How to solve the limit $lim_{k to infty} frac{(2k)!}{2^{2k} (k!)^2}$.











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How to solve this limit??




$$lim_{k to infty} frac{(2k)!}{2^{2k} (k!)^2}$$




It's a limit, not a series






calculus limits factorial







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  • I found that the answer is 0, however, I need to prove it.
    – S. Yoo
    Nov 22 at 5:26






  • 1




    HINT: note that $2^{2k}(k!)^2=left(prod_{j=1}^k 2jright)^2=(2n(2n-2)cdots 2)^2$
    – Masacroso
    Nov 22 at 5:30












  • en.wikipedia.org/wiki/Stirling%27s_approximation
    – lab bhattacharjee
    Nov 22 at 5:35










  • The SAME limit is here: math.stackexchange.com/questions/788096/…
    – Robert Z
    Nov 22 at 5:38












  • Possible duplicate of How to prove that $lim_{n toinfty} frac{(2n-1)!!}{(2n)!!}=0$
    – Nosrati
    Nov 22 at 5:48















up vote
2
down vote

favorite
2












How to solve this limit??




$$lim_{k to infty} frac{(2k)!}{2^{2k} (k!)^2}$$




It's a limit, not a series






calculus limits factorial







share|cite|improve this question
























  • I found that the answer is 0, however, I need to prove it.
    – S. Yoo
    Nov 22 at 5:26






  • 1




    HINT: note that $2^{2k}(k!)^2=left(prod_{j=1}^k 2jright)^2=(2n(2n-2)cdots 2)^2$
    – Masacroso
    Nov 22 at 5:30












  • en.wikipedia.org/wiki/Stirling%27s_approximation
    – lab bhattacharjee
    Nov 22 at 5:35










  • The SAME limit is here: math.stackexchange.com/questions/788096/…
    – Robert Z
    Nov 22 at 5:38












  • Possible duplicate of How to prove that $lim_{n toinfty} frac{(2n-1)!!}{(2n)!!}=0$
    – Nosrati
    Nov 22 at 5:48













up vote
2
down vote

favorite
2









up vote
2
down vote

favorite
2






2





How to solve this limit??




$$lim_{k to infty} frac{(2k)!}{2^{2k} (k!)^2}$$




It's a limit, not a series






calculus limits factorial







share|cite|improve this question















How to solve this limit??




$$lim_{k to infty} frac{(2k)!}{2^{2k} (k!)^2}$$




It's a limit, not a series






calculus limits factorial




calculus limits factorial






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 5:33









Nosrati

26.3k62353




26.3k62353










asked Nov 22 at 5:24









S. Yoo

453




453












  • I found that the answer is 0, however, I need to prove it.
    – S. Yoo
    Nov 22 at 5:26






  • 1




    HINT: note that $2^{2k}(k!)^2=left(prod_{j=1}^k 2jright)^2=(2n(2n-2)cdots 2)^2$
    – Masacroso
    Nov 22 at 5:30












  • en.wikipedia.org/wiki/Stirling%27s_approximation
    – lab bhattacharjee
    Nov 22 at 5:35










  • The SAME limit is here: math.stackexchange.com/questions/788096/…
    – Robert Z
    Nov 22 at 5:38












  • Possible duplicate of How to prove that $lim_{n toinfty} frac{(2n-1)!!}{(2n)!!}=0$
    – Nosrati
    Nov 22 at 5:48


















  • I found that the answer is 0, however, I need to prove it.
    – S. Yoo
    Nov 22 at 5:26






  • 1




    HINT: note that $2^{2k}(k!)^2=left(prod_{j=1}^k 2jright)^2=(2n(2n-2)cdots 2)^2$
    – Masacroso
    Nov 22 at 5:30












  • en.wikipedia.org/wiki/Stirling%27s_approximation
    – lab bhattacharjee
    Nov 22 at 5:35










  • The SAME limit is here: math.stackexchange.com/questions/788096/…
    – Robert Z
    Nov 22 at 5:38












  • Possible duplicate of How to prove that $lim_{n toinfty} frac{(2n-1)!!}{(2n)!!}=0$
    – Nosrati
    Nov 22 at 5:48
















I found that the answer is 0, however, I need to prove it.
– S. Yoo
Nov 22 at 5:26




I found that the answer is 0, however, I need to prove it.
– S. Yoo
Nov 22 at 5:26




1




1




HINT: note that $2^{2k}(k!)^2=left(prod_{j=1}^k 2jright)^2=(2n(2n-2)cdots 2)^2$
– Masacroso
Nov 22 at 5:30






