1. ## limt with factorial

find the limit of $\frac{(2n - 1)!}{(2n + 1)!}$ as n approaches infinity.

L'hop wont work b.c you cant take the derivative of a factorial. I wrote out the first three terms and they appear to be decreasing and approaching zero.

2. Hi,

We have $(2n+1)!=(2n+1)\times(2n)\times (2n-1)!$ so
$\frac{(2n-1)!}{(2n+1)!}=\frac{(2n-1)!}{(2n+1)\times(2n)\times (2n-1)!}=\frac{1}{(2n+1)\times(2n)}$

Now you can compute the limit, can't you?

3. Originally Posted by diroga
find the limit of $\frac{(2n - 1)!}{(2n + 1)!}$ as n approaches infinity.

L'hop wont work b.c you cant take the derivative of a factorial. I wrote out the first three terms and they appear to be decreasing and approaching zero.

$lim_{n\rightarrow\infty} \frac{(2n-1)!}{(2n+1)!}=lim_{n\rightarrow\infty} \frac{(2n-1)!}{(2n+1)(2n)(2n-1)!}$ $=\frac{{\color{red}(2n-1)!}}{(2n+1)(2n){\color{red}(2n-1)!}}=lim_{n\rightarrow\infty}\frac{1}{(2n+1)(2n)} =0$

4. I dont see how you could do this
$
(2n+1)!=(2n+1)\times(2n)\times (2n-1)!
$

I guess when it comes to limits you got be good with algebra and property tricks

5. $0!=1$

$1!=1$

$2!=2\cdot 1$

$3!=3\cdot 2 \cdot 1= 3 \cdot 2!$

$n!=n \cdot (n-1) \cdot (n-2) \cdot ... \cdot 3 \cdot 2 \cdot 1$

$(2n+1)!=(2n+1) \cdot (2n) \cdot (2n - 1) \cdot ... \cdot 1 = (2n+1) \cdot (2n) \cdot (2n-1)!$

6. Originally Posted by diroga
I dont see how you could do this
$
(2n+1)!=(2n+1)\times(2n)\times (2n-1)!
$

I guess when it comes to limits you got be good with algebra and property tricks
Or learn the definitions?

7. Originally Posted by Ruun
$0!=1$

$1!=1$

$2!=2\cdot 1$

$3!=3\cdot 2 \cdot 1= 3 \cdot 2!$

$n!=n \cdot (n-1) \cdot (n-2) \cdot ... \cdot 3 \cdot 2 \cdot 1$

$(2n+1)!=(2n+1) \cdot (2n) \cdot (2n - 1) \cdot ... \cdot 1 = (2n+1) \cdot (2n) \cdot (2n-1)!$
thanks

Originally Posted by HallsofIvy
Or learn the definitions?