Find the sum $\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\sqrt{41+\cdots}}}}}$ Okay so this can be written as $$\sqrt{5+\sqrt{(5+6)+\sqrt{(5+6+8)+\sqrt{(5+6+8+10)+\sqrt{(5+6+8+10+12)\cdots}}}}}$$
Putting it as $y$ and squaring both sides doesn't seem to help, and I don't know what else can be done.

This is a slightly more rigorous form of @Pablo_'s excellent insight. @Sangchul Lee covers the full, analytic answer.
Set $a_n = n^2 + 5n + 5$ for $n \geq 0$. This sequence gives the coefficients of the "infinite radical." Rather than consider the full infinite radical, consider the "partial radicals," defined as $$r_n = \sqrt{a_0 + \sqrt{a_1 + \cdots + \sqrt{a_n + (4 + n)}}}.$$
As Pablo_Lee notes, $r_n = 3$ for all $n$. To see this, observe that $a_n + (n + 4) = (n + 3)^2$. This allows us to "unroll" the radical back to $a_0$. For example, $$a_{n - 1} + \sqrt{a_n + (n + 4)} = a_{n - 1} + n + 3 = a_{n - 1} + ((n - 1) + 4) = ((n - 1) + 3)^2.$$ Therefore,
$$
\begin{align*}
    r_n &= \sqrt{a_0 + \sqrt{a_1 + \cdots + \sqrt{a_{n - 1} + \sqrt{a_n + (n + 4)}}}} \\
        &= \sqrt{a_0 + \sqrt{a_1 + \cdots + \sqrt{a_{n - 1} + ((n - 1) + 4)}}} \\
        &= \sqrt{a_0 + \sqrt{a_1 + \cdots + \sqrt{a_{n - 2} + ((n - 2) + 4)}}} \\
        &\vdots \\
        &= \sqrt{a_0 + \sqrt{a_1 + 5}} \\
        &= \sqrt{a_0 + 4} \\
        &= \sqrt{(0 + 3)^2} \\
        &= 3.
\end{align*}
$$
(There is likely a snappy way to do this by induction, but I don't see it yet.)
If we are willing to define the full radical as $\lim_{n \to \infty} r_n$, then this should also be an acceptable answer.
Now canonically, the expression $\sqrt{a_0 + \sqrt{a_1 + \sqrt{a_2 + \cdots}}}$ (note that the ellipsis appears in a sum) would refer to the sequence $$r'_n = \sqrt{a_0 + \sqrt{a_1 + \cdots + \sqrt{a_n + 0}}}.$$  But since $r'_n \le r_n$, convergence follows immediately from the monotone convergence theorem.

Edit: For any integer $r \geq 2$, setting $p_n = n^2 + (2r - 1)n + r^2 - r - 1$ and $q_n = n + r + 1$ should yield, through the same arguments, $$r = \sqrt{p_0 + \sqrt{p_1 + \cdots + \sqrt{p_n + q_n}}}$$ for all $n \geq 0$. Note that $p_n$ is merely a shifted form of the Fibonacci polynomial $n^2 - n - 1$ at integer values.
For example, $$4 = \sqrt{11 + \sqrt{19 + \sqrt{29 + \sqrt{41 + 8}}}}.$$
