Product of sequence?

[kaelbu]

The gist of this problem is to prove that $\displaystyle \sqrt{\pi} = \lim_{x\to\infty}2^{2n}(n!)^2/{((2n)!\sqrt{n})}$ using wallis' product. I understand the general process of this, however there is one step I don't understand:
$\displaystyle P_n= (2^n n!)^2/{(1*3*...*2n-1)(2n+1)} = (2^n n!)^2 ((2^n n!)^2/{((2n)!(2n+1)))}$ which seems to indicate that \(\displaystyle 1/(1*3*...*2n-1) = (2^n n!)^2/{((2n)!)\) How can I determine this?

 

[galactus]

$\displaystyle \prod_{k=1}^{n}\frac{1}{2k-1}=\frac{1}{1\cdot 3\cdot \cdot 5 \cdot \cdot (2n-1)}=\frac{n!\cdot 2^{n}}{(2n)!}$

 

[kaelbu]

Yes, that is evident from the solution guide I have, and I plugged in some numbers to verify it...but HOW can I know this without the benefit of an answer guide? Is this just a fact that I must memorize? Or is it derived from another fact that I ought to have memorized?

 

[galactus]

If we take $\displaystyle \frac{n!\cdot 2^{n}}{(2n)!}$ and just write it out, we can see what is left is the reciprocal of the odd terms:

$\displaystyle \prod_{k=1}^{n}2k=2\cdot 4\cdot 6\cdot\cdot\cdot 2n=2(1\cdot 2\cdot 3\cdot\cdot\cdot n)=n!\cdot 2^{n}$

The denominator is $\displaystyle 1\cdot 2\cdot 3\cdot\cdot\cdot (2n)$

So, we get:

$\displaystyle \frac{2\cdot 4\cdot 6\cdot\cdot\cdot 2n}{1\cdot 2\cdot 3\cdot\cdot\cdot 2n}$

Note, the even terms in the numerator cancel with those in the denominator and we are left with:

$\displaystyle \prod_{k=1}^{n}\frac{1}{2k-1}=\frac{1}{1\cdot 3\cdot 5\cdot\cdot\cdot (2n-1)}=\frac{n!\cdot 2^{n}}{(2n)!}$