Show that the sequences defined by:
$\displaystyle a_1 = 2$ ; $\displaystyle a_{n+1} = \frac{1}{3-a_n}$
satisfies $\displaystyle 0 < a_n <= 2$ and is decreasing. Deduce that the sequence is convergent and find it's limit.
Put $\displaystyle \phi=\frac{3-\sqrt{5}}{2}$ then $\displaystyle \phi$ is a root of $\displaystyle x^2-3x+1=0$.
(we will also require the other root $\displaystyle \phi^*=\frac{3+\sqrt{5}}{2}$ later so we may as well anounce it here)
Now suppose $\displaystyle a_n>\phi$, then let $\displaystyle a_{n+1}=\phi+\varepsilon$, and we have:
$\displaystyle
a_{n+1}=\phi+\varepsilon=\frac{1}{3-a_n}>\frac{1}{1-\phi}
$
So:
$\displaystyle
[\phi(3-\phi)-1] + \varepsilon (3-\phi) >0
$
But $\displaystyle [\phi(3-\phi)-1]=-\phi^2+3\phi-1=0$, so:
$\displaystyle \varepsilon (3-\phi)>0$
and as $\displaystyle 3-\phi>0$, we have $\displaystyle \varepsilon>0$, and so $\displaystyle a_{n+1}>\phi$.
Hence as $\displaystyle a_1>\phi$, $\displaystyle a_n>\phi$ for all $\displaystyle n \in \mathbb{Z}_+ $.
Now consider:
$\displaystyle \frac{a_{n-1}}{a_n}=a_{n-1}(3-a_{n-1})$
This is greater than $\displaystyle 1$ for $\displaystyle a_{n-1}$ between the roots of $\displaystyle -x^2+3x-1$ that is between $\displaystyle \frac{3 \pm \sqrt{5}}{2} $.
That is $\displaystyle \{a_n \}$ is decreasing while $\displaystyle a_n<\frac{3+\sqrt{5}}{2}=\phi^*$ and $\displaystyle a_n>\frac{3-\sqrt{5}}{2}=\phi$ and as $\displaystyle 2< \frac{3+\sqrt{5}}{2}$, $\displaystyle \{a_n\}$ is a decreasing sequence and also bounded below, so converges.
Now if $\displaystyle \{a_n\}$ converges the limit $\displaystyle a$ satisfies:
$\displaystyle
a=\frac{1}{3-a}
$
which has roots $\displaystyle \phi$ and $\displaystyle \phi^*.$ But it cannot converge on $\displaystyle \phi^*$ as the sequence is decreasing so the limit is $\displaystyle \phi$.
RonL