Limit involving binomial coefficients without Stirling's formula












6












$begingroup$


I have this question from a friend who is taking college admission exam, evaluate: $$lim_{ntoinfty} frac{binom{4n}{2n}}{4^nbinom{2n}{n}}$$ The only way I could do this is by using Stirling's formula:$$ n! sim sqrt{2 pi n} (frac{n}{e})^n$$ after rewriting as $$lim_{ntoinfty} frac{(4n)!(n!)^2}{4^n(2n)!^3}$$ and it simplifies really satisfying to $frac{1}{sqrt2}$.



However Stirling's formula is not in the syllabus nor taught in high school, is there an elementary approach to this limit?






limits binomial-coefficients







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$endgroup$








  • 1




    $begingroup$
    I think expecting high student to know more than the syllabus, in order for a college to admit 'good' student, may not be quite surprising...
    $endgroup$
    – Tony Ma
    Apr 1 '18 at 7:56
















6












$begingroup$


I have this question from a friend who is taking college admission exam, evaluate: $$lim_{ntoinfty} frac{binom{4n}{2n}}{4^nbinom{2n}{n}}$$ The only way I could do this is by using Stirling's formula:$$ n! sim sqrt{2 pi n} (frac{n}{e})^n$$ after rewriting as $$lim_{ntoinfty} frac{(4n)!(n!)^2}{4^n(2n)!^3}$$ and it simplifies really satisfying to $frac{1}{sqrt2}$.



However Stirling's formula is not in the syllabus nor taught in high school, is there an elementary approach to this limit?






limits binomial-coefficients







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I think expecting high student to know more than the syllabus, in order for a college to admit 'good' student, may not be quite surprising...
    $endgroup$
    – Tony Ma
    Apr 1 '18 at 7:56














6












6








6


3



$begingroup$


I have this question from a friend who is taking college admission exam, evaluate: $$lim_{ntoinfty} frac{binom{4n}{2n}}{4^nbinom{2n}{n}}$$ The only way I could do this is by using Stirling's formula:$$ n! sim sqrt{2 pi n} (frac{n}{e})^n$$ after rewriting as $$lim_{ntoinfty} frac{(4n)!(n!)^2}{4^n(2n)!^3}$$ and it simplifies really satisfying to $frac{1}{sqrt2}$.



However Stirling's formula is not in the syllabus nor taught in high school, is there an elementary approach to this limit?






limits binomial-coefficients







share|cite|improve this question











$endgroup$




I have this question from a friend who is taking college admission exam, evaluate: $$lim_{ntoinfty} frac{binom{4n}{2n}}{4^nbinom{2n}{n}}$$ The only way I could do this is by using Stirling's formula:$$ n! sim sqrt{2 pi n} (frac{n}{e})^n$$ after rewriting as $$lim_{ntoinfty} frac{(4n)!(n!)^2}{4^n(2n)!^3}$$ and it simplifies really satisfying to $frac{1}{sqrt2}$.



However Stirling's formula is not in the syllabus nor taught in high school, is there an elementary approach to this limit?






limits binomial-coefficients




limits binomial-coefficients






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 1 '18 at 22:16









Jack D'Aurizio

290k33282662




290k33282662










asked Apr 1 '18 at 7:48









ZackyZacky

6,6851958




6,6851958








  • 1




    $begingroup$
    I think expecting high student to know more than the syllabus, in order for a college to admit 'good' student, may not be quite surprising...
    $endgroup$
    – Tony Ma
    Apr 1 '18 at 7:56














  • 1




    $begingroup$
    I think expecting high student to know more than the syllabus, in order for a college to admit 'good' student, may not be quite surprising...
    $endgroup$
    – Tony Ma
    Apr 1 '18 at 7:56








1




1




$begingroup$
I think expecting high student to know more than the syllabus, in order for a college to admit 'good' student, may not be quite surprising...
$endgroup$
– Tony Ma
Apr 1 '18 at 7:56




$begingroup$
I think expecting high student to know more than the syllabus, in order for a college to admit 'good' student, may not be quite surprising...
$endgroup$
– Tony Ma
Apr 1 '18 at 7:56










