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?
 A: 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
$$ \frac{99!!}{100!!} \leq \frac{1}{\sqrt{50 \pi} } < \frac{1}{\sqrt{144}} = \frac{1}{12} \,.$$
A: We know that $\dfrac{99!}{100!}=\dfrac{1}{100}$
We rewrite $\dfrac{1}{100}=\dfrac{99!}{100!}=\dfrac{\color{blue}{1}\cdot\color{red}{2}\cdot\color{blue}{3}\cdot\color{red}{4}\cdots\color{red}{98}\cdot\color{blue}{99}}{\color{blue}{1}\cdot\color{red}{2}\cdot\color{blue}{3}\cdot\color{red}{4}\cdots\color{blue}{99}\cdot\color{red}{100}}=\dfrac{99!!\cdot 98!!}{100!!\cdot 99!!}$
We know trivially that both $\frac{99!!}{100!!}$ and $\frac{98!!}{99!!}$ are positive.
Now., there are four possibilities


*

*both $\frac{99!!}{100!!}$ and $\frac{98!!}{99!!}$ are less than $\frac{1}{10}$, which would lead to a contradiction as their product would be less than $\frac{1}{100}$

*both $\frac{99!!}{100!!}$ and $\frac{98!!}{99!!}$ are greater than or equal to $\frac{1}{10}$, which if at least one is strictly greater than $\frac{1}{10}$ would yield a contradiction as their product would be greater than $\frac{1}{100}$.  If both are equal to $\frac{1}{10}$, then using a prime factorization argument yields a contradiction.

*$\frac{99!!}{100!!}$ is less than $\frac{1}{10}$ and $\frac{98!!}{99!!}$ is greater than $\frac{1}{10}$ (this is what we want to show must be the case)

*$\frac{99!!}{100!!}$ is greater than $\frac{1}{10}$ and $\frac{98!!}{99!!}$ is less than $\frac{1}{10}$. (we want to show this can't be the case)
So, the question has become proving whether or not $\frac{(n-1)!!}{n!!}$ must be less than $\frac{n!!}{(n+1)!!}$
A: $$ \ln \left( \frac{99!!}{100!!} \right) = \ln \left(\frac{99}{100} \cdot \frac{97}{98} \cdot \ldots \cdot \frac{1}{2}\right) = \sum_{k=1}^{50} (\ln(2k-1)- \ln(2k)) $$
with $f(k) = \ln(2k-1)-\ln(2k)$ a concave function of $k$ for $k \ge 1$.
Thus 
$$\eqalign{ \ln \left( \frac{99!!}{100!!} \right) &= \frac{f(1)}{2} + \frac{f(50)}{2} + \sum_{k=1}^{49} \frac{f(k)+f(k+1)}{2}\cr
&\le \frac{f(1)}{2} + \frac{f(50)}{2} + \int_1^{50} f(t)\; dt\cr
&= -199 \frac{\ln(2)}{2}-100 \ln(5)+99 \ln(3)+99 \frac{\ln(11)}{2}+\frac{\ln(99)}{2}-\ln(10)\cr &< \ln(1/10)} $$
A: $$
(2k-1)!! = \frac{(2k-1)!}{2^{k-1}(k-1)!}
$$
$$
(2k)!! = 2^k k!
$$
$$
\frac{(2k-1)!!}{(2k)!!} = \frac{1}{2^{2k-1}}\frac{(2k-1)!}{k!(k-1)!}=\frac{1}{2^{2k-1}}\frac12\frac{2k(2k-1)!}{k!k(k-1)!}=\frac{1}{2^{2k}}\frac{(2k)!}{k!k!}
= \frac{1}{2^{2k}}\binom{2k}{k}
$$
We can express that central binomial coefficient in terms of Catalan numbers:
$$
\frac{(2k-1)!!}{(2k)!!} =  \frac{1}{2^{2k}} (k+1)C_k
$$
The asymptotic form of $C_k$ for $k>2$ is 
$$
\frac{4^k}{k^{3/2}\sqrt{\pi}} (1-\frac{9}{8k}) < C_k <\frac{4^k}{k^{3/2}\sqrt{\pi}} 
$$
So 
$$
\frac{(2k-1)!!}{(2k)!!} < \frac{1}{2^{2k}} (k+1) \frac{4^k}{k^{3/2}\sqrt{\pi}}
= (1+\frac1k)\frac1{\sqrt{k\pi}}
$$
If you plug in $k=33$ this says 
$$\frac{65!!}{66!!} < \frac{34}{33\sqrt{33\pi}} \sim 0.0996 < 0.1$$
And of course $\frac{99!!}{100!!}$ is less than that.
