Nested radicals like Ramanujan's infinite radicals

$$2=\sqrt{1+\sqrt9}=\sqrt{1+\sqrt{5+\sqrt{16}}}=\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{25}}}}=$$ $$=\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{36}}}}}=\sqrt{1+\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\sqrt{49}}}}}}=...$$ $$=\sqrt{1+\sqrt{5+\sqrt{11+...+\sqrt{n^2+n-1+\sqrt{(n+2)^2}}}}}=$$ $$=\sqrt{1+\sqrt{5+\sqrt{11+...+\sqrt{n^2+n-1+\sqrt{(n+1)^2+(n+1)-1+\sqrt{(n+3)^2}}}}}}=...$$

We can provide bounds by 'unwinding' from the back. Let $b_{m,n}$ be the 'partial root' $\displaystyle\sqrt{t_m+\sqrt{t_{m+1}+\sqrt{\ldots+\sqrt{t_n}}}}$, where $t_m=m^2-m-1$ is the $m$'th term in the original series; your $a_n=b_{2,n}$. Then the key insight is that we have $\lim_{n\to\infty} b_{m,n}=m$, and that approximation 'gets better' for a fixed $n$ as $m$ gets smaller. Let's start with $b_{n,n}=\sqrt{n^2-n-1}=n\sqrt{1-\frac{1+n}{n^2}}$ $\gt n-\epsilon_{n,n}$, where $\epsilon_{n,n}=\frac{1+n}{n}$. Now, we can look at $b_{(n-1),n}$: $b_{(n-1),n}=\sqrt{(n-1)^2-(n-1)-1+b_{n,n}}$ $\gt\sqrt{(n-1)^2-\epsilon_{n,n}}$ $=\displaystyle (n-1)\sqrt{1-\frac{\epsilon_{n,n}}{(n-1)^2}}$ $\gt (n-1)-\epsilon_{(n-1),n}$ with $\epsilon_{(n-1),n}=\dfrac{\epsilon_{n.n}}{n-1}$ $=\dfrac{1+n}{n(n-1)}$. Going on, we can see that $b_{m,n}\gt m-\epsilon_{m,n}$ where $\epsilon_{m,n}=\dfrac{1+n}{n(n-1)\ldots m}$. Finally, $b_{2,n}\gt 2-\dfrac{1+n}{n!}$. Since $m$ is an upper bound for $b_{m,n}$ (and in particular, $2$ is an upper bound for your radicals — this can be proved by induction), this gives the limit you're after.