The positive sequence: $a_{n+1} =\sqrt{1 + \frac{a^{2}_{n}}{4}}$. Is strictly increasing and bounded above?

Question

How can I prove that this positive sequence, with $0 \leq a_{0} < \frac{2}{\sqrt{3}}$:

$a_{n+1} =\sqrt{1 + \frac{a^{2}_{n}}{4}}$ is strictly increasing and bounded above?

Answer 1

Bounded: We need to show that for any $n \ge 0$, $0 \le a_n < \frac{2}{\sqrt{3}}$. This can be done by induction. Our base case is given in the prompt. Now assume that $0 \le a_k < \frac{2}{\sqrt{3}}$. Then $$0 \le a_k < \frac{2}{\sqrt{3}} \implies 0< 1\le\sqrt{1 + \frac{a_k^2}{4}} <\sqrt{1+\frac{1}{3}}= \frac{2}{\sqrt{3}}.$$ Therefore, $0 \le a_{k+1} < \frac{2}{\sqrt{3}}$. Therefore, the induction hypothesis holds and $0 \le a_n < \frac{2}{\sqrt{3}}$ for any $n \ge 0$.

Strictly increasing: Show that $a_{n+1} > a_n$. By applying the definition of $a_{n+1}$, $$a_{n+1} =\sqrt{1 + \frac{a_n^2}{4}} >a_n \underset{a_n\ge0}{\iff} 1+\frac{a_n^2}{4} > a_n^2 \iff \frac{3a_n^2}{4}<1 \underset{a_n\ge0}{\iff} a_n <\frac{2}{\sqrt{3}}.$$ Therefore, $\{a_n\}$ is strictly increasing.

Therefore, $\{a_n\}$ is convergent. Let $a_n \to L$ as $n \to \infty$. Then $$\lim_{n \to \infty} a_n = \lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \sqrt{1 + \frac{a_n^2}{4}} = \sqrt{1 + \frac{L^2}{4}} = L \implies L = \frac{2}{\sqrt{3}}.$$ Therefore, $$\lim_{n \to \infty} a_n = \frac{2}{\sqrt{3}}.$$

Answer 2

Hint: By induction $a_n < \frac 2 {\sqrt 3}$ for all $n$ and this implies $a_n^{2} <1+\frac {a_n^{2}} 4$ which gives $a_n <a_{n+1}$.