Prove $ \frac{1\cdot 3 \cdot 5 \cdot 7 \cdot \dots \cdot 2007}{2 \cdot 4 \cdot 6 \cdot \dots \cdot 2008}< \frac{1}{40} $ Prove:
$$
\frac{1\cdot 3 \cdot 5 \cdot 7  \cdot  \dots  \cdot 2007}{2 \cdot 4 \cdot 6 \cdot  \dots  \cdot 2008}< \frac{1}{40}
$$
i have tried to write $1/40$ as $(1/40^{1/2007})^{2007}$ and prove $(1/40^{1/2007})^{2007}$ to be greater than $2007/2008$ but i quickly found out this is not true. Is there another way to manipulate this product? I think telescoping is possible but I just don't know how to do it; maybe split the fractions and work it out? Any help will be appreciated thank you.

Let 
$$
A=\frac{1\cdot 3 \cdot 5 \cdot 7  \cdot  \dots  \cdot 2007}{2 \cdot 4 \cdot 6 \cdot  \dots  \cdot 2008}
$$
$$
B=\frac{2 \cdot 4 \cdot 6 \cdot  \dots  \cdot 2008}{1\cdot 3 \cdot 5 \cdot 7  \cdot  \dots  \cdot 2009}
$$
Then $$A<B$$
Then $$A^2<AB=\frac1{2009}$$
Then $$A<\frac1{\sqrt{2009}}<\frac1{\sqrt{1600}}=\frac1{40}$$
Generally:
$$ {1\over 2} \cdot {3\over 4} \cdot {5 \over 6} \cdot \ldots \cdot {2n-1 \over 2n} < {1 \over \sqrt{2n+1}}$$
$$A={1\over 2}\cdot{3\over 4}\cdot{5 \over 6}\cdot\ldots\cdot{2n-1\over 2n}, \quad B={2 \over 3} \cdot{4\over 5}\cdot{6\over 7}\cdot\ldots\cdot{2n \over 2n+1}.$$
$$A < B$$
$$A^2 < AB = {1 \over 2n+1}.$$
