Find the limit of the given sequence as $\displaystyle n \to \infty$.
$\displaystyle \frac{(n!)^2}{(2n)!}$
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.
Find the limit of the given sequence as $\displaystyle n \to \infty$.
$\displaystyle \frac{(n!)^2}{(2n)!}$
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!
$\displaystyle (2n)!\neq 2!n!$
Let $\displaystyle a_n=\frac{(n!)^2}{(2n)!}$
Then apply the ratio test:
$\displaystyle \lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\in fty}\frac{\left((n+1)!\right)^2}{(2(n+1))!}\cdot\f rac{(2n)!}{(n!)^2}=\lim_{n\to\infty}\frac{\left[\left(n+1\right)n!\right]^2}{(2n+2)(2n+1)(2n)!}\cdot\frac{(2n)!}{(n!)^2}$ $\displaystyle =\lim_{n\to\infty}\frac{(n+1)^2}{(2n+2)(2n+1)}$
Can you take it from here?
Yes, I can take it from there. And I didn't mean to imply that
$\displaystyle (2n)! = 2!n!$
I actually didn't even know how to begin the problem, the only help I had received was that:
$\displaystyle (2n)! = 1(2)(3)...(n-2)(n-1)(n)...(2n) = n!(n+1)(n+2)...(2n)$
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.