3 Answers
3






active

oldest

votes


















3












$begingroup$

Actually you don't even need $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pi n}}$ which can be proved, for instance, through
$$ frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}cos(t)^{2n},dt sim frac{2}{pi}int_{0}^{+infty}e^{-nt^2},dt. $$
Here you have to consider
$$ lim_{nto +infty}prod_{k=1}^{n}frac{kcdot kcdot(4k-3)cdot(4k-2)cdot(4k-1)cdot 4k}{4cdot(2k-1)^3cdot (2k)^3}=lim_{nto +infty}prod_{k=1}^{n}left(1-frac{1}{4(2k-1)^2}right) $$
and
$$ prod_{kgeq 0}left(1-frac{1}{4(2k+1)^2}right) = lim_{zto pi/4}prod_{kgeq 0}left(1-frac{4z^2}{(2k+1)^2pi^2}right)=lim_{zto pi/4}cos(z)=frac{1}{sqrt{2}} $$
by the Weierstrass product for the cosine function. Notice that the final outcome ($frac{1}{sqrt{2}}$) is unaffected if we know in advance that $lim_{nto +infty}frac{sqrt{n}}{4^n}binom{2n}{n}=C$ for some mysterious and non-explicit constant $C$. By squaring and telescoping, this boils down to the fact that $prod_{kgeq 1}left(1-frac{1}{(2k+1)^2}right)$ is convergent, which on its turn boils down to the fact that $sum_{kgeq 1}frac{1}{(2k+1)^2}$ is convergent.






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    $$l=lim_{nto infty} frac{(4n)!(n!)^2}{4^n(2n)!^3}=lim_{ntoinfty}frac{n!cdot (2cdot 2n)(4n-1)(2cdot (2n-1))...(2n+1)}{2^n2^n(2n)!cdot (2n)(2n-1)...(n+1)}$$ And with some adjustments results in: $$l=lim_{nto infty}frac{(2n+1)(2n+3)...(4n-1)}{(2n+2)(2n+4)...(4n)}$$ Taking logarithm on both sides gives:
    $$ln l=lim_{ntoinfty}sum_{k=1}^{n}lnleft(frac{2n+2k-1}{2n+2k}right)$$ And using the fact that $displaystyle{lim_{xto0} ln(1+x)=x}$ we get that: $$ln l=lim_{ntoinfty}sum_{k=1}^{n}frac{-1}{2n+2k}=-frac{1}{2}lim_{ntoinfty} frac{1}{n}sum_{k=1}^{n}frac{1}{1+frac{k}{n}}$$ Which evaluates as a Riemann sum:$$ln l=-frac{1}{2}int_0^1frac{1}{1+x}dx=-frac{1}{2}ln 2Rightarrow l=frac{1}{sqrt2}$$






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      If you look here, you will see that



      "The Wallis product can be written in asymptotic form for the central binomial coefficient"



      $${2m choose m} sim frac{4^m}{sqrt{pi ,m}}$$



      In the same page, they also give useful upper and lower bounds of it.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        that is nice! But is this elementary? I mean do they learn about asymptotics form until college?
        $endgroup$
        – Zacky
        Apr 1 '18 at 8:30










      • $begingroup$
        Isn't striling formula as 'elementary' as this or even more 'elementarier'?
        $endgroup$
        – Tony Ma
        Apr 1 '18 at 8:33






      • 2




        $begingroup$
        @Zacky. I totally agree with you. When I was young (long time ago), this is what we were given for the central binomial coefficient
        $endgroup$
        – Claude Leibovici
        Apr 1 '18 at 8:34










      • $begingroup$
        @TonyMa. It is more "elementary" than Stirling but still ...
        $endgroup$
        – Claude Leibovici
        Apr 1 '18 at 8:36











