<s>[INST] 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.
 [/INST] $$\frac{1\cdot 3 \cdot 5 \cdot 7  \cdot  \dots  \cdot 2007}{2 \cdot 4 \cdot 6 \cdot  \dots  \cdot 2008}=\sqrt{\frac{3\cdot (3 \cdot 5)\cdot(5 \cdot 7)\cdot...  \cdot   (2005\cdot 2007)\cdot2007}{4\cdot4^2 \cdot 6^2 \cdot  ...  \cdot2006^2\cdot 2008^2}}<\sqrt{\frac{3\cdot2007}{4\cdot2008^2}}<\frac{1}{40}.$$
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