Determine the convergence of a recursive sequence

**elim** · September 26th 2010, 03:38 PM

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

Determine the convergence of the sequence.

**Plato** · September 26th 2010, 04:02 PM

Can you show that this sequence is decreasing and bounded?

**elim** · September 26th 2010, 09:41 PM

Thanks Plato.

We have $0<x_2 = 1 = \frac{1}{4-3}<3=x_1<4$ . Suppose $0<x_n<\cdots < x_1<4$
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 $0<x_{n+1}<x_n<\cdots < x_1<4$ 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 $L<x_2 =1$ implies that $L = 2-\sqrt{3}$

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