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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Actually you don't even need $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pi n}}$ which can be proved, for instance, through
      $$ frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}cos(t)^{2n},dt sim frac{2}{pi}int_{0}^{+infty}e^{-nt^2},dt. $$
      Here you have to consider
      $$ lim_{nto +infty}prod_{k=1}^{n}frac{kcdot kcdot(4k-3)cdot(4k-2)cdot(4k-1)cdot 4k}{4cdot(2k-1)^3cdot (2k)^3}=lim_{nto +infty}prod_{k=1}^{n}left(1-frac{1}{4(2k-1)^2}right) $$
      and
      $$ prod_{kgeq 0}left(1-frac{1}{4(2k+1)^2}right) = lim_{zto pi/4}prod_{kgeq 0}left(1-frac{4z^2}{(2k+1)^2pi^2}right)=lim_{zto pi/4}cos(z)=frac{1}{sqrt{2}} $$
      by the Weierstrass product for the cosine function. Notice that the final outcome ($frac{1}{sqrt{2}}$) is unaffected if we know in advance that $lim_{nto +infty}frac{sqrt{n}}{4^n}binom{2n}{n}=C$ for some mysterious and non-explicit constant $C$. By squaring and telescoping, this boils down to the fact that $prod_{kgeq 1}left(1-frac{1}{(2k+1)^2}right)$ is convergent, which on its turn boils down to the fact that $sum_{kgeq 1}frac{1}{(2k+1)^2}$ is convergent.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        Actually you don't even need $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pi n}}$ which can be proved, for instance, through
        $$ frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}cos(t)^{2n},dt sim frac{2}{pi}int_{0}^{+infty}e^{-nt^2},dt. $$
        Here you have to consider
        $$ lim_{nto +infty}prod_{k=1}^{n}frac{kcdot kcdot(4k-3)cdot(4k-2)cdot(4k-1)cdot 4k}{4cdot(2k-1)^3cdot (2k)^3}=lim_{nto +infty}prod_{k=1}^{n}left(1-frac{1}{4(2k-1)^2}right) $$
        and
        $$ prod_{kgeq 0}left(1-frac{1}{4(2k+1)^2}right) = lim_{zto pi/4}prod_{kgeq 0}left(1-frac{4z^2}{(2k+1)^2pi^2}right)=lim_{zto pi/4}cos(z)=frac{1}{sqrt{2}} $$
        by the Weierstrass product for the cosine function. Notice that the final outcome ($frac{1}{sqrt{2}}$) is unaffected if we know in advance that $lim_{nto +infty}frac{sqrt{n}}{4^n}binom{2n}{n}=C$ for some mysterious and non-explicit constant $C$. By squaring and telescoping, this boils down to the fact that $prod_{kgeq 1}left(1-frac{1}{(2k+1)^2}right)$ is convergent, which on its turn boils down to the fact that $sum_{kgeq 1}frac{1}{(2k+1)^2}$ is convergent.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          Actually you don't even need $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pi n}}$ which can be proved, for instance, through
          $$ frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}cos(t)^{2n},dt sim frac{2}{pi}int_{0}^{+infty}e^{-nt^2},dt. $$
          Here you have to consider
          $$ lim_{nto +infty}prod_{k=1}^{n}frac{kcdot kcdot(4k-3)cdot(4k-2)cdot(4k-1)cdot 4k}{4cdot(2k-1)^3cdot (2k)^3}=lim_{nto +infty}prod_{k=1}^{n}left(1-frac{1}{4(2k-1)^2}right) $$
          and
          $$ prod_{kgeq 0}left(1-frac{1}{4(2k+1)^2}right) = lim_{zto pi/4}prod_{kgeq 0}left(1-frac{4z^2}{(2k+1)^2pi^2}right)=lim_{zto pi/4}cos(z)=frac{1}{sqrt{2}} $$
          by the Weierstrass product for the cosine function. Notice that the final outcome ($frac{1}{sqrt{2}}$) is unaffected if we know in advance that $lim_{nto +infty}frac{sqrt{n}}{4^n}binom{2n}{n}=C$ for some mysterious and non-explicit constant $C$. By squaring and telescoping, this boils down to the fact that $prod_{kgeq 1}left(1-frac{1}{(2k+1)^2}right)$ is convergent, which on its turn boils down to the fact that $sum_{kgeq 1}frac{1}{(2k+1)^2}$ is convergent.






