Prove the positive sequence $a_{n+1} = \sqrt{1+\frac{a^2_n}{4}} $ is strictly increasing for $0 \leq a_0<\frac{2}{\sqrt{3}}$ My attempt: 
$a_{n+1} - a_n=  \sqrt{1+\frac{a^2_n}{4}} -a_n > \frac{a_n}{2}-a_n = -\frac{1}{2}a_n$, which doesn't tell me any thing. How do I prove that this sequence is strickly increasing? Thank you! 

$a_n^2=1+\frac{a_{n-1}^2}{4}=\frac{1}{4}\left(4+a_{n-1}^2\right)=\frac{1}{4^2}\left(4^2+1\cdot4+a_{n-2}^2\right)=\frac{1}{4^3}\left(4^3+4^2+4+a_{n-3}^2\right)$
Hence, a recursive formula can be written as follows: $$ a_{n+1}^2=\frac{1}{4^n}\left(4^n+4^{n-1}+...4+a_1\right)=\frac{4}{3}\cdot\frac{4^n-1}{4^n}+\frac{a_1^2}{4^n}$$.
$\therefore\lim\limits_{n\to\infty}a_n=\frac{2}{\sqrt3}$ irrespective of the choice of $a_1$. 
Also $a_{n+1}^2-a_n^2=\frac{4-3a_n^2}{4}>0 \forall a_n<\frac{2}{\sqrt3}$. 
Can you do the rest?
