Show that the sequence {$a_n$}, such that $a_1 =4$ and $a_{n+1}=3-{{2}\over\ a_n}$ is convergent to 2. Show that the sequence {$a_n$}, such that
$a_1 =4$ and $a_{n+1}=3-{{2}\over\ a_n}$ is convergent to 2.
I show that the sequence is bounded but I cannot show that is monotone.
 A: To show it's monotone, consider the ratio,
$$\frac {a_{n+1}}{a_n}=\frac{3-\frac 2{a_n}}{a_n}=\frac{3a_n-2}{a_n^2}<1$$
So, for $a_1=4$, the sequence $\left\{a_n\right\}_{n=1}^{\infty}$ is decreasing and thus monotone. 
A: If $2<a_n<3$ then $2<a_{n+1}<3$. But $2<a_2<3$ so by induction $2<a_i<3$ for $i\geq 2$. Hence $a_n$ is positive, so if $a_{i-1}>a_i$ then $3-\frac{2}{a_{i-1}}>3-\frac{2}{a_i}$ and $a_1>a_2$ so by induction the series is monotone descending.
Note this answers only your question about monotonicity and shows only that the sequence converges to a value $L\geq 2$.
A: the recurrence relation implies that if $a_n  \gt 2$ then $a_{n+1} \gt 2$
also if $a_n \gt 2$
$$
a_{n+1}-a_n = \frac{(a_n-1)(2-a_n)}{a_n} \lt 0
$$
A: We can establish that each $a_n$ lies within the interval $[2, 4]$ by induction. The base case is clear. Assuming some $a_n \in [2, 4]$, then we have 
$2 = 3 - \frac{2}{2} \leq 3 - \frac{2}{a_n} =  a_{n+1} \leq 3 - \frac{2}{4} < 4$
Thus, it is bounded. Now we need to show $a_{n+1} = 3 - \frac{2}{a_n} \leq a_n$. This demonstrates monotonicity. 
It's sufficient to show that $x + \frac{2}{x} - 3 \geq 0$ in $[2,4]$. This is simply a quadratic inequality, and I leave it to you to show that the inequality holds. 
Since it is bounded and monotone, it is convergent. Now showing what it converges to is easy. 
A: Let's solve the difference equation.
Find the first few terms:
$a_2=3-\frac{2}{4}=\frac{5}{2}=\left(2+\frac{1}{2}\right)$
$a_3=3-\frac{4}{5}=\frac{11}{5}=\left(2+\frac{1}{5}\right)$
$a_4=3-\frac{10}{11}=\frac{23}{11}=\left(2+\frac{1}{11}\right)$
$a_5=3-\frac{22}{23}=\frac{47}{23}=\left(2+\frac{1}{23}\right)$
...
$a_{n+1}=2+\frac{1}{3\cdot2^{n-1}-1}$.
Thus, $\lim \limits_{n\to\infty}a_{n+1}=2.$
Note: $3\cdot2^{n-1}-1$ is a solution of the difference equation: $b_{n+1}=2b_n+1,b_1=2.$
