1. ## Sequence Proof

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.

2. Hello,

Use induction to prove it...

$\displaystyle a_1 \geq 2$

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

It's the same for $\displaystyle a_n > 0$

3. Originally Posted by FalconPUNCH!
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