real analysis - Prove that the sequence $a_n=frac{1cdot3cdot5cdots(2n-1)}{2cdot4cdot6cdots(2n)}$ is monotonically decreasing sequence

Prove that the sequence $$a_n=\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}$$ is monotonically decreasing sequence. I tried $a_{n+1} - a_{n} < 0$, but i was not able to do it.

Help will be appreciated.

Thanks!

Answer

Notice, we have $$a_n=\frac{1.3.5\dots (2n-1)}{2.4.6\ldots (2n)}$$
$$=\frac{[1.3.5\dots (2n-1)]\cdot [2.4.6\ldots (2n)]}{[2.4.6\ldots (2n)]\cdot[(2.4.6\ldots (2n)]}$$ $$=\frac{1.2.3.4.5\dots (2n-1)(2n)}{[2.4.6\ldots (2n)]^2}$$ $$=\frac{(2n)!}{[2^n(1.2.3\ldots (n))]^2}$$
$$=\frac{(2n)!}{[2^n(n!)]^2}$$
$$a_n=\frac{(2n)!}{2^{2n}(n!)^2}\tag 1$$ $$\implies a_{n+1}=\frac{(2n+2)!}{2^{2n+2}((n+1)!)^2}$$ $$\implies a_{n+1}=\frac{(2n+2)(2n+1)(2n)!}{4\cdot 2^{2n}((n+1)n!)^2}$$

$$=\frac{2(n+1)(2n+1)(2n)!}{4(n+1)^2\cdot 2^{2n}(n!)^2}$$
$$a_{n+1}=\frac{(2n+1)(2n)!}{2(n+1)\cdot 2^{2n}(n!)^2}\tag 2$$

Now, dividing (2) by (1), we get $$\frac{a_{n+1}}{a_n}=\frac{\frac{(2n+1)(2n)!}{2(n+1)\cdot 2^{2n}(n!)^2}}{\frac{(2n)!}{2^{2n}(n!)^2}}$$ $$=\frac{2n+1}{2(n+1)}$$$$\frac{a_{n+1}}{a_n}=\frac{2n+1}{2n+2}<1$$