Here I'd like to introduce:
$6-\sqrt{6-\sqrt{6-\sqrt{6-\sqrt{6}}}}$
If continuing the pattern indefinitely, the limit is $4$. The values are successively higher and lower than $4$.
In a recursive operation, this would be:
$a(n+1)=6-\sqrt{a(n)}$, with $a(1)=6-\sqrt6$.
How to find $a(n)$ in a compact form?
I am not sure that the limit of the sequence $a_1, a_2, a_3,\cdots$ is equal to the infinite nested radical you wrote at the beginning. It completely depends on how we define "infinite nested radicals", and I do not know anything about them. But I can show you that the recurrence relation $a_{n+1}=6-\sqrt{a_n},\quad a_1=6-\sqrt{6}$ defines a convergent sequence. I'll give you the outline of my solution with some missing details. Please let me know if anything is not clear.
First, note that $a_1\lt 4$ (and $a_2\gt 4$).Then, assuming $a_n\lt 4$ we can deduce that $a_{n+1}>4$ and $a_{n+2}<4.$ By mathematical induction, this is true for any $n\in\mathbb{N}^+.$
Also, the above result shows that we should look at the subsequences of odd and even indices separately. In this direction, it would be helpful to notice that $$a_{n+2}=6-\sqrt{6-\sqrt{a_n}}.$$ Now, it is an easy exercise to deduce $a_{n+4}\lessgtr a_{n+2}$ provided that $a_{n+2}\lessgtr a_{n},$ where $\lessgtr$ means one inequality at a time. Since $$a_1\approx3.5505, \qquad a_2\approx 4.1157, \qquad a_3\approx 3.9712, \qquad, a_4\approx 4.0071,$$ by mathematical induction we can claim that the subsequence of odd indices is increasing while the subsequence of even indices is decreasing.
By the Monotone convergent theorem, both subsequences are convergent but may have different limits. Those limits are given by the real-valued solutions to the equation $$x=6-\sqrt{6-\sqrt{x}}.$$ Guess what, the only real solution to this equation is $x=4.$