Can Inequality with Factorial Be Proven without Induction?

[cupcakes]

Homework Statement

$\frac{1^2*3^2*5^2...(2n-1)^2}{2^2*4^2*6^2...(2n)^2}<\frac{1}{2n+1}$

Edit: Must be proven without using induction.

Homework Equations

The Attempt at a Solution

I understand the LHS is the same thing as

$$\frac{(2n-1)!}{(2n)!}$$

And (2n)! = $k!2^k$ & (2n-1)! = $\frac{(2k)!}{k!2^k}$

I've tried substituting and it doesn't seem to help. Any ideas? Thanks.

 

[scurty]

I was able to prove this by induction, try it out!

 

[SammyS]

Homework Statement

$\frac{1^2*3^2*5^2...(2n-1)^2}{2^2*4^2*6^2...(2n)^2}<\frac{1}{2n+1}$

Homework Equations

The Attempt at a Solution

I understand the LHS is the same thing as

$$\frac{(2n-1)!}{(2n)!}$$

And (2n)! = $k!2^k$ & (2n-1)! = $\frac{(2k)!}{k!2^k}$

I've tried substituting and it doesn't seem to help. Any ideas? Thanks.

There are some typos there.

$\displaystyle (2k)!=k!2^k$

$\displaystyle (2k-1)!=\frac{(2k-1)!\ (2k)!}{(2k)!}=\frac{(2k)!}{k!2^k}$

 

[cupcakes]

I was able to prove this by induction, try it out!

I forgot to mention that the problem states that it must be proven without using induction. :(

There are some typos there.

$\displaystyle (2k)!=k!2^k$

$\displaystyle (2k-1)!=\frac{(2k-1)!\ (2k)!}{(2k)!}=\frac{(2k)!}{k!2^k}$

Thanks Sammy
Does anyone have any other idea or hint that does not involve induction?

 

[cupcakes]

I've solved it.