What is the asymptotic order of $sum_{k=0}^n {nchoose k}^2$?












3












$begingroup$


What is the asymptotic order of $sum_{k=0}^n {nchoose k}^2$? That is, find $g(n)$ such that $$lim_{nto infty}frac{sum_{k=0}^n {nchoose k}^2}{g(n)}=1$$



We can expand the binomial coefficient and use Stirling's approximation but I can not determine g(n).






probability asymptotics







share|cite|improve this question









$endgroup$








  • 6




    $begingroup$
    Were you able to simplify the summation to $binom{2n}{n}$?
    $endgroup$
    – JimmyK4542
    Dec 5 '18 at 2:27










  • $begingroup$
    From here: math.stackexchange.com/questions/3025663/… you should now be able to solve it much easier
    $endgroup$
    – Henry Lee
    Dec 5 '18 at 2:28










  • $begingroup$
    I see, thank you. So $g(n)=frac{4^n}{sqrt{pi n}}$
    $endgroup$
    – Probability student
    Dec 5 '18 at 2:37






  • 1




    $begingroup$
    What do we get if we have $sum_{k=0}^n{n choose k}{n+1 choose k}$?
    $endgroup$
    – Probability student
    Dec 5 '18 at 2:44










  • $begingroup$
    See here math.stackexchange.com/questions/74651/…
    $endgroup$
    – leonbloy
    Dec 5 '18 at 3:27
















3












$begingroup$


What is the asymptotic order of $sum_{k=0}^n {nchoose k}^2$? That is, find $g(n)$ such that $$lim_{nto infty}frac{sum_{k=0}^n {nchoose k}^2}{g(n)}=1$$



We can expand the binomial coefficient and use Stirling's approximation but I can not determine g(n).






probability asymptotics







share|cite|improve this question









$endgroup$








  • 6




    $begingroup$
    Were you able to simplify the summation to $binom{2n}{n}$?
    $endgroup$
    – JimmyK4542
    Dec 5 '18 at 2:27










  • $begingroup$
    From here: math.stackexchange.com/questions/3025663/… you should now be able to solve it much easier
    $endgroup$
    – Henry Lee
    Dec 5 '18 at 2:28










  • $begingroup$
    I see, thank you. So $g(n)=frac{4^n}{sqrt{pi n}}$
    $endgroup$
    – Probability student
    Dec 5 '18 at 2:37






  • 1




    $begingroup$
    What do we get if we have $sum_{k=0}^n{n choose k}{n+1 choose k}$?
    $endgroup$
    – Probability student
    Dec 5 '18 at 2:44










  • $begingroup$
    See here math.stackexchange.com/questions/74651/…
    $endgroup$
    – leonbloy
    Dec 5 '18 at 3:27














3












3








3


1



$begingroup$


What is the asymptotic order of $sum_{k=0}^n {nchoose k}^2$? That is, find $g(n)$ such that $$lim_{nto infty}frac{sum_{k=0}^n {nchoose k}^2}{g(n)}=1$$



We can expand the binomial coefficient and use Stirling's approximation but I can not determine g(n).






probability asymptotics







share|cite|improve this question









$endgroup$




What is the asymptotic order of $sum_{k=0}^n {nchoose k}^2$? That is, find $g(n)$ such that $$lim_{nto infty}frac{sum_{k=0}^n {nchoose k}^2}{g(n)}=1$$



We can expand the binomial coefficient and use Stirling's approximation but I can not determine g(n).






probability asymptotics




probability asymptotics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 5 '18 at 2:17









Probability studentProbability student

213




213








  • 6




    $begingroup$
    Were you able to simplify the summation to $binom{2n}{n}$?
    $endgroup$
    – JimmyK4542
    Dec 5 '18 at 2:27










  • $begingroup$
    From here: math.stackexchange.com/questions/3025663/… you should now be able to solve it much easier
    $endgroup$
    – Henry Lee
    Dec 5 '18 at 2:28










