contractive and find its limit The real sequence $\{P_n\}$ is defined as $p_1$=2, $p_{n+1}$=$\frac{2}{1+p_n}$, n $\in N$. prove that $\{p_n\}$ is contractive, deduce that it converges, and find its limit.
first we need to show that $\frac{2}{3}$ $\leq$ $p_n$$\leq$2
pf:
as $p_1$ = 2, $p_2$=$\frac{2}{3}$,.. so that  $\frac{2}{3}$ $\leq$ 1 $\leq$2
now we have $\vert \frac{2}{1+p_{n+1}} - \frac{2}{1+p_n} \vert$ =...some crazy math here..=
since $p_1$=2 and $p_n\lt$1, for all$ $ $n\geq2$ by induction we have 2$\frac{\vert p_n - p_{n+1}\vert}{(1+p_n)(1+p_{n+1})}$ $\lt$ $\frac{2}{2}$* $\frac{2}{2}$ $\vert p_n - p_{n+1} \vert$ = $\vert p_n - p_{n+1} \vert$ therefore $p_n$ is a contractive sequence. 
is that good or need to fix my $\lt$ $\frac{2}{2}$* $\frac{2}{2}$

You are quite close. If we can show that $p_k\ge \frac{2}{3}$ for all $k$, we will have, by a calculation similar to yours, that
$$|p_{n+1}-p_n|=\left|\frac{2}{1+p_n}-\frac{2}{1+p_{n-1}}\right|=\frac{2|p_n-p_{n-1}|}{(1+p_n)(1+p_{n-1})}.$$
But since $p_k\ge \frac{2}{3}$ for all $k$, it follows that $(1+p_k)(1+p_{n-1})\ge \frac{25}{9}$. This implies that
$$|p_{n+1}-p_n|\le \frac{18}{25}|p_n-p_{n-1}|,$$
which proves contractivenes.

To show that $p_k\ge \frac{2}{3}$ for all $k$, we show the stronger  result $\frac{2}{3}\le p_k\le 2$. This is done by induction.
Suppose that $p_k\le 2$. We show that $p_{k+1}\ge \frac{2}{3}$. This is obvious.
Suppose that $\frac{2}{3}\le p_k$. Then $p_{k+1}\le \frac{2}{1+\frac{2}{3}}=\frac{6}{5}\le 2$. 
This completes the induction step.
