Induction and convergence of an inequality: $frac{1cdot3cdot5cdots(2n-1)}{2cdot4cdot6cdots(2n)}leq frac{1}{sqrt{2n+1}}$

Problem statement:
Prove that $\frac{1*3*5*...*(2n-1)}{2*4*6*...(2n)}\leq \frac{1}{\sqrt{2n+1}}$ and that there exists a limit when $n \to \infty$.
, $n\in \mathbb{N}$

My progress

LHS is equivalent to $\frac{(2n-1)!}{(2n)!}=\frac{(2n-1)(2n-2)(2n-3)\cdot ....}{(2n)(2n-1)(2n-2)\cdot ....}=\frac{1}{2n}$ So we can rewrite our inequality as:

$\frac{1}{2n}\leq \frac{1}{\sqrt{2n+1}}$ Let's use induction:

For $n=1$ it is obviously true. Assume $n=k$ is correct and show that $n=k+1$ holds.

$\frac{1}{2k+2}\leq \frac{1}{\sqrt{2k+3}}\Leftrightarrow 2k+2\geq\sqrt{2k+3}\Leftrightarrow 4(k+\frac{3}{4})^2-\frac{5}{4}$ after squaring and completing the square. And this does not hold for all $n$

About convergence: Is it not enough to check that $\lim_{n \to \infty}\frac{1}{2n}=\infty$ and conclude that it does not converge?

Answer

First, note that $2\cdot 4 \cdot 6\cdots (2n)=2^n(n!)$. Next, note that if we multiplied $1\cdot 3\cdot 5\cdots (2n-1)$ by $2\cdot 4\cdot 6\cdots (2n)$, that would exactly fill the gaps and produce $(2n)!$. Hence, the denominator of the LHS is $2^nn!$, while the numerator of the LHS is $\frac{(2n)!}{2^nn!}$ Combining, the LHS equals $$\frac{1}{2^{2n}}\frac{(2n)!}{n!n!}=2^{-2n}{2n\choose n}$$

This is a central binomial coefficient, which are well-studied. For example, one bound given is that ${2n \choose n}\le \frac{4^n}{\sqrt{3n+1}}$; applying it in this case gives $$LHS=4^{-n}{2n\choose n}\le \frac{1}{\sqrt{3n+1}}\le \frac{1}{\sqrt{2n+1}}$$