          share|cite|improve this answer











          $endgroup$



          Actually you don't even need $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pi n}}$ which can be proved, for instance, through
          $$ frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}cos(t)^{2n},dt sim frac{2}{pi}int_{0}^{+infty}e^{-nt^2},dt. $$
          Here you have to consider
          $$ lim_{nto +infty}prod_{k=1}^{n}frac{kcdot kcdot(4k-3)cdot(4k-2)cdot(4k-1)cdot 4k}{4cdot(2k-1)^3cdot (2k)^3}=lim_{nto +infty}prod_{k=1}^{n}left(1-frac{1}{4(2k-1)^2}right) $$
          and
          $$ prod_{kgeq 0}left(1-frac{1}{4(2k+1)^2}right) = lim_{zto pi/4}prod_{kgeq 0}left(1-frac{4z^2}{(2k+1)^2pi^2}right)=lim_{zto pi/4}cos(z)=frac{1}{sqrt{2}} $$
          by the Weierstrass product for the cosine function. Notice that the final outcome ($frac{1}{sqrt{2}}$) is unaffected if we know in advance that $lim_{nto +infty}frac{sqrt{n}}{4^n}binom{2n}{n}=C$ for some mysterious and non-explicit constant $C$. By squaring and telescoping, this boils down to the fact that $prod_{kgeq 1}left(1-frac{1}{(2k+1)^2}right)$ is convergent, which on its turn boils down to the fact that $sum_{kgeq 1}frac{1}{(2k+1)^2}$ is convergent.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Apr 1 '18 at 22:36

























          answered Apr 1 '18 at 22:25









          Jack D'AurizioJack D'Aurizio

          290k33282662




          290k33282662























              3












              $begingroup$

              $$l=lim_{nto infty} frac{(4n)!(n!)^2}{4^n(2n)!^3}=lim_{ntoinfty}frac{n!cdot (2cdot 2n)(4n-1)(2cdot (2n-1))...(2n+1)}{2^n2^n(2n)!cdot (2n)(2n-1)...(n+1)}$$ And with some adjustments results in: $$l=lim_{nto infty}frac{(2n+1)(2n+3)...(4n-1)}{(2n+2)(2n+4)...(4n)}$$ Taking logarithm on both sides gives:
              $$ln l=lim_{ntoinfty}sum_{k=1}^{n}lnleft(frac{2n+2k-1}{2n+2k}right)$$ And using the fact that $displaystyle{lim_{xto0} ln(1+x)=x}$ we get that: $$ln l=lim_{ntoinfty}sum_{k=1}^{n}frac{-1}{2n+2k}=-frac{1}{2}lim_{ntoinfty} frac{1}{n}sum_{k=1}^{n}frac{1}{1+frac{k}{n}}$$ Which evaluates as a Riemann sum:$$ln l=-frac{1}{2}int_0^1frac{1}{1+x}dx=-frac{1}{2}ln 2Rightarrow l=frac{1}{sqrt2}$$






              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

                $$l=lim_{nto infty} frac{(4n)!(n!)^2}{4^n(2n)!^3}=lim_{ntoinfty}frac{n!cdot (2cdot 2n)(4n-1)(2cdot (2n-1))...(2n+1)}{2^n2^n(2n)!cdot (2n)(2n-1)...(n+1)}$$ And with some adjustments results in: $$l=lim_{nto infty}frac{(2n+1)(2n+3)...(4n-1)}{(2n+2)(2n+4)...(4n)}$$ Taking logarithm on both sides gives:
                $$ln l=lim_{ntoinfty}sum_{k=1}^{n}lnleft(frac{2n+2k-1}{2n+2k}right)$$ And using the fact that $displaystyle{lim_{xto0} ln(1+x)=x}$ we get that: $$ln l=lim_{ntoinfty}sum_{k=1}^{n}frac{-1}{2n+2k}=-frac{1}{2}lim_{ntoinfty} frac{1}{n}sum_{k=1}^{n}frac{1}{1+frac{k}{n}}$$ Which evaluates as a Riemann sum:$$ln l=-frac{1}{2}int_0^1frac{1}{1+x}dx=-frac{1}{2}ln 2Rightarrow l=frac{1}{sqrt2}$$






