# Thread: Series Problem With Factorials

1. ## Series Problem With Factorials

I have redone this problem many times, and can't seem to figure it out!

The problem is to use the ratio test to find the limit and convergence of this series:

$\displaystyle \sum_{i=1}^\infty \frac{(2n)!}{n!n+3)!}$

Now, using the ratio test, I come up with this:

$\displaystyle \lim_{n \rightarrow \infty} \frac{n!(n+3)!(2(n+1))!}{(2n)!(n+1)!(n+4)!}$

How do I expand this to cancel the series out? This is what I tried:

$\displaystyle \lim_{n \rightarrow \infty} \frac{(1*2...n)(1*2...(n+1)(n+2)(n+3)) (24*720...(2(n+1)) )}{(2*24*720...(2n)) (1*2...(n+1)) (1*2...(n+1)(n+2)(n+3)(n+4))}$

However, when I cancel everything out, I get $\displaystyle \lim_{n \rightarrow \infty} \frac{2n}{4n(n+4)}$ which results in the limit being 0.

But 0 is not the correct limit.

Any help would be very appreciated!

2. Originally Posted by lysserloo
I have redone this problem many times, and can't seem to figure it out!

The problem is to use the ratio test to find the limit and convergence of this series:

$\displaystyle \sum_{i=1}^\infty \frac{(2n)!}{n!n+3)!}$

Now, using the ratio test, I come up with this:

$\displaystyle \lim_{n \rightarrow \infty} \frac{n!(n+3)!(2(n+1))!}{(2n)!(n+1)!(n+4)!}$
Try cancelling!

3. Haha yes I know. I didn't mean to post when I did, I wasn't quite finished typing out my problem. Thank you for your quick reply though!

4. Originally Posted by lysserloo
I have redone this problem many times, and can't seem to figure it out!

The problem is to use the ratio test to find the limit and convergence of this series:

$\displaystyle \sum_{i=1}^\infty \frac{(2n)!}{n!n+3)!}$

Now, using the ratio test, I come up with this:

$\displaystyle \lim_{n \rightarrow \infty} \frac{n!(n+3)!(2(n+1))!}{(2n)!(n+1)!(n+4)!}$
$\displaystyle \frac{(2(n+1))!}{(n+1)!(n+4)!}\cdot\frac{n!(n+3)!} {(2n)!}=\frac{(2n+2)(2n+1)(2n)!}{(n+1)(n+4)n!(n+3) !}\cdot\frac{n!(n+3)!}{(2n)!}$$\displaystyle =\frac{(2n+2)(2n+1)}{(n+1)(n+4)}\to 4 How do I expand this to cancel the series out? This is what I tried: \displaystyle \lim_{n \rightarrow \infty} \frac{(1*2...n)(1*2...(n+1)(n+2)(n+3)) (24*720...(2(n+1)) )}{(2*24*720...(2n)) (1*2...(n+1)) (1*2...(n+1)(n+2)(n+3)(n+4))} However, when I cancel everything out, I get \displaystyle \lim_{n \rightarrow \infty} \frac{2n}{4n(n+4)} which results in the limit being 0. But 0 is not the correct limit. Any help would be very appreciated! Originally Posted by lysserloo Haha yes I know. I didn't mean to post when I did, I wasn't quite finished typing out my problem. Thank you for your quick reply though! 5. Originally Posted by Drexel28 \displaystyle \frac{(2(n+1))!}{(n+1)!(n+4)!}\cdot\frac{n!(n+3)!} {(2n)!}=\frac{(2n+2)(2n+1)(2n)!}{(n+1)(n+4)n!(n+3) !}\cdot\frac{n!(n+3)!}{(2n)!}$$\displaystyle =\frac{(2n+2)(2n+1)}{(n+1)(n+4)}\to 4$
I do not follow what you did here at all : (

I think the problem is that I really don't understand factorials very well...I'm not sure.