proof of the convergence of a recursive sequence

Question

The following recursive sequence seems to converge towards 8. How could you prove this?

$$a_{0} = 0$$ $$a_{n+1} = 4\cdot\sqrt[4]{8+a_{n}}$$

Appreciate any responses.

Answer 1

Claim: The sequence is bounded above by 8.

Proof by induction:

This holds for $a_0$ trivially. Now, assuming $a_n < 8$, we obtain

$$a_{n+1} = 4\cdot \sqrt[4]{8+a_{n}} < 4\cdot \sqrt[4]{8+8} = 4\cdot 2=8$$

Hence, $a_n<8$ for every term.

Now, we show that $a_{n+1}-a_{n}\geq 0$:

$$\iff 4\cdot \sqrt[4]{8+a_{n}}-a_{n}\geq0$$

$$\iff 4^4(8+a_n)\geq a_{n}^{4}$$

$$\iff a_{n}^{4}-4^4a_n-8\cdot 4^4 \leq0$$

($a_n=0$ is a solution to this inequality.)

We quickly verify that $x=8$ solves the polynomial equation $x^{4}-4^4x-8\cdot 4^4 =0$ and that its derivative $4x^3-256$ has only one root ($x=4$). Furthermore, the derivative is negative given $x<4$, hence the polynomial is decreasing through $0\leq x < 4$. The derivative is positive given $x>4$, therefore the polynomial is increasing through $4<x<8$. As previously noted, $x=0 \Rightarrow x^{4}-4^4x-8\cdot 4^4 \leq 0$, so the inequality holds given $0\leq x \leq 4$ because the function is decreasing. For similar reasons, the inequality also holds given $4<x\leq8$. Therefore, the inequality holds given $0\leq x \leq 8$.

Thus, for any $0\leq a_{n}\leq 8$ we have $a_{n+1}-a_{n}\geq 0$. Because we start initially at $a_0=0$, the sequence is increasing.

The only solution to $4\cdot \sqrt[4]{8+a_{n}}-a_{n}=0$ in $\mathbb{R}^{+}$ is $a_{n}=8$, so it must be the least upper bound.

For any bounded increasing sequence, its limit is equal to its least upper bound.

Therefore the sequence $\{a_{n}\}$ converges to 8.

QED.