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?

Suppose $$\frac{2(k+1)-1}{2(k+1)}.\frac{1}{\sqrt{3k+1}}\geq \frac{1}{\sqrt{3(k+1)+1}}$$
$$ \frac{2k+1}{2(k+1)}.\frac{1}{\sqrt{3k+1}}\geq \frac{1}{\sqrt{3(k+1)+1}}$$
$$\frac{(2k+1)^2}{4(k+1)^2 (3k+1)}\geq \frac{1}{3(k+1)+1}$$
$$ 3(k+1)(2k+1)^2+(2k+1)^2 \geq 4(k^2+2k+1)(3k+1)$$
$$ (3k+3)(4k^2+4k+1)+4k^2+4k+1 \geq (k^2+2k+1)(12k+4)$$
$$12k^3+12k^2+12k^2+12k+3k+3+4k^2+4k+1 \geq 12k^3+24k^2+12k+4k^2+8k+4$$
$$19k \geq 20k$$
$$ 1\geq k$$ 
But .....
