# Thread: Determine the convergence of a recursive sequence

1. ## Determine the convergence of a recursive sequence

Let $x_1 = 3$ and $x_{n+1} = (4-x_n)^{-1}$

Determine the convergence of the sequence.

2. Can you show that this sequence is decreasing and bounded?

3. Thanks Plato.

We have $0. Suppose $0
then $x_{n+1} = 1/(4-x_n) > 0$ and $x_{n+1}-x_n = \frac{1}{4-x_n} -\frac{1}{4-x_{n-1}}=\frac{x_n-x_{n-1}}{(4-x_n)(4-x_{n-1})}<0$
Thus $0 and so $\{x_n\}$ is bounded and decreasing.
Therefore it's convergent and $L=\lim_{n \to \infty} x_n = \frac{1}{4-L}$. Now $L implies that $L = 2-\sqrt{3}$