1. ## Sequence Proof

Show that the sequences defined by:

$a_1 = 2$ ; $a_{n+1} = \frac{1}{3-a_n}$

satisfies $0 < a_n <= 2$ and is decreasing. Deduce that the sequence is convergent and find it's limit.

2. Hello,

Use induction to prove it...

$a_1 \geq 2$

If $a_n \geq 2$ would $a_{n+1}=\frac{1}{3-a_n}$ be $\geq 2$?

It's the same for $a_n > 0$

3. Originally Posted by FalconPUNCH!
Show that the sequences defined by:

$a_1 = 2$ ; $a_{n+1} = \frac{1}{3-a_n}$

satisfies $0 < a_n <= 2$ and is decreasing. Deduce that the sequence is convergent and find it's limit.
Put $\phi=\frac{3-\sqrt{5}}{2}$ then $\phi$ is a root of $x^2-3x+1=0$.
(we will also require the other root $\phi^*=\frac{3+\sqrt{5}}{2}$ later so we may as well anounce it here)

Now suppose $a_n>\phi$, then let $a_{n+1}=\phi+\varepsilon$, and we have:

$
a_{n+1}=\phi+\varepsilon=\frac{1}{3-a_n}>\frac{1}{1-\phi}
$

So:

$
[\phi(3-\phi)-1] + \varepsilon (3-\phi) >0
$

But $[\phi(3-\phi)-1]=-\phi^2+3\phi-1=0$, so:

$\varepsilon (3-\phi)>0$

and as $3-\phi>0$, we have $\varepsilon>0$, and so $a_{n+1}>\phi$.

Hence as $a_1>\phi$, $a_n>\phi$ for all $n \in \mathbb{Z}_+$.

Now consider:

$\frac{a_{n-1}}{a_n}=a_{n-1}(3-a_{n-1})$

This is greater than $1$ for $a_{n-1}$ between the roots of $-x^2+3x-1$ that is between $\frac{3 \pm \sqrt{5}}{2}$.

That is $\{a_n \}$ is decreasing while $a_n<\frac{3+\sqrt{5}}{2}=\phi^*$ and $a_n>\frac{3-\sqrt{5}}{2}=\phi$ and as $2< \frac{3+\sqrt{5}}{2}$, $\{a_n\}$ is a decreasing sequence and also bounded below, so converges.

Now if $\{a_n\}$ converges the limit $a$ satisfies:

$
a=\frac{1}{3-a}
$

which has roots $\phi$ and $\phi^*.$ But it cannot converge on $\phi^*$ as the sequence is decreasing so the limit is $\phi$.

RonL