How to prove that $ \frac{(99)!!}{(100)!!} < \frac{1}{10}$ How to prove that $ \dfrac{(99)!!}{(100)!!}=\dfrac{1\cdot3\cdot5\cdot7\cdot9 \cdots99}{2\cdot4\cdot6\cdot8\cdot10\cdots100} < \dfrac{1}{10}$
Any hint to prove it?

In general the ratio
$$ r_n = \frac{(2n-1)!!}{(2n)!!}$$
is related to the central binomial coefficient for which many results are known.
In particular, we have that
$$ (2n-1) (2n-3) \cdots 1 = \frac{(2n) (2n-1)\cdots 1}{(2n)(2n-2) \cdots 2} =2^{-n} \frac{(2n) (2n-1)\cdots 1}{n(n-1) \cdots 1} = 2^{-n} \frac{(2n)!}{n!}$$
and thus
$$ r_n = \frac{(2n-1)(2n-3) \cdots 1}{(2n) (2n-2) \cdots 2}=2^{-n}\frac{(2n-1)(2n-3) \cdots 1}{n (n-1) \cdots 1}=4^{-n} \frac{(2n)!}{(n!)^2}  = 4^{-n} \binom{2n}{n}\,.$$
It can be shown that
$$ \binom{2n}{n} \leq \frac{4^n}{\sqrt{\pi n}}$$
for $n\geq 1$.
Thus, we have that
$$r_n \leq \frac{1}{\sqrt{\pi n}}.$$
Specialising for $n=50$, we obtain
$$ \frac{99!!}{100!!} \leq \frac{1}{\sqrt{50 \pi} } < \frac{1}{\sqrt{144}} = \frac{1}{12} \,.$$