                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  $$l=lim_{nto infty} frac{(4n)!(n!)^2}{4^n(2n)!^3}=lim_{ntoinfty}frac{n!cdot (2cdot 2n)(4n-1)(2cdot (2n-1))...(2n+1)}{2^n2^n(2n)!cdot (2n)(2n-1)...(n+1)}$$ And with some adjustments results in: $$l=lim_{nto infty}frac{(2n+1)(2n+3)...(4n-1)}{(2n+2)(2n+4)...(4n)}$$ Taking logarithm on both sides gives:
                  $$ln l=lim_{ntoinfty}sum_{k=1}^{n}lnleft(frac{2n+2k-1}{2n+2k}right)$$ And using the fact that $displaystyle{lim_{xto0} ln(1+x)=x}$ we get that: $$ln l=lim_{ntoinfty}sum_{k=1}^{n}frac{-1}{2n+2k}=-frac{1}{2}lim_{ntoinfty} frac{1}{n}sum_{k=1}^{n}frac{1}{1+frac{k}{n}}$$ Which evaluates as a Riemann sum:$$ln l=-frac{1}{2}int_0^1frac{1}{1+x}dx=-frac{1}{2}ln 2Rightarrow l=frac{1}{sqrt2}$$






                  share|cite|improve this answer











                  $endgroup$



                  $$l=lim_{nto infty} frac{(4n)!(n!)^2}{4^n(2n)!^3}=lim_{ntoinfty}frac{n!cdot (2cdot 2n)(4n-1)(2cdot (2n-1))...(2n+1)}{2^n2^n(2n)!cdot (2n)(2n-1)...(n+1)}$$ And with some adjustments results in: $$l=lim_{nto infty}frac{(2n+1)(2n+3)...(4n-1)}{(2n+2)(2n+4)...(4n)}$$ Taking logarithm on both sides gives:
                  $$ln l=lim_{ntoinfty}sum_{k=1}^{n}lnleft(frac{2n+2k-1}{2n+2k}right)$$ And using the fact that $displaystyle{lim_{xto0} ln(1+x)=x}$ we get that: $$ln l=lim_{ntoinfty}sum_{k=1}^{n}frac{-1}{2n+2k}=-frac{1}{2}lim_{ntoinfty} frac{1}{n}sum_{k=1}^{n}frac{1}{1+frac{k}{n}}$$ Which evaluates as a Riemann sum:$$ln l=-frac{1}{2}int_0^1frac{1}{1+x}dx=-frac{1}{2}ln 2Rightarrow l=frac{1}{sqrt2}$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 23 '18 at 19:04

























                  answered Apr 6 '18 at 14:48









                  ZackyZacky

                  6,6851958




                  6,6851958























                      1












                      $begingroup$

                      If you look here, you will see that



                      "The Wallis product can be written in asymptotic form for the central binomial coefficient"



                      $${2m choose m} sim frac{4^m}{sqrt{pi ,m}}$$



                      In the same page, they also give useful upper and lower bounds of it.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        that is nice! But is this elementary? I mean do they learn about asymptotics form until college?
                        $endgroup$
                        – Zacky
                        Apr 1 '18 at 8:30










                      • $begingroup$
                        Isn't striling formula as 'elementary' as this or even more 'elementarier'?
                        $endgroup$
                        – Tony Ma
                        Apr 1 '18 at 8:33






                      • 2




                        $begingroup$
                        @Zacky. I totally agree with you. When I was young (long time ago), this is what we were given for the central binomial coefficient
                        $endgroup$
                        – Claude Leibovici
                        Apr 1 '18 at 8:34










                      • $begingroup$
                        @TonyMa. It is more "elementary" than Stirling but still ...
                        $endgroup$
                        – Claude Leibovici
                        Apr 1 '18 at 8:36
















                      1












                      $begingroup$

                      If you look here, you will see that



                      "The Wallis product can be written in asymptotic form for the central binomial coefficient"



                      $${2m choose m} sim frac{4^m}{sqrt{pi ,m}}$$



                      In the same page, they also give useful upper and lower bounds of it.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        that is nice! But is this elementary? I mean do they learn about asymptotics form until college?
                        $endgroup$
                        – Zacky
                        Apr 1 '18 at 8:30










                      • $begingroup$
                        Isn't striling formula as 'elementary' as this or even more 'elementarier'?
                        $endgroup$
                        – Tony Ma
                        Apr 1 '18 at 8:33






                      • 2




                        $begingroup$
                        @Zacky. I totally agree with you. When I was young (long time ago), this is what we were given for the central binomial coefficient
                        $endgroup$
                        – Claude Leibovici
                        Apr 1 '18 at 8:34










