The problem is the following:
$$ a_1 = \sqrt6 \ \text{ and } \ a_{n+1}=\sqrt{6 +a_n} \ \\ \ \\ \text{Find} \ \lim_{n\to \infty}a_n $$
I tried dividing $a_{n+1}$ by $a_n$, which is bigger than one. I assume one can use the comparison test to conclude that this limit goes to infinity? But how should I go about finding the limit if I don't have a calculator to see that $\dfrac{a_{n+1}}{a_n}$ = 1,18... ?
We can prove by induction that $a_n < 3$ for any positive integer $n$. Indeed $a_1 =\sqrt{6}<3$ and if $a_k<3$, then $a_{k+1} = \sqrt{a_k+6}<\sqrt{3+6}=3$, for some $k$. It is also obvious that $a_n \geq \sqrt{6}$ for any $n$. Therefore the sequence is bounded.
From the boundedness, we can also prove that it is increasing because
$$a_n < a_{n+1} \Leftrightarrow a_n < \sqrt{6+a_n} \Leftrightarrow a_n^2 < 6+a_n\Leftrightarrow (a_n-3)(a_n+2)<0 $$
which is obvious from the boundedness. Therefore the sequence is convergent and denote $l$ its limit. Now pass to limit in the recurrence formula to get:
$$l=\sqrt{6+l} \Rightarrow l = 3$$
If this series is ever to converge to a value $L$, then we'd have:
$$ L = \sqrt{6+L} $$ Which implies $L>0$ and then $L=3$.
Because any $a_n$ is non-negative the sequence is limited, it remains to show that it does converge, you can check it by seeing that it is monotonic.
