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?

About the original attempt:
Your first "rewrite" (immediately above (1))is incorrect:
$$\frac{1}{\sqrt{3n+1}} \times \frac{2(n+1)-1}{2(n+1)} \leq \frac{1}{\sqrt{3(n+1)+1}}$$
is NOT equivalent to, nor does it entail the rewrite I referenced:
$$\underbrace{\frac{\sqrt{3n+1}}{3n+1}}_{\large =\,\frac 1{\sqrt{3n+1}}} \times \frac{2(n+1)-1}{2(n+1)} \leq \frac{\sqrt{3n+1}}{\sqrt{3(n+1)+1}}$$
I presume you meant to multiply both sides of the inequality with $\sqrt{3n+1}$, which is valid, but note that in doing so, $\sqrt{3n+1}\cdot \dfrac 1{\sqrt{3n+1}} = 1 \neq \dfrac {\sqrt{3n+1}}{3n+1}$.

Suggestion on your revised work:
You've done fine so far. So we can pick up from $(3)$:
Rewrite $(3)$ as follows: $$\sqrt{3n+4}(2n+1) \leq (2n+2)\sqrt{3n+1} \iff  \sqrt{(3n+4)(2n+1)^2} \leq \sqrt{(2n+2)^2(3n+1)}$$ Expand the factors under each square root sign, and unless I made a mistake, you should end up with the true statement:
$$\sqrt{12n^3 + 28n^2 + \color{blue}{\bf 19n} +4}\leq \sqrt{12n^3 + 28n^2 + \color{blue}{\bf 20n} + 4}$$
