prove that $\{a_n\}_{n=1}^{\infty}$ is bounded and monotone increasing. I am trying to prove the sequence $\{a_n\}_{n=1}^{\infty}$, which is defined by $a_{n+1} = \frac {2(a_n+1)}{a_n+2}$, and $a_1=1$.
(1) prove that it is monotone increasing. ($a_{n+1} \ge a_n$)
proof by induction. 
$P(1)$ : $a_2= \frac {2(1+1)}{2+1} = \frac 43 \ge a_1 = 1$
Suppose $P(k)$ holds true : $a_k \le a_{k+1}$. 
$P(k+1)$ : $a_{k+2}= \frac {2(a_k+1)}{a_k+2}$. Then, how can I proceed from here??
(2) prove that it is bounded ($a^2_n<2$)
$P(1): a^2_1=1<2 $
Suppose $P(k)$ holds true: $a^2_k<2$
$P(k+1): a^2_{k+1}= (\frac {2(a_k+1)}{a_k+2})^2= \frac {2a_k^2+4a_k+2}{a^2_k+4a_k+4}$ I also don't know how to proceed from here. 
Thank you in advance !!

Prove first it is bounded: as it is defined recursively   by the function
$$ f(x)=\frac{2(x+1)}{x+2}=2-\frac 2{x+2} $$
we see, since $ 0 < a_1 < 2$, that for all $n:\quad$ 
(i) $\;a_n>0$;$\quad$(ii) $\;a_n<2$.
Next, observe $f$ is a increasing function on  $[0,2]$. Therefore $(a_n)$ is a monotonic sequence, and its monotonicity is determined by what happens between $a_1$ and $a_2$:
$$a_2=\frac{2(1+1)}{1+2}=\frac43>a_1,$$
hence it is an increasing sequence.