  • $begingroup$
    I see, thank you. So $g(n)=frac{4^n}{sqrt{pi n}}$
    $endgroup$
    – Probability student
    Dec 5 '18 at 2:37






  • 1




    $begingroup$
    What do we get if we have $sum_{k=0}^n{n choose k}{n+1 choose k}$?
    $endgroup$
    – Probability student
    Dec 5 '18 at 2:44










  • $begingroup$
    See here math.stackexchange.com/questions/74651/…
    $endgroup$
    – leonbloy
    Dec 5 '18 at 3:27














  • 6




    $begingroup$
    Were you able to simplify the summation to $binom{2n}{n}$?
    $endgroup$
    – JimmyK4542
    Dec 5 '18 at 2:27










  • $begingroup$
    From here: math.stackexchange.com/questions/3025663/… you should now be able to solve it much easier
    $endgroup$
    – Henry Lee
    Dec 5 '18 at 2:28










  • $begingroup$
    I see, thank you. So $g(n)=frac{4^n}{sqrt{pi n}}$
    $endgroup$
    – Probability student
    Dec 5 '18 at 2:37






  • 1




    $begingroup$
    What do we get if we have $sum_{k=0}^n{n choose k}{n+1 choose k}$?
    $endgroup$
    – Probability student
    Dec 5 '18 at 2:44










  • $begingroup$
    See here math.stackexchange.com/questions/74651/…
    $endgroup$
    – leonbloy
    Dec 5 '18 at 3:27








6




6




$begingroup$
Were you able to simplify the summation to $binom{2n}{n}$?
$endgroup$
– JimmyK4542
Dec 5 '18 at 2:27




$begingroup$
Were you able to simplify the summation to $binom{2n}{n}$?
$endgroup$
– JimmyK4542
Dec 5 '18 at 2:27












$begingroup$
From here: math.stackexchange.com/questions/3025663/… you should now be able to solve it much easier
$endgroup$
– Henry Lee
Dec 5 '18 at 2:28




$begingroup$
From here: math.stackexchange.com/questions/3025663/… you should now be able to solve it much easier
$endgroup$
– Henry Lee
Dec 5 '18 at 2:28












$begingroup$
I see, thank you. So $g(n)=frac{4^n}{sqrt{pi n}}$
$endgroup$
– Probability student
Dec 5 '18 at 2:37




$begingroup$
I see, thank you. So $g(n)=frac{4^n}{sqrt{pi n}}$
$endgroup$
– Probability student
Dec 5 '18 at 2:37




1




1




$begingroup$
What do we get if we have $sum_{k=0}^n{n choose k}{n+1 choose k}$?
$endgroup$
– Probability student
Dec 5 '18 at 2:44




$begingroup$
What do we get if we have $sum_{k=0}^n{n choose k}{n+1 choose k}$?
$endgroup$
– Probability student
Dec 5 '18 at 2:44












$begingroup$
See here math.stackexchange.com/questions/74651/…
$endgroup$
– leonbloy
Dec 5 '18 at 3:27




$begingroup$
See here math.stackexchange.com/questions/74651/…
$endgroup$
– leonbloy
Dec 5 '18 at 3:27










3 Answers
3






active

oldest

votes


















1












$begingroup$

$$sum_{k=0}^n binom{n}{k}binom{n+m}{k} = sum_{k=0}^n binom{n}{n-k}binom{n+m}{k}.$$



This counts the number of ways to choose $n$ objects from a set of $n+2m$. To see this, let there be $n$ red ones and $n+m$ blue ones, and do casework on how many red and how many blue we choose. So:



$$sum_{k=0}^n binom{n}{k}binom{n+m}{k} = binom{2n+m}{n}.$$



I think, using Stirling's formula, you can probably do the rest.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks Carl. I think it should be $${2n+m choose n}$$
    $endgroup$
    – Probability student
    Dec 5 '18 at 3:19










