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.
 A: You want to prove that $\frac{n-1}{n}<\frac{n}{n+1}$. Mutiply both sides by $n(n+1)$ and it is pretty obvious.
A: If $a_n = \frac{n-1}{n}$,
$$a_{n+1} - a_n = \frac{n}{n+1} - \frac{n-1}{n} = \frac{n^2 - (n-1)(n+1)}{n(n+1)} = \frac{1}{n(n+1)} > 0$$

Also,
$$a_n = \frac{n-1}{n} = 1 - \frac1 n$$
It should be clear that $1/n$ is decreasing as $n$ increases, and so this sequence increases towards $1$.
A: put $f(x)=\frac{x}{x+1}$  it is an increasing function.
A: Here's an idea. In general you want to show that
$$\frac{a}{a+1} < \frac{a+1}{a+2}.$$
What happens if you multiply $\frac{a}{a+1}$ by $\frac{(a+1)^2}{a+2}$? What do you know about the number $\frac{(a+1)^2}{a+2}$?
A: If $n>1$ is an integer we have
\begin{align*}
n-1&<n\\
\implies \frac{1}{n-1}&>\frac{1}{n}\\
\iff 1-\frac{1}{n-1}&<1-\frac{1}{n}\\
\iff \frac{n-2}{n-1}&<\frac{n-1}{n}
\end{align*}
A: Subtracting everything from $1$, this is the same as trying to prove that:
$$\frac12>\frac13>\frac14>\frac15>\dotsb$$
Taking the reciprocal, we see that it's the same as trying to prove that:
$$2<3<4<5<\dotsb$$
which is obviously true.
I used the fact that "subtracting from $1$" and "taking the reciprocal" both reverse inequalities.

Technically, we need to be careful with the second one, since $-2<3$ but $-\frac12<\frac13$; in fact, reciprocals only reverse the inequality when the two sides have the same sign (both positive or both negative). This isn't a problem here, though, since everything is positive.

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