right)$
It would seem that I should take $[a,b] = [0,1]$ and $f(x) = ln(1+x)$, but I am pretty certain that I have made a mistake somewhere. Can anyone point out my mistake?






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  • In the third line, you have the indeterminate form $ln L = -infty + infty$.
    – angryavian
    Nov 27 at 17:57















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1
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I need to comupte this limit by treating it as a Riemann sum
$$L := lim_{ntoinfty} frac{1}{n}sqrt[n]{(n+1)(n+2)...(n+n)}$$
So first I took the natural log from both sides to get
$$ ln L = lim_{ntoinfty} left(lnfrac{1}{n} + frac{1}{n}left(ln(n+1)+ln(n+2)+...+ln(n+n)right)right)$$
which can then be written as
$$ ln L = lim_{ntoinfty} lnfrac{1}{n} + lim_{ntoinfty}left(frac{ln(n+1)+ln(n+2)+...+ln(n+n)}{n}right)$$
And using the sigma notation
$$ ln L = lim_{ntoinfty} lnfrac{1}{n} + lim_{ntoinfty}sum_{k=1}^{n}left(frac{1}{n}ln(n+k)right)$$
Further analysing this second bit I got that since $ln(n+k) = lnleft(nleft(1+frac{k}{n}right)right) = ln(n) +lnleft(1+frac{k}{n}right)$
It would seem that I should take $[a,b] = [0,1]$ and $f(x) = ln(1+x)$, but I am pretty certain that I have made a mistake somewhere. Can anyone point out my mistake?






calculus real-analysis







share|cite|improve this question






















  • In the third line, you have the indeterminate form $ln L = -infty + infty$.
    – angryavian
    Nov 27 at 17:57













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I need to comupte this limit by treating it as a Riemann sum
$$L := lim_{ntoinfty} frac{1}{n}sqrt[n]{(n+1)(n+2)...(n+n)}$$
So first I took the natural log from both sides to get
$$ ln L = lim_{ntoinfty} left(lnfrac{1}{n} + frac{1}{n}left(ln(n+1)+ln(n+2)+...+ln(n+n)right)right)$$
which can then be written as
$$ ln L = lim_{ntoinfty} lnfrac{1}{n} + lim_{ntoinfty}left(frac{ln(n+1)+ln(n+2)+...+ln(n+n)}{n}right)$$
And using the sigma notation
$$ ln L = lim_{ntoinfty} lnfrac{1}{n} + lim_{ntoinfty}sum_{k=1}^{n}left(frac{1}{n}ln(n+k)right)$$
Further analysing this second bit I got that since $ln(n+k) = lnleft(nleft(1+frac{k}{n}right)right) = ln(n) +lnleft(1+frac{k}{n}right)$
It would seem that I should take $[a,b] = [0,1]$ and $f(x) = ln(1+x)$, but I am pretty certain that I have made a mistake somewhere. Can anyone point out my mistake?






calculus real-analysis







share|cite|improve this question













I need to comupte this limit by treating it as a Riemann sum
$$L := lim_{ntoinfty} frac{1}{n}sqrt[n]{(n+1)(n+2)...(n+n)}$$
So first I took the natural log from both sides to get
$$ ln L = lim_{ntoinfty} left(lnfrac{1}{n} + frac{1}{n}left(ln(n+1)+ln(n+2)+...+ln(n+n)right)right)$$
which can then be written as
$$ ln L = lim_{ntoinfty} lnfrac{1}{n} + lim_{ntoinfty}left(frac{ln(n+1)+ln(n+2)+...+ln(n+n)}{n}right)$$
And using the sigma notation
$$ ln L = lim_{ntoinfty} lnfrac{1}{n} + lim_{ntoinfty}sum_{k=1}^{n}left(frac{1}{n}ln(n+k)right)$$
Further analysing this second bit I got that since $ln(n+k) = lnleft(nleft(1+frac{k}{n}right)right) = ln(n) +lnleft(1+frac{k}{n}right)$
It would seem that I should take $[a,b] = [0,1]$ and $f(x) = ln(1+x)$, but I am pretty certain that I have made a mistake somewhere. Can anyone point out my mistake?






calculus real-analysis




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asked Nov 27 at 17:55









Markus Punnar

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  • In the third line, you have the indeterminate form $ln L = -infty + infty$.
    – angryavian
    Nov 27 at 17:57


