  • $begingroup$
    @Probabilitystudent Thanks -- I've fixed it.
    $endgroup$
    – Carl Schildkraut
    Dec 5 '18 at 3:47



















0












$begingroup$

This is a comment regarding
the comment of Probability student:



According to Wolfy,



$sum_{k=0}^n({n choose k}{n+1choose k})
= dfrac{2^{2n+1}(n + 1/2)!}{sqrt{π} (n + 1)!}
$



$sum_{k=0}^n({n choose k}{n+2choose k})
= dfrac{2^{2n + 2} (n + 1) (n + 1/2)!}{sqrt{π} (n + 2)!}
$



Looks like it might be possible
to get a closed form for
$sum_{k=0}^n({n choose k}{n+mchoose k})$.



And Wolfy says that
$sum_{k=0}^n({n choose k}{n+mchoose k})
=dfrac{(m + 2 n)!}{n! (m + n)!}
$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    thank you but what is Wolfy? How do you get the first result, for instance? Thanks!
    $endgroup$
    – Probability student
    Dec 5 '18 at 3:04










  • $begingroup$
    "Wolfy" is my pet term for Wolfram Alpha. I often throw expressions at it to see what happens.
    $endgroup$
    – marty cohen
    Dec 5 '18 at 3:09



















0












$begingroup$

If you use marty cohen's answer, taking the logarithm and then using Stirling approximation, you should get something like
$$log left(frac{(m+2 n)!}{n! (m+n)!}right)=2 n log (2)+log left(frac{2^m}{sqrt{pi n}}right)-frac{2 m^2+2 m+1}{8
n}+Oleft(frac{1}{n^2}right)$$ and then, for large values of $n$,
$$frac{(m+2 n)!}{n! (m+n)!} sim frac{2^{2n+m}} {sqrt{pi n}}$$ What is interesting (at least to me) is the next term which shows how is approached the asymptotics.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! I think it should be $$sim frac{2^{2n+m}e^{m}}{sqrt{pi n}}$$
    $endgroup$
    – Probability student
    Dec 5 '18 at 10:24













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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

$$sum_{k=0}^n binom{n}{k}binom{n+m}{k} = sum_{k=0}^n binom{n}{n-k}binom{n+m}{k}.$$



This counts the number of ways to choose $n$ objects from a set of $n+2m$. To see this, let there be $n$ red ones and $n+m$ blue ones, and do casework on how many red and how many blue we choose. So:



$$sum_{k=0}^n binom{n}{k}binom{n+m}{k} = binom{2n+m}{n}.$$



I think, using Stirling's formula, you can probably do the rest.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks Carl. I think it should be $${2n+m choose n}$$
    $endgroup$
    – Probability student
    Dec 5 '18 at 3:19










  • $begingroup$
    @Probabilitystudent Thanks -- I've fixed it.
    $endgroup$
    – Carl Schildkraut
    Dec 5 '18 at 3:47
















1












$begingroup$

$$sum_{k=0}^n binom{n}{k}binom{n+m}{k} = sum_{k=0}^n binom{n}{n-k}binom{n+m}{k}.$$



This counts the number of ways to choose $n$ objects from a set of $n+2m$. To see this, let there be $n$ red ones and $n+m$ blue ones, and do casework on how many red and how many blue we choose. So:



$$sum_{k=0}^n binom{n}{k}binom{n+m}{k} = binom{2n+m}{n}.$$



I think, using Stirling's formula, you can probably do the rest.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks Carl. I think it should be $${2n+m choose n}$$
    $endgroup$
    – Probability student
    Dec 5 '18 at 3:19










  • $begingroup$
    @Probabilitystudent Thanks -- I've fixed it.
    $endgroup$
    – Carl Schildkraut
    Dec 5 '18 at 3:47














