To show for following sequence $\lim_{n \to \infty} a_n = 0$ where $a_n$ = $1.3.5 ... (2n-1)\over 2.4.6...(2n)$

Using $\boldsymbol{1+x\le e^x}$

Since $1+x\le e^x$ for all $x\in\mathbb{R}$, $$ \begin{align} a_n &=\prod_{k=1}^n\frac{2k-1}{2k}\\ &=\prod_{k=1}^n\left(1-\frac1{2k}\right)\\ &\le\prod_{k=1}^ne^{-\large\frac1{2k}}\\[3pt] &=e^{-\frac12H_n}\tag{1} \end{align} $$ where $H_n$ are the Harmonic Numbers. Since the Harmonic Series diverges, $(1)$ tends to $0$.

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Using Gautschi's Inequality $$ \begin{align} a_n &=\prod_{k=1}^n\frac{2k-1}{2k}\\ &=\prod_{k=1}^n\frac{k-\frac12}{k}\\ &=\frac{\Gamma\!\left(n+\frac12\right)}{\sqrt\pi\,\Gamma(n+1)}\\[6pt] &\le\frac1{\sqrt{\pi n}}\tag{2} \end{align} $$

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Commentary

Gautschi's Inequality also says that $$ \begin{align} a_n &=\frac{\Gamma\!\left(n+\frac12\right)}{\sqrt\pi\,\Gamma(n+1)}\\ &\ge\frac1{\sqrt{\pi\left(n+\frac12\right)}}\tag{3} \end{align} $$ Therefore, $$ \lim_{n\to\infty}\sqrt{n}\,a_n=\frac1{\sqrt\pi}\tag{4} $$ Putting the second line of $(1)$ and the limit in $(4)$ together, we can show that $$ \gamma+\sum_{n=2}^\infty\frac{\zeta(n)}{n2^{n-1}}=\log(\pi)\tag{5} $$ where $\gamma$ is the Euler-Mascheroni Constant and $\zeta$ is the Riemann Zeta Function.

Multiply the numerator by the denominator; so $$a_n=\frac{1\times3\times5\times ... \times(2n-1)}{ 2\times4\times6\times...\times(2n)}=\frac{1\times2\times3\times ... \times(2n)}{\Big(2\times4\times6\times...\times(2n)\Big)^2}=\frac{(2n)!}{4^n(n!)^2}$$ If now you use Stirling approximation $$m! \approx \sqrt{2 \pi } e^{-m} m^{m+\frac{1}{2}}$$ and then $$a_n \approx \frac{1}{\sqrt{\pi n} }$$ A more detailed approach would show for the asymptotic behavior $$a_n \approx \frac{1}{\sqrt{\pi n} }\Big(1-\frac{1}{8n}\Big)$$

Let's prove using induction that $$a_n\le\frac{1}{\sqrt{3n+1}}.$$ For $n=1$ it is true. Now we just need to prove that$$\frac{(2n+1)^2}{(2n+2)^2}\le\frac{3n+1}{3n+4}$$or $$(4n^2+4n+1)(3n+4)\le(4n^2+8n+4)(3n+1)$$or$$12n^3+28n^2+19n+4\le12n^3+28n^2+20n+4.$$