Prove $\frac{2k+2}{2k+3}>\frac{2k}{2k+1}, k\in\mathbb N$ Prove $\frac{2k+2}{2k+3}>\frac{2k}{2k+1}, k\in\mathbb N$
The book I am using asserts a non-trivial way of proving this inequality, but I cannot see why this cannot be proven by rearranging the statement equally as rigorously.
Let
$$\frac{2k+2}{2k+3}=\frac{2k}{2k+1}, k\in\mathbb N$$
$$(2k+2)(2k+1)=2k(2k+3)$$
$$2k^2+3k+1=2k^2+3k$$
$$1=0$$
Contradiction, therefore no solution to where the two sides are equal
Let k=1
$$\frac{2+2}{2+3}= 0.8$$
$$\frac{2}{3}=0.66666$$
Since there are no points of intersection, and for k=1 LHS>RHS, and both sides are continuous on k>0, inequality must hold.

$\frac{2x+2}{2x+3}>\frac{2x}{2x+1}, X\in\mathbb R^+$
$ f: x \rightarrow \frac{2x}{2x+1} = f: x \rightarrow 1- \frac{1}{2x+1}$ is easily demonstrable to be strictly increasing  
$x > y \rightarrow 2x+1 > 2y+1 \rightarrow \frac{1}{2x+1} <\frac{1}{2y+1} \rightarrow 1- \frac{1}{2x+1} > 1-\frac{1}{2y+1}$ therefore $f(k+1) > f(k) $
i.e : 
$\frac{2k+2}{2k+3}>\frac{2k}{2k+1}, k\in\mathbb N \subset \mathbb R^+$
