# limt with factorial

• Jun 9th 2009, 10:34 AM
diroga
limt with factorial
find the limit of $\displaystyle \frac{(2n - 1)!}{(2n + 1)!}$ 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.
• Jun 9th 2009, 10:53 AM
flyingsquirrel
Hi,

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

Now you can compute the limit, can't you?
• Jun 9th 2009, 11:01 AM
Amer
Quote:

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

$\displaystyle lim_{n\rightarrow\infty} \frac{(2n-1)!}{(2n+1)!}=lim_{n\rightarrow\infty} \frac{(2n-1)!}{(2n+1)(2n)(2n-1)!}$$\displaystyle =\frac{{\color{red}(2n-1)!}}{(2n+1)(2n){\color{red}(2n-1)!}}=lim_{n\rightarrow\infty}\frac{1}{(2n+1)(2n)} =0$
• Jun 9th 2009, 02:06 PM
diroga
I dont see how you could do this
$\displaystyle (2n+1)!=(2n+1)\times(2n)\times (2n-1)!$

I guess when it comes to limits you got be good with algebra and property tricks
• Jun 9th 2009, 02:14 PM
Ruun
$\displaystyle 0!=1$

$\displaystyle 1!=1$

$\displaystyle 2!=2\cdot 1$

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

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

$\displaystyle (2n+1)!=(2n+1) \cdot (2n) \cdot (2n - 1) \cdot ... \cdot 1 = (2n+1) \cdot (2n) \cdot (2n-1)!$
• Jun 9th 2009, 02:23 PM
HallsofIvy
Quote:

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

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

Or learn the definitions?
• Jun 9th 2009, 08:39 PM
diroga
Quote:

Originally Posted by Ruun
$\displaystyle 0!=1$

$\displaystyle 1!=1$

$\displaystyle 2!=2\cdot 1$

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

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

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

thanks

Quote:

Originally Posted by HallsofIvy
Or learn the definitions?

(Crying)