limit of a sequence of binomialcoefficents

[Gobard]

Hallo, I am writing a paper on voting and I am interested in the limit of the following sequence that is defined for all even integers: (if you want you can multiply n by 2 and have it as a sequence for all integers)

S_n=binom(n,n/2)/2^n

where binom(n,n/2)=n*(n-1)*..*(n/2+1)/((n/2)!)

L'Hopital rule does not help I feel. According to the computer, this sequence seemingly converges very slowly to 0 but the numbers get too big at some point to handle for the computer. Anyways, I need an analytical result.

Can someone help me?

 

[pka]

I am interested in the limit of the following sequence that is defined for all even integers: (if you want you can multiply n by 2 and have it as a sequence for all integers)
S_n=binom(n,n/2)/2^n
where binom(n,n/2)=n*(n-1)*..*(n/2+1)/((n/2)!)

I don't really follow what you are asking.
Do you have that formula written correctly?
It should be: \(\displaystyle \dbinom{2n}{n}=\dfrac{(2n)!}{(n!)^2}\)

It is well know that \(\displaystyle (2n)!=2^n(n!)\left[(2n-1)(2n-3)\cdots3\cdot1\right]\).

 

[Gobard]

Alright, thanks.

Again, lets define S_n=binom(2n,n)/2^(2n). Then, as you said, this is just (2n)!/[(n!)^2 * 2^(2n)], and according to your post, this is equal to

[2^n * n! *(2n-1)*..*3]/[(n!)^2 * 2^(2n)]=[(2n-1)*..*3]/[n! * 2^n].

How do I go on from here?

 

[pka]

Again, lets define S_n=binom(2n,n)/2^(2n). Then, as you said, this is just (2n)!/[(n!)^2 * 2^(2n)], and according to your post, this is equal to
[2^n * n! *(2n-1)*..*3]/[(n!)^2 * 2^(2n)]=[(2n-1)*..*3]/[n! * 2^n].

That sequence does converge to zero (very slowly).
The proof is tedious. See the above.

 

[Gobard]

wow, great, I think I could not have figured that out. Thanks a lot.

edit: I feel kind of stupid but how do you proof that [(2n+1)/(2n+2)]^n does converge to zero?

 

[Gobard]

you are sure that this last sequence converges to zero? The first 100000 elements stay above 0.6!

 

[pka]

Take a look at the webpage.
That is a very tedious process as you can see.

In a way, I am sorry to have drawn this out.

However \(\displaystyle \displaystyle\lim _{n \to \infty } \left( {\frac{{2n + 1}}{{2n + 2}}} \right)^n = \frac{1}{{\sqrt e }}\)