  • In the third line, you have the indeterminate form $ln L = -infty + infty$.
    – angryavian
    Nov 27 at 17:57
















In the third line, you have the indeterminate form $ln L = -infty + infty$.
– angryavian
Nov 27 at 17:57




In the third line, you have the indeterminate form $ln L = -infty + infty$.
– angryavian
Nov 27 at 17:57










2 Answers
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$$ ln L = lim_{ntoinfty} left(lnfrac{1}{n} + frac{1}{n}left(ln(n+1)+ln(n+2)+...+ln(n+n)right)right)=\lim_{ntoinfty} left( frac{1}{n}ln(1+frac{1}{n})+frac{1}{n}ln(1+frac{2}{n})+...+frac{1}{n}ln(1+frac{n}{n})right)=\lim_{nto infty}sum_{i=1}^{n}{1over n}ln (1+{iover n})=int_{0}^{1}ln (1+x) dx=(1+x)ln (1+x)-(1+x)Bigg|_{0}^{1}=2ln 2-1$$therefore $$L={4over e}$$






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    $ln L = -ln n + frac 1n sum_limits{i= n+1}^{2n} ln i$



    Consider a Riemann sum with $n$ intervals of step 1



    $sum_limits{i= n}^{2n-1} ln i < int_limits{n}^{2n} ln x dx <sum_limits{i= n+1}^{2n} ln i$



    $intlimits_n^{2n} ln x dx = xln x - x|_{n}^{2n}\
    (2nln 2n - 2n) - (nln n - n)$



    This is Stirling's approximation.



    or more precisely Stirling's approximation says



    $ln n! approx nln n - n$



    Plugged into



    $L = frac 1n (frac {2n!}{n!})^frac 1n\
    ln L = -ln n + frac 1n ln 2n! - frac 1n ln n!$



    $ln L = -ln n + 2ln 2n -2 - ln n + 1\
    ln L = 2ln 2 - 1\
    L = 4e^{-1}$






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      2 Answers
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      2 Answers
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      $$ ln L = lim_{ntoinfty} left(lnfrac{1}{n} + frac{1}{n}left(ln(n+1)+ln(n+2)+...+ln(n+n)right)right)=\lim_{ntoinfty} left( frac{1}{n}ln(1+frac{1}{n})+frac{1}{n}ln(1+frac{2}{n})+...+frac{1}{n}ln(1+frac{n}{n})right)=\lim_{nto infty}sum_{i=1}^{n}{1over n}ln (1+{iover n})=int_{0}^{1}ln (1+x) dx=(1+x)ln (1+x)-(1+x)Bigg|_{0}^{1}=2ln 2-1$$therefore $$L={4over e}$$






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        $$ ln L = lim_{ntoinfty} left(lnfrac{1}{n} + frac{1}{n}left(ln(n+1)+ln(n+2)+...+ln(n+n)right)right)=\lim_{ntoinfty} left( frac{1}{n}ln(1+frac{1}{n})+frac{1}{n}ln(1+frac{2}{n})+...+frac{1}{n}ln(1+frac{n}{n})right)=\lim_{nto infty}sum_{i=1}^{n}{1over n}ln (1+{iover n})=int_{0}^{1}ln (1+x) dx=(1+x)ln (1+x)-(1+x)Bigg|_{0}^{1}=2ln 2-1$$therefore $$L={4over e}$$






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          up vote
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          up vote
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          accepted






          $$ ln L = lim_{ntoinfty} left(lnfrac{1}{n} + frac{1}{n}left(ln(n+1)+ln(n+2)+...+ln(n+n)right)right)=\lim_{ntoinfty} left( frac{1}{n}ln(1+frac{1}{n})+frac{1}{n}ln(1+frac{2}{n})+...+frac{1}{n}ln(1+frac{n}{n})right)=\lim_{nto infty}sum_{i=1}^{n}{1over n}ln (1+{iover n})=int_{0}^{1}ln (1+x) dx=(1+x)ln (1+x)-(1+x)Bigg|_{0}^{1}=2ln 2-1$$therefore $$L={4over e}$$






          share|cite|improve this answer












          $$ ln L = lim_{ntoinfty} left(lnfrac{1}{n} + frac{1}{n}left(ln(n+1)+ln(n+2)+...+ln(n+n)right)right)=\lim_{ntoinfty} left( frac{1}{n}ln(1+frac{1}{n})+frac{1}{n}ln(1+frac{2}{n})+...+frac{1}{n}ln(1+frac{n}{n})right)=\lim_{nto infty}sum_{i=1}^{n}{1over n}ln (1+{iover n})=int_{0}^{1}ln (1+x) dx=(1+x)ln (1+x)-(1+x)Bigg|_{0}^{1}=2ln 2-1$$therefore $$L={4over e}$$







