Proving that $\frac{1}{2}<\frac{2}{3}<\frac{3}{4}<$...$<\frac{n-1}{n}$ In an attempt to find a pattern, I did this:
Let a,b,c,d be non-zero consecutive numbers. Then we have:
$a=a$
$b=a+1$
$c=a+2$
$d=a+3$
This implies:
$\frac{a}{b}=\frac{a}{a+1}$
$\frac{b}{c}=\frac{a+1}{a+2}$
$\frac{c}{d}=\frac{a+2}{a+3}$
I don't know how that helps. I'm greatly seeking your help. Thank you very much.

Now that I understand better, another simple way to prove the inequality is:
$$\begin{align}& n^{2}-1<n^{2}\\
& \implies\dfrac{n^{2}-1}{n(n+1)}<\dfrac{n^2}{n(n+1)}\\
&\implies \dfrac{(n-1)(n+1)}{n(n+1)}<\dfrac{n^{2}}{n(n+1)}\\
&\implies \dfrac{n-1}{n}<\dfrac{n}{n+1}\\
\end {align}$$