1












1








1





$begingroup$

$$sum_{k=0}^n binom{n}{k}binom{n+m}{k} = sum_{k=0}^n binom{n}{n-k}binom{n+m}{k}.$$



This counts the number of ways to choose $n$ objects from a set of $n+2m$. To see this, let there be $n$ red ones and $n+m$ blue ones, and do casework on how many red and how many blue we choose. So:



$$sum_{k=0}^n binom{n}{k}binom{n+m}{k} = binom{2n+m}{n}.$$



I think, using Stirling's formula, you can probably do the rest.






share|cite|improve this answer











$endgroup$



$$sum_{k=0}^n binom{n}{k}binom{n+m}{k} = sum_{k=0}^n binom{n}{n-k}binom{n+m}{k}.$$



This counts the number of ways to choose $n$ objects from a set of $n+2m$. To see this, let there be $n$ red ones and $n+m$ blue ones, and do casework on how many red and how many blue we choose. So:



$$sum_{k=0}^n binom{n}{k}binom{n+m}{k} = binom{2n+m}{n}.$$



I think, using Stirling's formula, you can probably do the rest.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 5 '18 at 3:47

























answered Dec 5 '18 at 3:14









Carl SchildkrautCarl Schildkraut

11.2k11441




11.2k11441












  • $begingroup$
    Thanks Carl. I think it should be $${2n+m choose n}$$
    $endgroup$
    – Probability student
    Dec 5 '18 at 3:19










  • $begingroup$
    @Probabilitystudent Thanks -- I've fixed it.
    $endgroup$
    – Carl Schildkraut
    Dec 5 '18 at 3:47


















  • $begingroup$
    Thanks Carl. I think it should be $${2n+m choose n}$$
    $endgroup$
    – Probability student
    Dec 5 '18 at 3:19










  • $begingroup$
    @Probabilitystudent Thanks -- I've fixed it.
    $endgroup$
    – Carl Schildkraut
    Dec 5 '18 at 3:47
















$begingroup$
Thanks Carl. I think it should be $${2n+m choose n}$$
$endgroup$
– Probability student
Dec 5 '18 at 3:19




$begingroup$
Thanks Carl. I think it should be $${2n+m choose n}$$
$endgroup$
– Probability student
Dec 5 '18 at 3:19












$begingroup$
@Probabilitystudent Thanks -- I've fixed it.
$endgroup$
– Carl Schildkraut
Dec 5 '18 at 3:47




$begingroup$
@Probabilitystudent Thanks -- I've fixed it.
$endgroup$
– Carl Schildkraut
Dec 5 '18 at 3:47











0












$begingroup$

This is a comment regarding
the comment of Probability student:



According to Wolfy,



$sum_{k=0}^n({n choose k}{n+1choose k})
= dfrac{2^{2n+1}(n + 1/2)!}{sqrt{π} (n + 1)!}
$



$sum_{k=0}^n({n choose k}{n+2choose k})
= dfrac{2^{2n + 2} (n + 1) (n + 1/2)!}{sqrt{π} (n + 2)!}
$



Looks like it might be possible
to get a closed form for
$sum_{k=0}^n({n choose k}{n+mchoose k})$.



And Wolfy says that
$sum_{k=0}^n({n choose k}{n+mchoose k})
=dfrac{(m + 2 n)!}{n! (m + n)!}
$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    thank you but what is Wolfy? How do you get the first result, for instance? Thanks!
    $endgroup$
    – Probability student
    Dec 5 '18 at 3:04










  • $begingroup$
    "Wolfy" is my pet term for Wolfram Alpha. I often throw expressions at it to see what happens.
    $endgroup$
    – marty cohen
    Dec 5 '18 at 3:09
















0












$begingroup$

This is a comment regarding
the comment of Probability student:



According to Wolfy,



$sum_{k=0}^n({n choose k}{n+1choose k})
= dfrac{2^{2n+1}(n + 1/2)!}{sqrt{π} (n + 1)!}
$



$sum_{k=0}^n({n choose k}{n+2choose k})
= dfrac{2^{2n + 2} (n + 1) (n + 1/2)!}{sqrt{π} (n + 2)!}
$



Looks like it might be possible
to get a closed form for
$sum_{k=0}^n({n choose k}{n+mchoose k})$.