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          answered Nov 27 at 18:11









          Mostafa Ayaz

          13.5k3836




          13.5k3836






















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              down vote













              $ln L = -ln n + frac 1n sum_limits{i= n+1}^{2n} ln i$



              Consider a Riemann sum with $n$ intervals of step 1



              $sum_limits{i= n}^{2n-1} ln i < int_limits{n}^{2n} ln x dx <sum_limits{i= n+1}^{2n} ln i$



              $intlimits_n^{2n} ln x dx = xln x - x|_{n}^{2n}\
              (2nln 2n - 2n) - (nln n - n)$



              This is Stirling's approximation.



              or more precisely Stirling's approximation says



              $ln n! approx nln n - n$



              Plugged into



              $L = frac 1n (frac {2n!}{n!})^frac 1n\
              ln L = -ln n + frac 1n ln 2n! - frac 1n ln n!$



              $ln L = -ln n + 2ln 2n -2 - ln n + 1\
              ln L = 2ln 2 - 1\
              L = 4e^{-1}$






              share|cite|improve this answer

























                up vote
                0
                down vote













                $ln L = -ln n + frac 1n sum_limits{i= n+1}^{2n} ln i$



                Consider a Riemann sum with $n$ intervals of step 1



                $sum_limits{i= n}^{2n-1} ln i < int_limits{n}^{2n} ln x dx <sum_limits{i= n+1}^{2n} ln i$



                $intlimits_n^{2n} ln x dx = xln x - x|_{n}^{2n}\
                (2nln 2n - 2n) - (nln n - n)$



                This is Stirling's approximation.



                or more precisely Stirling's approximation says



                $ln n! approx nln n - n$



                Plugged into



                $L = frac 1n (frac {2n!}{n!})^frac 1n\
                ln L = -ln n + frac 1n ln 2n! - frac 1n ln n!$



                $ln L = -ln n + 2ln 2n -2 - ln n + 1\
                ln L = 2ln 2 - 1\
                L = 4e^{-1}$






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  $ln L = -ln n + frac 1n sum_limits{i= n+1}^{2n} ln i$



                  Consider a Riemann sum with $n$ intervals of step 1



                  $sum_limits{i= n}^{2n-1} ln i < int_limits{n}^{2n} ln x dx <sum_limits{i= n+1}^{2n} ln i$



                  $intlimits_n^{2n} ln x dx = xln x - x|_{n}^{2n}\
                  (2nln 2n - 2n) - (nln n - n)$



                  This is Stirling's approximation.



                  or more precisely Stirling's approximation says



                  $ln n! approx nln n - n$



                  Plugged into



                  $L = frac 1n (frac {2n!}{n!})^frac 1n\
                  ln L = -ln n + frac 1n ln 2n! - frac 1n ln n!$



                  $ln L = -ln n + 2ln 2n -2 - ln n + 1\
                  ln L = 2ln 2 - 1\
                  L = 4e^{-1}$






                  share|cite|improve this answer












                  $ln L = -ln n + frac 1n sum_limits{i= n+1}^{2n} ln i$



                  Consider a Riemann sum with $n$ intervals of step 1



                  $sum_limits{i= n}^{2n-1} ln i < int_limits{n}^{2n} ln x dx <sum_limits{i= n+1}^{2n} ln i$



                  $intlimits_n^{2n} ln x dx = xln x - x|_{n}^{2n}\
                  (2nln 2n - 2n) - (nln n - n)$



                  This is Stirling's approximation.



                  or more precisely Stirling's approximation says



                  $ln n! approx nln n - n$



                  Plugged into



                  $L = frac 1n (frac {2n!}{n!})^frac 1n\
                  ln L = -ln n + frac 1n ln 2n! - frac 1n ln n!$



                  $ln L = -ln n + 2ln 2n -2 - ln n + 1\
                  ln L = 2ln 2 - 1\
                  L = 4e^{-1}$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 27 at 18:50









                  Doug M

                  43.5k31854




                  43.5k31854






























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