Limit of this recursive sequence and convergence

Question

$$a_{n+1}=\sqrt{4a_n+3}$$ $a_1=5$
I can solve simpler but I get stuck here because I cant find an upper bound or roots of the quadratic equation $a_{n+1} -a_n= \frac{4a_n+3 - a_n^2}{\sqrt{4a_n +3}+a_n}...$ to find monotony.
I tried this generic aproach but have difficulties Convergence and limit of a recursive sequence

Answer 1

Prove it using induction. Whether the sequence is increasing or decreasing depends on the value of $a_1$. Observe that $$a_n \le a_{n+1} \implies 4a_n + 3 \le 4a_{n+1} + 3 \implies a_{n+1} \le a_{n+2}$$ and similarly $$a_n \ge a_{n+1} \implies 4a_n + 3 \ge 4a_{n+1} + 3 \implies a_{n+1} \ge a_{n+2}.$$ Thus if $a_1 \le a_2$ the full sequence is nondecreasing and if $a_1 \ge a_2$ the full sequence is nonincreasing.

Answer 2

If you assume that the limit exists, then $$ L=\lim_{n\to +\infty}a_n $$ must satisfy: $$ L^2 = 4L+3,\qquad L\geq 0, $$ hence the limit, if existing, is $2+\sqrt{7}$. Consider now that @Umberto P.'s answer gives that the sequence $\{a_n\}_{n\in\mathbb{N}}$ is monotonic.

Answer 3

By the continuity of our recursive definition: if this sequence converges to a limit $a$, then that limit will be a solution to the equation $$ a = \sqrt{4a + 3} $$ So, solving for $a$, we have $$ a^2 - 4a - 3 = 0 \implies\\ a = \frac{4 \pm\sqrt{4^2 +4\cdot 3}}{2} = 2 \pm \sqrt{7} $$ Plugging into the original equation, we find that $2 - \sqrt{7}$ is an extraneous root, so that the only possibility would be $a = 2 + \sqrt{7}$.

It remains to be shown that the sequence does, in fact, converge to this limit (which the other answers have covered).

An idea to add to your attempts: try using the fact that $a_n$ will always be in an interval of the form $(a,a + \epsilon)$.