HINT: note that $2^{2k}(k!)^2=left(prod_{j=1}^k 2jright)^2=(2n(2n-2)cdots 2)^2$
– Masacroso
Nov 22 at 5:30














en.wikipedia.org/wiki/Stirling%27s_approximation
– lab bhattacharjee
Nov 22 at 5:35




en.wikipedia.org/wiki/Stirling%27s_approximation
– lab bhattacharjee
Nov 22 at 5:35












The SAME limit is here: math.stackexchange.com/questions/788096/…
– Robert Z
Nov 22 at 5:38






The SAME limit is here: math.stackexchange.com/questions/788096/…
– Robert Z
Nov 22 at 5:38














Possible duplicate of How to prove that $lim_{n toinfty} frac{(2n-1)!!}{(2n)!!}=0$
– Nosrati
Nov 22 at 5:48




Possible duplicate of How to prove that $lim_{n toinfty} frac{(2n-1)!!}{(2n)!!}=0$
– Nosrati
Nov 22 at 5:48










2 Answers
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begin{align}
lim_{ktoinfty} frac{(2k)!}{2^{2k}cdot(k!)^2}
&=lim_{ktoinfty} frac{(2k)!}{2^k cdot 2^k cdot k! cdot k!} \
&=lim_{ktoinfty} frac{(2k)!}{(2^k cdot k!)^2} \
&=lim_{ktoinfty} frac{(2k)(2k-1)cdots(2)(1)}{(2k)^2 (2k-2)^2 cdots (4)^2 (2)^2} \
&=lim_{ktoinfty} frac{(2k-1)(2k-3)cdots(1)}{(2k)(2k-2)cdots(2)} \
&=0
end{align}



In the last step, you can think of the fraction as the infinite product of fractions less than $1$ ($frac{1}{2} timesfrac{3}{4} times frac{5}{6} timesfrac{7}{8}...$), which will decrease to $0$.






share|cite|improve this answer



















  • 3




    Not all such products converge to zero. This one may but it isn't automatic just from factors less than $1.$ Actually the product is of terms $1-(1/(2k))$ and so since the sum of the subtracted part goes to infinity you do have product zero here.
    – coffeemath
    Nov 22 at 7:16




















up vote
0
down vote













This is the probability of having $n$ heads and $n$ tails in $2n$ times of a fair coin toss. This probability then simply goes to 0. You can also use the Stirling's approximation to attain this.






share|cite|improve this answer





















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    2 Answers
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    up vote
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    begin{align}
    lim_{ktoinfty} frac{(2k)!}{2^{2k}cdot(k!)^2}
    &=lim_{ktoinfty} frac{(2k)!}{2^k cdot 2^k cdot k! cdot k!} \
    &=lim_{ktoinfty} frac{(2k)!}{(2^k cdot k!)^2} \
    &=lim_{ktoinfty} frac{(2k)(2k-1)cdots(2)(1)}{(2k)^2 (2k-2)^2 cdots (4)^2 (2)^2} \
    &=lim_{ktoinfty} frac{(2k-1)(2k-3)cdots(1)}{(2k)(2k-2)cdots(2)} \
    &=0
    end{align}



    In the last step, you can think of the fraction as the infinite product of fractions less than $1$ ($frac{1}{2} timesfrac{3}{4} times frac{5}{6} timesfrac{7}{8}...$), which will decrease to $0$.






    share|cite|improve this answer



















    • 3




      Not all such products converge to zero. This one may but it isn't automatic just from factors less than $1.$ Actually the product is of terms $1-(1/(2k))$ and so since the sum of the subtracted part goes to infinity you do have product zero here.
      – coffeemath
      Nov 22 at 7:16

















    up vote
    2
    down vote













    begin{align}
    lim_{ktoinfty} frac{(2k)!}{2^{2k}cdot(k!)^2}
    &=lim_{ktoinfty} frac{(2k)!}{2^k cdot 2^k cdot k! cdot k!} \
    &=lim_{ktoinfty} frac{(2k)!}{(2^k cdot k!)^2} \
    &=lim_{ktoinfty} frac{(2k)(2k-1)cdots(2)(1)}{(2k)^2 (2k-2)^2 cdots (4)^2 (2)^2} \
    &=lim_{ktoinfty} frac{(2k-1)(2k-3)cdots(1)}{(2k)(2k-2)cdots(2)} \
    &=0
    end{align}



