Possible Duplicate:
We define a sequence of rational numbers {$a_n$} by putting $a_1=3$ and $a_{n+1}=4-\frac{2}{a_n}$ for all natural numbers. Put $\alpha = 2 + \sqrt{2}$
Could someone help me with a proof for the following.
Prove by induction on $n$ that $3 \leq a_n < 4$, where $a_1 = 3$ and $a_{n+1} = 4- 2/a_n$.