And Wolfy says that
$sum_{k=0}^n({n choose k}{n+mchoose k})
=dfrac{(m + 2 n)!}{n! (m + n)!}
$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    thank you but what is Wolfy? How do you get the first result, for instance? Thanks!
    $endgroup$
    – Probability student
    Dec 5 '18 at 3:04










  • $begingroup$
    "Wolfy" is my pet term for Wolfram Alpha. I often throw expressions at it to see what happens.
    $endgroup$
    – marty cohen
    Dec 5 '18 at 3:09














0












0








0





$begingroup$

This is a comment regarding
the comment of Probability student:



According to Wolfy,



$sum_{k=0}^n({n choose k}{n+1choose k})
= dfrac{2^{2n+1}(n + 1/2)!}{sqrt{π} (n + 1)!}
$



$sum_{k=0}^n({n choose k}{n+2choose k})
= dfrac{2^{2n + 2} (n + 1) (n + 1/2)!}{sqrt{π} (n + 2)!}
$



Looks like it might be possible
to get a closed form for
$sum_{k=0}^n({n choose k}{n+mchoose k})$.



And Wolfy says that
$sum_{k=0}^n({n choose k}{n+mchoose k})
=dfrac{(m + 2 n)!}{n! (m + n)!}
$.






share|cite|improve this answer









$endgroup$



This is a comment regarding
the comment of Probability student:



According to Wolfy,



$sum_{k=0}^n({n choose k}{n+1choose k})
= dfrac{2^{2n+1}(n + 1/2)!}{sqrt{π} (n + 1)!}
$



$sum_{k=0}^n({n choose k}{n+2choose k})
= dfrac{2^{2n + 2} (n + 1) (n + 1/2)!}{sqrt{π} (n + 2)!}
$



Looks like it might be possible
to get a closed form for
$sum_{k=0}^n({n choose k}{n+mchoose k})$.



And Wolfy says that
$sum_{k=0}^n({n choose k}{n+mchoose k})
=dfrac{(m + 2 n)!}{n! (m + n)!}
$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 3:00









marty cohenmarty cohen

73.2k549128




73.2k549128








  • 1




    $begingroup$
    thank you but what is Wolfy? How do you get the first result, for instance? Thanks!
    $endgroup$
    – Probability student
    Dec 5 '18 at 3:04










  • $begingroup$
    "Wolfy" is my pet term for Wolfram Alpha. I often throw expressions at it to see what happens.
    $endgroup$
    – marty cohen
    Dec 5 '18 at 3:09














  • 1




    $begingroup$
    thank you but what is Wolfy? How do you get the first result, for instance? Thanks!
    $endgroup$
    – Probability student
    Dec 5 '18 at 3:04










  • $begingroup$
    "Wolfy" is my pet term for Wolfram Alpha. I often throw expressions at it to see what happens.
    $endgroup$
    – marty cohen
    Dec 5 '18 at 3:09








1




1




$begingroup$
thank you but what is Wolfy? How do you get the first result, for instance? Thanks!
$endgroup$
– Probability student
Dec 5 '18 at 3:04




$begingroup$
thank you but what is Wolfy? How do you get the first result, for instance? Thanks!
$endgroup$
– Probability student
Dec 5 '18 at 3:04












$begingroup$
"Wolfy" is my pet term for Wolfram Alpha. I often throw expressions at it to see what happens.
$endgroup$
– marty cohen
Dec 5 '18 at 3:09




$begingroup$
"Wolfy" is my pet term for Wolfram Alpha. I often throw expressions at it to see what happens.
$endgroup$
– marty cohen
Dec 5 '18 at 3:09











