What is the non-inductive proof of this inequality? $$\dfrac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} < \dfrac{1}{\sqrt{3n+1}}.$$
However I've non-inductive proof of $\dfrac{1 \cdot 3 \cdot 5 \cdots(2n-1)}{2 \cdot 4 \cdot 6 \cdots(2n)} < \dfrac{1}{\sqrt{2n+1}}$, but I can't prove it for $3n+1$. It is obvious is to see $\dfrac{1 \cdot 3 \cdot 5 \cdots(2n-1)}{2 \cdot 4 \cdot 6 \cdots(2n)} < \dfrac{1}{\sqrt{3n+1}}$ is stronger inequality.
Let $S=\dfrac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} < \dfrac{2 \cdot 4 \cdot 6 \cdots(2n)}{3 \cdot 5 \cdot 7 \cdots(2n+1)}$ [$\frac{1}{2}<\frac{2}{3}  $ and so on.]
So, $S<\dfrac{1}{S(2n+1)}$ Implies, $S<\dfrac{1}{\sqrt{2n+1}}$ 

$$S = \prod_{k=1}^n \frac{2k - 1}{2k} = \prod_{k=1}^n \frac{(2k)(2k - 1)}{(2k)^2} = \frac{(2n)!}{2^{2n}(n!)^2} = \frac{1}{4^n} {2n \choose n} $$
According to Wikipedia, ${2n \choose n} \le \frac{4^n}{\sqrt{3n + 1}}$
And the result follows.
(This is a proof by Wikipedia, rather than induction).
