I have to use the sequence defined by: $$x_{n+1} = \frac{ 1 }{ 4-x_n }$$ and $$x_1 = 3$$ in a proof, but before that I need to show this sequence converges. I assume the way to do this is (given where we are at in the book) to show that this sequence is monotone, and that is where I am stuck.
I've done some test values, and it appears to be decreasing, how do I prove this?

Question

I have to use the sequence defined by: $$x_{n+1} = \frac{ 1 }{ 4-x_n }$$ and $$x_1 = 3$$ in a proof, but before that I need to show this sequence converges. I assume the way to do this is (given where we are at in the book) to show that this sequence is monotone, and that is where I am stuck. 
I've done some test values, and it appears to be decreasing, how do I prove this?

anonymous · Answer

Any suggestions?

anonymous · Answer

Ok so, I tried using induction. So I wanted to show$$x_{n+1} \le x_n$$
so $$\frac{ 1 }{ 4-x_n } \le x_n$$
manipulating the inequality gave me a quadratic: $$x _{n}^2-4x_n+1 \le 0$$

anonymous · Answer

Do I need to apply the quadratic formula or something or am I completely off base?

anonymous · Answer

So that gives me roots at 2+/- sqrt(3)