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.
 A: $$\frac{1\cdot 3 \cdot 5 \cdot 7  \cdot  \dots  \cdot 2007}{2 \cdot 4 \cdot 6 \cdot  \dots  \cdot 2008}=\sqrt{\frac{3\cdot (3 \cdot 5)\cdot(5 \cdot 7)\cdot...  \cdot   (2005\cdot 2007)\cdot2007}{4\cdot4^2 \cdot 6^2 \cdot  ...  \cdot2006^2\cdot 2008^2}}<\sqrt{\frac{3\cdot2007}{4\cdot2008^2}}<\frac{1}{40}.$$
A: I would use Stirling's approximation.
$$\frac{1\cdot 3 \cdot 5 \cdot 7  \cdot  \dots  \cdot 2007}{2 \cdot 4 \cdot 6 \cdot  \dots  \cdot 2008}=\frac {2008!}{(1004!)^22^{2008}}$$
Now if you apply Stirling everything cancels except $$\frac {\sqrt {2\pi\cdot 2008}}{2\pi\cdot 1004}\approx 0.017\lt \frac 1{40}$$
I leave checking the error bounds to you, but Stirling is very close at this range and we have a good bit of margin.
A: 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:
$$ {1\over 2} \cdot {3\over 4} \cdot {5 \over 6} \cdot \ldots \cdot {2n-1 \over 2n} < {1 \over \sqrt{2n+1}}$$
$$A={1\over 2}\cdot{3\over 4}\cdot{5 \over 6}\cdot\ldots\cdot{2n-1\over 2n}, \quad B={2 \over 3} \cdot{4\over 5}\cdot{6\over 7}\cdot\ldots\cdot{2n \over 2n+1}.$$
$$A < B$$
$$A^2 < AB = {1 \over 2n+1}.$$
A: $$\frac{(2n-1)!!}{(2n)!!}=\frac{1}{4^n}\binom{2n}{n}=\frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n}(x)\,dx\stackrel{(*)}{\leq}\frac{2}{\pi}\int_{0}^{\pi/2}e^{-nx^2}\,dx\leq\frac{1}{\sqrt{\pi n}} $$
evaluated at $n=1004$ immediately gives that the LHS is $\leq \color{red}{\frac{1}{56}}$.

Proof of $(*)$: for any $z\in(0,\pi/2)$ we have $\tan z\geq z$. For any $x\in(0,\pi/2)$, by integrating both sides on the interval $(0,x)$ we get $-\log\cos x\geq \frac{x^2}{2}$ and $\cos(x)\leq e^{-x^2/2}$ by exponentiating both sides.
