<s>[INST] inequality $\prod\limits_{k=1}^n\frac{2k-1}{2k}\lt\frac1{\sqrt{3n}}$ $n$ is  a positive integer, then 
$$\prod_{k=1}^n\frac{2k-1}{2k}\lt\frac1{\sqrt{3n}}$$

with mathematical induction, we can prove this. 
But I would love to find a wonderful method without mathematical induction
Thank you!
 [/INST] Building on André Nicolas' idea, you may rewrite $\prod_{k=1}^n\frac{2k-1}{2k}$ as
$$\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{7}{8}\cdots\frac{2n-1}{2n}=\frac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdots (2n-1)2n}{(2\cdot 4 \cdot 6\cdot 8\cdots 2n)^2}=\frac{(2n)!}{(2^n\cdot 1\cdot 2\cdot 3\cdots n)^2}=\frac{1}{4^n}\binom{2n}{n}.$$
Now, $\binom{2n}{n}$ is the central binomial coefficient. Its asymptotic is both well-known and easily determined as
$$\binom{2n}{n}\sim \frac{4^n}{\sqrt{\pi n}},$$
so that asymptotically,
$$\prod_{k=1}^n \frac{2k-1}{2k}\sim \frac{1}{\sqrt{\pi n}}<\frac{1}{\sqrt{3n}}\quad(\pi=3.1416...).$$
Moreover, wikipedia also lists a property of central binomial coefficients which directly proves your result, namely,
$$\binom{2n}{n}\le \frac{4^n}{\sqrt{3n+1}}<\frac{4^n}{\sqrt{3n}}.$$
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