sequences and series - Let $x_1 = 3, x_{n+1} = frac{1}{4-x_n}, text{for } n geq 1$. Use induction to show that $0 < x_{n+1} < x_n < 4, forall n in mathbb{N}$.

Let

$$x_1 = 3, x_{n+1} = \frac{1}{4-x_n}, \text{for } n \geq 1$$

Use induction to show that $0 < x_{n+1} < x_n < 4, \forall n \in \mathbb{N}$.

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Let $S(n)$ be the proposition that $0 < x_{n+1} < x_n < 4$.

Base Case:

$S(1)$, and thus $x_1 = 3$, $x_2 = \frac{1}{4-3} = 1$. We have $ 0 < 1 < 3 < 4$, thus $S(1)$ holds.

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Inductive Hypothesis (strong):

Assume that $S(n)$ holds for $1 \leq i \leq n$.

Assume $S(i)$ holds therefore. So, $0 < x_{i+1} < x_i < 4$

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Inductive Step:

Show that $S(i + 1)$ holds.

This is where I am stuck on the IS. Any suggestions?

Answer

What you really need to prove by induction is the stronger statement that $x_n$ is decreasing and:
$$S(n):2-\sqrt 3

$S(1)$ is obviously true. Then if $S(n-1)$ is true ($2-\sqrt 3

$$x_{n+1}=\frac{1}{4-x_n}>\frac{1}{4-(2-\sqrt 3)}=2-\sqrt 3$$

$$x_n-x_{n+1}=x_n-\frac{1}{4-x_n}=\frac{-x_n^2+4x_n-1}{4-x_n}=\frac{3-(x_n-2)^2}{4-x_n}>0$$

since ${4-x_n}>0$ (obviously), and $3-(x_n-2)^2>0$ is equivalent to $2-\sqrt 3

So you proved that $2-\sqrt 3