# Convergence of series

1. Oct 16, 2005

### thenewbosco

How do i show the following series from n=1 to infinity converges?
$$\sum\frac{(n!)^2}{(2n)!}$$
what i did was apply the ratio test so i ended up with
the limit as n--> infinity of
$$\frac{((n+1)^2)(n!)^2}{(2(n+1))!(2n)!}(\frac{((2n)!}{(n!)^2})$$

then after the cancelation of the factorial terms, this limit goes to infinity...
however this series converges by the answer i have for this problem. where did i go wrong?

thanks

Last edited: Oct 16, 2005
2. Oct 16, 2005

### AKG

For one, it appears you have an extra (2n)! in your denominator. You also appear to have an extra bracket in the top-right area, not that it matters. But I got that the limit goes to 1/4, so the series converges. I don't know how you got infinity. The extra (2n)! in the denominator should have made it converge even "more", so I'm not sure what kind of cancelling you did.

You can also compare this to $\sum _{n = 1} ^{\infty} 2^{-n}$ using induction to see that it converges. Note that (n!)²/(2n)! = ${{2n}\choose{n}} ^{-1}$.

Last edited: Oct 16, 2005
3. Oct 16, 2005

### Tom Mattson

Staff Emeritus
Ignore my initial response, I misread the problem.

You went wrong rewriting $(2(n+1))!$ as $2(n+1))!(2n)!$.

Go back to the definition of the factorial function. You will see that:

$(2(n+1))!=(2n+2)!=(2n+2)(2n+1)(2n)!$

4. Oct 16, 2005

### Hurkyl

Staff Emeritus
Your problem is probably that you tried to do several steps at once and messed up: try just writing the ratio first, then simplifying.