Can Inequality with Factorial Be Proven without Induction?

AI Thread Summary
The discussion centers on proving the inequality \(\frac{1^2*3^2*5^2...(2n-1)^2}{2^2*4^2*6^2...(2n)^2}<\frac{1}{2n+1}\) without using induction. Participants explore the relationship between the left-hand side (LHS) and factorials, noting that the LHS can be expressed as \(\frac{(2n-1)!}{(2n)!}\). Some users mention successfully proving the inequality through induction, while others seek alternative methods. There are mentions of typographical errors in the equations presented, which may affect clarity. The conversation highlights the challenge of finding a non-inductive proof for the stated inequality.
cupcakes
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Homework Statement


\frac{1^2*3^2*5^2...(2n-1)^2}{2^2*4^2*6^2...(2n)^2}&lt;\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.
 
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I was able to prove this by induction, try it out!
 
cupcakes said:

Homework Statement


\frac{1^2*3^2*5^2...(2n-1)^2}{2^2*4^2*6^2...(2n)^2}&lt;\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}
 
scurty said:
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. :(

SammyS said:
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 :smile:
Does anyone have any other idea or hint that does not involve induction?
 
I've solved it. :smile:
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
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