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 !!

I'll play around
and see what happens.
$a_{n+1} = \frac {2(a_n+1)}{a_n+2},
a_1=1
$
$\begin{array}\\
a_{n+1}^2-2 
&= \dfrac {4(a_n+1)^2}{(a_n+2)^2}-2\\
&= \dfrac {4a_n^2+8a_n+4-2(a_n^2+4a_n+4)}{(a_n+2)^2}\\
&= \dfrac {4a_n^2+8a_n+4-2a_n^2-8a_n-8}{(a_n+2)^2}\\
&= \dfrac {2a_n^2-4}{(a_n+2)^2}\\
&= 2\dfrac {a_n^2-2}{(a_n+2)^2}\\
\end{array}
$
So $a_{n+1}^2-2$
has the same sign as
$a_n^2-2$.
Since $a_1 = 1$,
all $a_n$ satisfy
$a_n^2 < 2$.
$\begin{array}\\
a_{n+1} -a_n
&= \dfrac {2(a_n+1)}{a_n+2}-a_n\\
&= \dfrac {2a_n+2-a_n^2-2a_n}{a_n+2}\\
&= \dfrac {2-a_n^2}{a_n+2}\\
&> 0
\qquad\text{since } a_n^2 < 2\\
\end{array}
$
Note that if
$a_1^2 > 2$
then
all $a_n^2 > 2$
and the $a_n$
are decreasing.
In either case,
the $a_n$ are bounded and
monotonic,
so they approach a limit.
If $L$ is this limit,
since
$a_{n+1} -a_n
= \dfrac {2-a_n^2}{a_n+2}
$,
$\dfrac {2-a_n^2}{a_n+2} \to 0$
so
$a_n^2 \to 2$.
