Recursively defined sequence (a) Consider the recursively defined sequence $y_1 = 1$,
$y_{n+1} = 3−y_n$,
and set $y = \lim y_{n}$. Because $(y_n)$ and $(y_{n+1})$ have the same limit, taking the limit across the recursive equation gives $y = 3−y$. Solving for y, we conclude $\lim y_n = 3/2$.
What is wrong with this argument? 
(b) This time set $y_1 = 1$ and $y_{n+1} = 3− 1/y_{n}$ . Can the strategy in (a) be applied to compute the limit of this sequence?
My solution:
(a) The argument incorrectly assumes that the limit of (yn) exists,
which is not true.
(b) I know that I have to show by induction that $y_{n+1}\geq y_n
\geq 1$ but I don't know how to illustrate this.
 A: (i) First we show  that $y_n\ge 1$ for all $n$. The result certainly holds at $n=1$. Suppose the result holds at $n=k$. We show the result holds at $n=k+1$. This is easy, for 
$$y_{k+1}=3-\frac{1}{y_k}\ge 3-\frac{1}{1}=2\ge 1.$$
(ii) Next we show that our sequence is bounded above by $3$. In principle this is done by induction. The result is true at $1$. If the result is true at $k$, it is true at $k+1$. For $y_{k+1}=3-\frac{1}{y_k}\le 3$ since $y_k$ is positive. Note that we did not even use the induction hypothesis! The only thing we used is the result of (i): because $y_k\ge 1$, in particular $y_k$ is positive.
(iii) Finally we show that the sequence $(y_n)$ is non-decreasing. It is clear that $y_2\ge y_1$. Suppose that for a certain $k$ we have $y_{k+1}\ge y_k$. We show that $y_{k+2}\ge y_{k+1}$.
We have 
$$y_{k+2}=3-\frac{1}{y_{k+1}}\ge 3-\frac{1}{y_k}=y_{k+1},$$
completing the induction step.
We have shown that our sequence is non-decreasing and bounded above. It follows that the sequence converges. 
Since the limit exists, and is non-zero, we can suppose it is $L$ and conclude that $L=3-\frac{1}{L}$. To find $L$ we need to solve a quadratic equation.
