<s>[INST] 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?
 [/INST] 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 $\frac{n!!}{(n+1)!!}$
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