1. ## Infinite Series

Find the limit of the given sequence as $n \to \infty$.

$\frac{(n!)^2}{(2n)!}$

Now, I know that in the both the numerator and the denominator, a factor n! can be pulled out... From there, I have no idea where to go.

2. Originally Posted by Aryth
Find the limit of the given sequence as $n \to \infty$.

$\frac{(n!)^2}{(2n)!}$

Now, I know that in the both the numerator and the denominator, a factor n! can be pulled out... From there, I have no idea where to go.
No, you can't!

$(2n)!\neq 2!n!$

Let $a_n=\frac{(n!)^2}{(2n)!}$

Then apply the ratio test:

$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\in fty}\frac{\left((n+1)!\right)^2}{(2(n+1))!}\cdot\f rac{(2n)!}{(n!)^2}=\lim_{n\to\infty}\frac{\left[\left(n+1\right)n!\right]^2}{(2n+2)(2n+1)(2n)!}\cdot\frac{(2n)!}{(n!)^2}$ $=\lim_{n\to\infty}\frac{(n+1)^2}{(2n+2)(2n+1)}$

Can you take it from here?

3. Yes, I can take it from there. And I didn't mean to imply that

$(2n)! = 2!n!$

I actually didn't even know how to begin the problem, the only help I had received was that:

$(2n)! = 1(2)(3)...(n-2)(n-1)(n)...(2n) = n!(n+1)(n+2)...(2n)$

I didn't know what to do with it... And I'm glad I don't need it, what you helped me with makes much more sense.

Thanks.

4. Use Stirling's approximation.

for large n $\frac{(n!)^2}{(2n)!} \approx \frac{(n^{n}e^{-n})^{2}}{(2n)^{2n}e^{-2n}} = \frac{n^{2n}e^{-2n}}{2^{2n}n^{2n}e^{-2n}} = \frac{1}{2^{2n}}$

$\lim_{n \to \infty} \frac{1}{2^{2n}} = 0$