    In the last step, you can think of the fraction as the infinite product of fractions less than $1$ ($frac{1}{2} timesfrac{3}{4} times frac{5}{6} timesfrac{7}{8}...$), which will decrease to $0$.






    share|cite|improve this answer



















    • 3




      Not all such products converge to zero. This one may but it isn't automatic just from factors less than $1.$ Actually the product is of terms $1-(1/(2k))$ and so since the sum of the subtracted part goes to infinity you do have product zero here.
      – coffeemath
      Nov 22 at 7:16















    up vote
    2
    down vote










    up vote
    2
    down vote









    begin{align}
    lim_{ktoinfty} frac{(2k)!}{2^{2k}cdot(k!)^2}
    &=lim_{ktoinfty} frac{(2k)!}{2^k cdot 2^k cdot k! cdot k!} \
    &=lim_{ktoinfty} frac{(2k)!}{(2^k cdot k!)^2} \
    &=lim_{ktoinfty} frac{(2k)(2k-1)cdots(2)(1)}{(2k)^2 (2k-2)^2 cdots (4)^2 (2)^2} \
    &=lim_{ktoinfty} frac{(2k-1)(2k-3)cdots(1)}{(2k)(2k-2)cdots(2)} \
    &=0
    end{align}



    In the last step, you can think of the fraction as the infinite product of fractions less than $1$ ($frac{1}{2} timesfrac{3}{4} times frac{5}{6} timesfrac{7}{8}...$), which will decrease to $0$.






    share|cite|improve this answer














    begin{align}
    lim_{ktoinfty} frac{(2k)!}{2^{2k}cdot(k!)^2}
    &=lim_{ktoinfty} frac{(2k)!}{2^k cdot 2^k cdot k! cdot k!} \
    &=lim_{ktoinfty} frac{(2k)!}{(2^k cdot k!)^2} \
    &=lim_{ktoinfty} frac{(2k)(2k-1)cdots(2)(1)}{(2k)^2 (2k-2)^2 cdots (4)^2 (2)^2} \
    &=lim_{ktoinfty} frac{(2k-1)(2k-3)cdots(1)}{(2k)(2k-2)cdots(2)} \
    &=0
    end{align}



    In the last step, you can think of the fraction as the infinite product of fractions less than $1$ ($frac{1}{2} timesfrac{3}{4} times frac{5}{6} timesfrac{7}{8}...$), which will decrease to $0$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 24 at 4:37









    Rócherz

    2,7412721




    2,7412721










    answered Nov 22 at 6:04









    Muchang Bahng

    562




    562








    • 3




      Not all such products converge to zero. This one may but it isn't automatic just from factors less than $1.$ Actually the product is of terms $1-(1/(2k))$ and so since the sum of the subtracted part goes to infinity you do have product zero here.
      – coffeemath
      Nov 22 at 7:16
















    • 3




      Not all such products converge to zero. This one may but it isn't automatic just from factors less than $1.$ Actually the product is of terms $1-(1/(2k))$ and so since the sum of the subtracted part goes to infinity you do have product zero here.
      – coffeemath
      Nov 22 at 7:16










    3




    3




    Not all such products converge to zero. This one may but it isn't automatic just from factors less than $1.$ Actually the product is of terms $1-(1/(2k))$ and so since the sum of the subtracted part goes to infinity you do have product zero here.
    – coffeemath
    Nov 22 at 7:16






    Not all such products converge to zero. This one may but it isn't automatic just from factors less than $1.$ Actually the product is of terms $1-(1/(2k))$ and so since the sum of the subtracted part goes to infinity you do have product zero here.
    – coffeemath
    Nov 22 at 7:16












    up vote
    0
    down vote













    This is the probability of having $n$ heads and $n$ tails in $2n$ times of a fair coin toss. This probability then simply goes to 0. You can also use the Stirling's approximation to attain this.






    share|cite|improve this answer

























      up vote
      0
      down vote













      This is the probability of having $n$ heads and $n$ tails in $2n$ times of a fair coin toss. This probability then simply goes to 0. You can also use the Stirling's approximation to attain this.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        This is the probability of having $n$ heads and $n$ tails in $2n$ times of a fair coin toss. This probability then simply goes to 0. You can also use the Stirling's approximation to attain this.






        share|cite|improve this answer












        This is the probability of having $n$ heads and $n$ tails in $2n$ times of a fair coin toss. This probability then simply goes to 0. You can also use the Stirling's approximation to attain this.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 22 at 6:39









        Mostafa Ayaz

        13.5k3836




        13.5k3836






























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