This problem is just for fun, meant as a challenge.

Calculate $\displaystyle \lim_{n \to \infty} \frac{2^{2n-1}((n-1)!)^2 \sqrt{n}}{(2n-1)!} $.

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- Dec 17th 2009, 01:03 PMchiph588@Just for Fun
This problem is just for fun, meant as a challenge.

Calculate $\displaystyle \lim_{n \to \infty} \frac{2^{2n-1}((n-1)!)^2 \sqrt{n}}{(2n-1)!} $. - Dec 17th 2009, 07:11 PMBruno J.
The answer is $\displaystyle \sqrt \pi$!

The solution is easy : Gamma duplication formula! - Dec 17th 2009, 07:29 PMchiph588@
Good job!

- Dec 17th 2009, 07:37 PMBruno J.
Thank you for the problem! (Cool)

- Dec 17th 2009, 07:41 PMchiph588@
How did you come about your solution? I thought it would be harder because a lot cancels out so it doesn't resemble the duplication formula.

- Dec 17th 2009, 08:31 PMBruno J.
I don't know, I just thought of it right away. I think $\displaystyle 2^{2n-1}$ gave it away!

- Dec 19th 2009, 01:46 AMNonCommAlg
i'm sure you guys know that applying Strling's formula will make the problem very straightforward.