# Thread: Ratio test with factorials

1. ## Ratio test with factorials

This problem is tripping me up:

$\displaystyle \sum_{n=1}^{\infty} \frac {(n!)^2}{(2n)!}$

I'm trying to use the ratio test to see if it converges/diverges and am getting:

$\displaystyle \lim_{n \rightarrow\ \infty} \frac{((n+1)!)^2}{(2(n+1))!}\times\frac{(2n)!}{(n! )^2}$

I'm assuming the work is correct so far, and that the ratio test is the right test to use, but I really have no idea where to go from here. I'd like to be able to cancel some stuff but I think I'm lacking a good understanding of how these factorials relate to each other and how to work with them squared. Any help appreciated.

2. Originally Posted by Shananay
This problem is tripping me up:

$\displaystyle \sum_{n=1}^{\infty} \frac {(n!)^2}{(2n)!}$

I'm trying to use the ratio test to see if it converges/diverges and am getting:

$\displaystyle \lim_{n \rightarrow\ \infty} \frac{((n+1)!)^2}{(2(n+1))!}\times\frac{(2n)!}{(n! )^2}$

I'm assuming the work is correct so far, and that the ratio test is the right test to use, but I really have no idea where to go from here. I'd like to be able to cancel some stuff but I think I'm lacking a good understanding of how these factorials relate to each other and how to work with them squared. Any help appreciated.
$\displaystyle n!=n(n-1)(n-2)...(3)(2)(1)$
$\displaystyle 5!=(5)(4)(3)(2)(1)=5(4!)$
$\displaystyle ((n+1)!)^2=((n+1)n!)^2=(n+1)^2(n!)^2$.
$\displaystyle (2(n+1))!=(2n+2)!=(2n+2)(2n+1)(2n)!$.
Now, You cancel them out.