<s>[INST] 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.
 [/INST] 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: $b_{n+1}=2b_n+1,b_1=2.$
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