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 $\alpha = 2 + \sqrt{2}$ 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 $n \in \mathbb N$. Put $\alpha = 2 + \sqrt{2}$
(a) Prove by induction on n, that $3 \le a_n \lt 4$ for all $n \in \mathbb N$.
(b) Show that $3\lt \alpha \lt 4$ and $\alpha=4-\frac{2}{\alpha}$
(c) Show that $a_{n+1}-\alpha=\frac{2(a_n-\alpha)}{\alpha a_n}$ for all $n \in \mathbb N$.
(d) Prove, by induction on n, that |$a_n-\alpha$|$\le\frac{|a_1-\alpha|}{4^{n-1}}$ for all $n \in \mathbb N$.
(e) Deduce that $a_n \rightarrow \alpha$ as $n \rightarrow \infty$
$a_2=3.33333$
$a_3=3.4$
$a_4=3.411764706$
$a_5=3.413793103$
$a_6=3.1414141414$
$\alpha=3.1414213562$
 A: Certainly $a_1 \le 3 \lt 4$, since $a_1=3$.
Suppose that for the specific integer $k$, we know that $3\le a_k \lt 4$.
We want to show that similar inequalities hold for the "next" term $a_{k+1}$. That is, we want to show that $3\le a_{k+1}\lt 4$.  
There are two inequalities to prove. Let us deal with them separately. First we show that (i) $3\le a_{k+1}$ and then that (ii) $a_{k+1}\lt 4$.  
(i) Note that $a_{k+1}=4-\frac{2}{a_k}$. We know that $a_k \ge 3$. So $\frac{2}{a_k}\le \frac{2}{3}$. It follows that $4-\frac{2}{a_k} \ge 4-\frac{2}{3}=\frac{10}{3} \ge 3$. 
(ii) Because $a_k$ is positive, $4-\frac{2}{a_k}\lt 4$. 
So we have shown that our inequalities hold at $n=1$, and that if they hold for some integer $k$, they hold for the next integer $k+1$. By the principle of mathematical induction, the inequality $3\le a_n \lt 4$ holds for every positive integer $n$. 
Added: Since the above answer was written, the question has quintupled in length. I hope that the above will at least help with the mechanics of induction. We make a few comments about the added parts.
The added question (b) is undoubtedly something you can handle. Showing that $3\lt \alpha \lt 4$ can be done by observing that $1\lt \sqrt{2}\lt 2$. To show that $\alpha=4-\frac{2}{\alpha}$, note that $\frac{2}{\alpha}=\frac{2}{2+\sqrt{2}}$. Multiply top and bottom by $2-\sqrt{2}$, and everything will collapse. Alternately, we want to show that $\alpha^2=4\alpha -2$. Solve the quadratic equation $x^2-4x+2=0$ using the Quadratic Formula. You will find that the roots are $2\pm\sqrt{2}$. 
For (c), the natural thing is to take the left-hand side $a_{n+1}-\alpha$, and replace $a_{n+1}$ by its value in terms of $a_n$. Also, we use the result of (b) as a hint, and replace $\alpha$ by $4-\frac{2}{\alpha}$. We get 
$$a_{n+1}-\alpha=\left(4-\frac{2}{a_n}\right)-\left(4-\frac{2}{\alpha}\right).$$
There is some nice cancellation: The expression on the right simplifies to $-\frac{2}{a_n}+\frac{2}{\alpha}$. Bring to a common denominator and we get
$$a_{n+1}-\alpha=\frac{2(a_n-\alpha)}{a_n \alpha}.$$
Now you be able to show (d). For the above formula gives us information about how close $a_{n+1}$ is to $\alpha$ in terms of how close $a_n$ is to $\alpha$. Use the fact that since $a_n \gt 3$ and $\alpha\gt 3$, we have $\frac{2}{a_n \alpha}\lt \frac{2}{9}\lt \frac{1}{4}$.   
A: Hint: Notice that $(2+\sqrt2)(2-\sqrt2)=2$, i.e. $\frac{2}{2+\sqrt2}=2-\sqrt2$. 
What can you say about $a_{n+1}$ if $a_n\le 2+\sqrt2$? What about the case $a_n\ge2+\sqrt2$?
EDIT: The above was written as the response to the original version of the question.
If you want to show also that $\lim\limits_{n\to\infty} a_n=\alpha$, a viable strategy seems to be:


*

*Showing that $f(x)=4-\frac2x$ is a contraction mapping on the interval $[3,4]$.

*By Banach fixed-point theorem there exists a limit of the iterations and it must fulfill the equation $f(x)=4-\frac2x=x$.


Another possibility could be playing with the sequences of odd and even terms of this sequence.
EDIT 2: The question was edited again, so I am adding two more hints:
(c) $a_{n+1}-\alpha=4-\frac2{a_n}-\alpha=\frac{(4-\alpha)a_n-2}{a_n}$
What do you get from this if you use $4-\alpha=\frac2\alpha$?
(d) You have $|a_{n+1}-\alpha|=\frac{2}{\alpha a_n}|a_n-\alpha|$.
From the preceding parts you have $\frac2{\alpha a_n}\le\frac2{3^2}=\frac29<\frac14$.