0












$begingroup$

If you use marty cohen's answer, taking the logarithm and then using Stirling approximation, you should get something like
$$log left(frac{(m+2 n)!}{n! (m+n)!}right)=2 n log (2)+log left(frac{2^m}{sqrt{pi n}}right)-frac{2 m^2+2 m+1}{8
n}+Oleft(frac{1}{n^2}right)$$ and then, for large values of $n$,
$$frac{(m+2 n)!}{n! (m+n)!} sim frac{2^{2n+m}} {sqrt{pi n}}$$ What is interesting (at least to me) is the next term which shows how is approached the asymptotics.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! I think it should be $$sim frac{2^{2n+m}e^{m}}{sqrt{pi n}}$$
    $endgroup$
    – Probability student
    Dec 5 '18 at 10:24


















0












$begingroup$

If you use marty cohen's answer, taking the logarithm and then using Stirling approximation, you should get something like
$$log left(frac{(m+2 n)!}{n! (m+n)!}right)=2 n log (2)+log left(frac{2^m}{sqrt{pi n}}right)-frac{2 m^2+2 m+1}{8
n}+Oleft(frac{1}{n^2}right)$$ and then, for large values of $n$,
$$frac{(m+2 n)!}{n! (m+n)!} sim frac{2^{2n+m}} {sqrt{pi n}}$$ What is interesting (at least to me) is the next term which shows how is approached the asymptotics.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! I think it should be $$sim frac{2^{2n+m}e^{m}}{sqrt{pi n}}$$
    $endgroup$
    – Probability student
    Dec 5 '18 at 10:24
















0












0








0





$begingroup$

If you use marty cohen's answer, taking the logarithm and then using Stirling approximation, you should get something like
$$log left(frac{(m+2 n)!}{n! (m+n)!}right)=2 n log (2)+log left(frac{2^m}{sqrt{pi n}}right)-frac{2 m^2+2 m+1}{8
n}+Oleft(frac{1}{n^2}right)$$ and then, for large values of $n$,
$$frac{(m+2 n)!}{n! (m+n)!} sim frac{2^{2n+m}} {sqrt{pi n}}$$ What is interesting (at least to me) is the next term which shows how is approached the asymptotics.






share|cite|improve this answer









$endgroup$



If you use marty cohen's answer, taking the logarithm and then using Stirling approximation, you should get something like
$$log left(frac{(m+2 n)!}{n! (m+n)!}right)=2 n log (2)+log left(frac{2^m}{sqrt{pi n}}right)-frac{2 m^2+2 m+1}{8
n}+Oleft(frac{1}{n^2}right)$$ and then, for large values of $n$,
$$frac{(m+2 n)!}{n! (m+n)!} sim frac{2^{2n+m}} {sqrt{pi n}}$$ What is interesting (at least to me) is the next term which shows how is approached the asymptotics.







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answered Dec 5 '18 at 4:20









Claude LeiboviciClaude Leibovici

120k1157132




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  • $begingroup$
    Thanks! I think it should be $$sim frac{2^{2n+m}e^{m}}{sqrt{pi n}}$$
    $endgroup$
    – Probability student
    Dec 5 '18 at 10:24




















  • $begingroup$
    Thanks! I think it should be $$sim frac{2^{2n+m}e^{m}}{sqrt{pi n}}$$
    $endgroup$
    – Probability student
    Dec 5 '18 at 10:24


















$begingroup$
Thanks! I think it should be $$sim frac{2^{2n+m}e^{m}}{sqrt{pi n}}$$
$endgroup$
– Probability student
Dec 5 '18 at 10:24






$begingroup$
Thanks! I think it should be $$sim frac{2^{2n+m}e^{m}}{sqrt{pi n}}$$
$endgroup$
– Probability student
Dec 5 '18 at 10:24




















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