                      • $begingroup$
                        @TonyMa. It is more "elementary" than Stirling but still ...
                        $endgroup$
                        – Claude Leibovici
                        Apr 1 '18 at 8:36














                      1












                      1








                      1





                      $begingroup$

                      If you look here, you will see that



                      "The Wallis product can be written in asymptotic form for the central binomial coefficient"



                      $${2m choose m} sim frac{4^m}{sqrt{pi ,m}}$$



                      In the same page, they also give useful upper and lower bounds of it.






                      share|cite|improve this answer









                      $endgroup$



                      If you look here, you will see that



                      "The Wallis product can be written in asymptotic form for the central binomial coefficient"



                      $${2m choose m} sim frac{4^m}{sqrt{pi ,m}}$$



                      In the same page, they also give useful upper and lower bounds of it.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Apr 1 '18 at 8:09









                      Claude LeiboviciClaude Leibovici

                      122k1157134




                      122k1157134












                      • $begingroup$
                        that is nice! But is this elementary? I mean do they learn about asymptotics form until college?
                        $endgroup$
                        – Zacky
                        Apr 1 '18 at 8:30










                      • $begingroup$
                        Isn't striling formula as 'elementary' as this or even more 'elementarier'?
                        $endgroup$
                        – Tony Ma
                        Apr 1 '18 at 8:33






                      • 2




                        $begingroup$
                        @Zacky. I totally agree with you. When I was young (long time ago), this is what we were given for the central binomial coefficient
                        $endgroup$
                        – Claude Leibovici
                        Apr 1 '18 at 8:34










                      • $begingroup$
                        @TonyMa. It is more "elementary" than Stirling but still ...
                        $endgroup$
                        – Claude Leibovici
                        Apr 1 '18 at 8:36


















                      • $begingroup$
                        that is nice! But is this elementary? I mean do they learn about asymptotics form until college?
                        $endgroup$
                        – Zacky
                        Apr 1 '18 at 8:30










                      • $begingroup$
                        Isn't striling formula as 'elementary' as this or even more 'elementarier'?
                        $endgroup$
                        – Tony Ma
                        Apr 1 '18 at 8:33






                      • 2




                        $begingroup$
                        @Zacky. I totally agree with you. When I was young (long time ago), this is what we were given for the central binomial coefficient
                        $endgroup$
                        – Claude Leibovici
                        Apr 1 '18 at 8:34










                      • $begingroup$
                        @TonyMa. It is more "elementary" than Stirling but still ...
                        $endgroup$
                        – Claude Leibovici
                        Apr 1 '18 at 8:36
















                      $begingroup$
                      that is nice! But is this elementary? I mean do they learn about asymptotics form until college?
                      $endgroup$
                      – Zacky
                      Apr 1 '18 at 8:30




                      $begingroup$
                      that is nice! But is this elementary? I mean do they learn about asymptotics form until college?
                      $endgroup$
                      – Zacky
                      Apr 1 '18 at 8:30












                      $begingroup$
                      Isn't striling formula as 'elementary' as this or even more 'elementarier'?
                      $endgroup$
                      – Tony Ma
                      Apr 1 '18 at 8:33




                      $begingroup$
                      Isn't striling formula as 'elementary' as this or even more 'elementarier'?
                      $endgroup$
                      – Tony Ma
                      Apr 1 '18 at 8:33




                      2




                      2




                      $begingroup$
                      @Zacky. I totally agree with you. When I was young (long time ago), this is what we were given for the central binomial coefficient
                      $endgroup$
                      – Claude Leibovici
                      Apr 1 '18 at 8:34




                      $begingroup$
                      @Zacky. I totally agree with you. When I was young (long time ago), this is what we were given for the central binomial coefficient
                      $endgroup$
                      – Claude Leibovici
                      Apr 1 '18 at 8:34












                      $begingroup$
                      @TonyMa. It is more "elementary" than Stirling but still ...
                      $endgroup$
                      – Claude Leibovici
                      Apr 1 '18 at 8:36




                      $begingroup$
                      @TonyMa. It is more "elementary" than Stirling but still ...
                      $endgroup$
                      – Claude Leibovici
                      Apr 1 '18 at 8:36


















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