Let $\displaystyle x_1 = 3$ and $\displaystyle x_{n+1} = (4-x_n)^{-1}$
Determine the convergence of the sequence.
Thanks Plato.
We have $\displaystyle 0<x_2 = 1 = \frac{1}{4-3}<3=x_1<4$. Suppose $\displaystyle 0<x_n<\cdots < x_1<4$
then $\displaystyle x_{n+1} = 1/(4-x_n) > 0$ and $\displaystyle x_{n+1}-x_n = \frac{1}{4-x_n} -\frac{1}{4-x_{n-1}}=\frac{x_n-x_{n-1}}{(4-x_n)(4-x_{n-1})}<0$
Thus $\displaystyle 0<x_{n+1}<x_n<\cdots < x_1<4$ and so $\displaystyle \{x_n\}$ is bounded and decreasing.
Therefore it's convergent and $\displaystyle L=\lim_{n \to \infty} x_n = \frac{1}{4-L}$. Now $\displaystyle L<x_2 =1$ implies that $\displaystyle L = 2-\sqrt{3}$