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.

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- Jun 9th 2009, 10:34 AMdirogalimt 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 AMflyingsquirrel
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 AMAmer
- Jun 9th 2009, 02:06 PMdiroga
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 PMRuun
$\displaystyle 0!=1$

$\displaystyle 1!=1$

$\displaystyle 2!=2\cdot 1$

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

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

$\displaystyle (2n+1)!=(2n+1) \cdot (2n) \cdot (2n - 1) \cdot ... \cdot 1 = (2n+1) \cdot (2n) \cdot (2n-1)!$ - Jun 9th 2009, 02:23 PMHallsofIvy
- Jun 9th 2009, 08:39 PMdiroga