Series Problem With Factorials

**lysserloo** · April 1st 2010, 05:13 PM

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:

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

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

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

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

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

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

But 0 is not the correct limit.

Any help would be very appreciated!

**Drexel28** · April 1st 2010, 05:16 PM

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:

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

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

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

Try cancelling!

**lysserloo** · April 1st 2010, 05:21 PM

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!

**Drexel28** · April 1st 2010, 05:26 PM

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:

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

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

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

$\frac{(2(n+1))!}{(n+1)!(n+4)!}\cdot\frac{n!(n+3)!} {(2n)!}=\frac{(2n+2)(2n+1)(2n)!}{(n+1)(n+4)n!(n+3) !}\cdot\frac{n!(n+3)!}{(2n)!}$ $=\frac{(2n+2)(2n+1)}{(n+1)(n+4)}\to 4$

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

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

However, when I cancel everything out, I get $\lim_{n \rightarrow \infty} \frac{2n}{4n(n+4)}$ 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!

**lysserloo** · April 1st 2010, 05:32 PM

Originally Posted by Drexel28

$\frac{(2(n+1))!}{(n+1)!(n+4)!}\cdot\frac{n!(n+3)!} {(2n)!}=\frac{(2n+2)(2n+1)(2n)!}{(n+1)(n+4)n!(n+3) !}\cdot\frac{n!(n+3)!}{(2n)!}$ $=\frac{(2n+2)(2n+1)}{(n+1)(n+4)}\to 4$

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.

Thank you for your reply though.

Edit: After looking at it, the only part I don't understand is after the first equals sign. Where did that n!(n+3)! on the bottom come from? The rest I understand.

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