<s>[INST] Induction and convergence of an inequality: $\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\leq \frac{1}{\sqrt{2n+1}}$ 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?
 [/INST] Suppose that $P(n)$ which is $\frac{1}{2}\frac{3}{4} \dots \frac{2n-1}{2n} \lt \frac{1}{\sqrt{3n+1}}$. Now $P(n + 1)$ would be  $\frac{1}{2}\frac{3}{4} \dots \frac{2n-1}{2n}\frac{2n+1}{2n+2} \overset{?}{\lt} \frac{1}{\sqrt{3n+4}} $. Using the induction hypothesis it would be enough to prove that $\frac{1}{\sqrt{3n+1}}\frac{2n+1}{2n+2} \lt \frac{1}{\sqrt{3n+4}} \leftrightarrow (2n+1)^2(3n+4) \lt (2n+2)^2(3n+1) \leftrightarrow 12n^3+28n^2+19n+4 $$\lt 12n^3+28n^2+20n+4$ which is true.
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