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Let $\{x_n\}$ be a sequence defined by $x_1 = 3, x_{n+1} = \frac{1}{4-x_n}$, for n $\geq 1$. prove it converges

(a) Prove that $\{x_n\}$ is decreasing and bounded (below by $0$ and above by $4$), i.e., prove that $0 < x_{n+1} < x_n < 4$, for all $n \in \mathbb N$.

(b) Does $\{x_n\}$ converge or diverge? Justify

My attempt: (Could someone check)

Claim: $\{x_n\}$ is bounded below 0 and bounded above by 4 (1), $\{x_n\}$ is strictly decreasing (2), for all $n \in \mathbb N$

Proof of (1)

WTS: $0 < x_n < 4, \forall n \in \mathbb N$

Since the function is not continuous at x = 4 I will find its limit for a better bound

$$x_n = \frac{1}{4-x_n} \leftrightarrow \left(x_{n} - \frac{4-\sqrt{12}}{2}\right)\left(x_n - \frac{4+\sqrt{12}}{2}\right)$$

Since the function is decreasing this will be the new bound $$0 < x_n < \frac{4-\sqrt{12}}{2}$$

I will prove this by induction:

Base case (n=1)

0 < 3 < $\frac{4-\sqrt{12}}{2} < 4$ holds

Inductive hypothesis: Let $k \in \mathbb N$ be arbitrary. Assume $0 < a_k < \frac{4-\sqrt{12}}{2}$

Inductive step: $\forall k \in \mathbb N$, $0 < a_k < \frac{4-\sqrt{12}}{2} \to 0 < a_{k+1} < \frac{4-\sqrt{12}}{2}$

$$a_{k+1} = \frac{1}{4-x_n} \text{ By definition}$$

$$< \frac{1}{4-\frac{4-\sqrt{12}}{2}} \text{ By I.H.}$$

$$= \frac{2}{4-\sqrt{12}}$$

$$= \frac{2(4+\sqrt{12})}{16-12}$$

$$= \frac{4+\sqrt{12}}{2} < 4$$

Now for bounded below

$$a_{k+1} = \frac{1}{4-x_n} \text{ By def}$$

$$> \frac{1}{4-0} \text{ By I.H.}$$

$$> 0$$

Therefore by induction $0 < x_n < \frac{4-\sqrt{12}}{2} < 4, \forall n \in \mathbb N$

Proof of (2)

WTS: $x_{n} > x_{n+1}, \forall n \in \mathbb N$

$$x_n > x_{n+1}$$

$$\leftrightarrow x_n > \frac{1}{4-x_n} \text{ By definition}$$

$$\leftrightarrow 0 > \left(x_{n} - \frac{4-\sqrt{12}}{2}\right)\left(x_n - \frac{4+\sqrt{12}}{2}\right)$$

By 1 we know that the first one is > 0 and the second < 0. Then (+)(-) = - < 0

Therefore it is decreasing. For (b) by the bounded monotone convergence theorem we know that this converges since it is bounded below and strictly decreasing.

Is this right?

I think I did the choosing the limit part wrong. If its decreasing I choose the smaller one as the upperbound right?

Tinler
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  • the Limit is $$2-\sqrt{3}$$ this is the solution of the equation$$ x=\frac{1}{4-x}$$ – Dr. Sonnhard Graubner Apr 19 '17 at 19:53
  • Why is $x_n - \dfrac{4 - \sqrt{12}}{2} > 0$? – Umberto P. Apr 19 '17 at 19:53
  • hmm. Would this be correct $$\frac{4-\sqrt{12}}{2} < x_n < \frac{4+\sqrt{12}}{2}$$ ? I used the wrong upperbound I think. I was thinking it like that when I was doing the decreasing part and this is obvious it was $(+)(-) = -$ – Tinler Apr 19 '17 at 19:57
  • It is simpler to observe that $0 \le x \le 1$ implies $3 \le 4-x \le 4$ and $0 < \frac 14 \le \frac{1}{4-x} \le \frac 13 < 1$. The second term of the sequence satisfies this inequality. – Umberto P. Apr 19 '17 at 20:01
  • As for monotonicity, observe that $x_1 > x_2$, and $$x_n > x_{n+1} \implies 4 - x_{n+1} > 4 - x_n \implies \frac{1}{4-x_n} > \frac{1}{4-x_{n+1}}$$ provided that each $x_n$ is less than $4$. – Umberto P. Apr 19 '17 at 20:02
  • Do you know how to choose the bounds though? Like If I take the limit of a function and it gives 2 values. If its decreasing then what would be the upper bound / lower bound. Also what to pick if increasing – Tinler Apr 19 '17 at 20:11

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