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.

Set
$ p(x) = x^2 - 3x + 1$
and observe that
$ p(x) + x = (x - 1)^2.$
Then for $r \ge 0$,
$$r = \sqrt{r^2} \\ 
= \sqrt{p(r + 1) + r + 1} \\
= \sqrt{p(r + 1) + \sqrt{(r + 1)^2}} \\
= \sqrt{p(r + 1) + \sqrt{p(r + 2) + r + 2}} \\
= \sqrt{p(r + 1) + \sqrt{p(r + 2) +  \sqrt{(r + 2)^2}}} \\
= \sqrt{p(r + 1) + \sqrt{p(r + 2) +  \sqrt{p(r + 3) + r + 3}}} \\
= \sqrt{p(r + 1) + \sqrt{p(r + 2) +  \sqrt{p(r + 3) + \sqrt{p(r + 4) + r + 4}}}} \\
= \sqrt{p(r + 1) + \sqrt{p(r + 2) +  \sqrt{p(r + 3) + \sqrt{p(r + 4) + \sqrt{p(r + 5) + r + 5}}}}} \\
= \ldots
$$
Now define for $q \ge 0$:
$$ F(r,q,x_0) = \begin{cases}
\sqrt{p(r + 1) + F(r + 1, q, x_0)}, & \mbox{if } r + 1 \le q\\
x_0, & \mbox{if }  r + 1 \gt q.
\end{cases}
$$
So in particular:
$$\begin{array}{llll} 
 & F(r, r + 1, 0) &=& \sqrt{p(r + 1) + 0} \\ 
<& F(r, r + 2, 0) &=& \sqrt{p(r + 1) + \sqrt{p(r + 2) + 0}} \\
<& F(r, r + 3, 0) &=& \sqrt{p(r + 1) + \sqrt{p(r + 2) + \sqrt{p(r + 3) + 0}}} \\
<& \lim_{n \rightarrow \infty} F(r, r + n, 0) &=_{\mbox{def}}& \sqrt{p(r + 1) + \sqrt{p(r + 2) + \sqrt{p(r + 3) + \ldots}}}
\end{array}$$
and also
$$\begin{array}{llll}
F(r, r + 1, 0) &< F(r, r + 1, r + 1) &= \sqrt{p(r + 1) + r + 1} &= r\\
F(r, r + 2, 0) &< F(r, r + 2, r + 2) &= \sqrt{p(r + 1) + \sqrt{p(r + 2) + r + 2}} &= r \\
F(r, r + 3, 0) &< F(r, r + 3, r + 3) &= \sqrt{p(r + 1) + \sqrt{p(r + 2) +  \sqrt{p(r + 3) + r + 3}}} &= r \\
F(r, r + n, 0) &< F(r, r + n, r + n) &= \ldots &= r.
\end{array}$$
So by the monotone convergence theorem, the limit
$$F(r, \infty, 0) =_{\mbox{def}} \sqrt{p(r + 1) + \sqrt{p(r + 2) + \sqrt{p(r + 3) + \ldots}}}$$
actually exists (and it's at most $r$). As it exists, it must satisfy the equation
$$F(r, \infty, 0) = \sqrt{p(r + 1) + F(r + 1, \infty, 0)}.$$
Now this is clearly satisfied for $F(r, \infty, 0) = r$, but that doesn't mean (yet) that $r$ is the answer. For $r = 0$, of course, we know
$$0 \le F(0,\infty,0) \le 0.$$
We also know
$$F(r + 1, \infty, 0) = F^2(r, \infty, 0) - p(r + 1).$$
If we use $F(r, \infty, 0) = r$ as an induction hypothesis, we obtain
$$\begin{array}{ll}
F(r + 1, \infty, 0) &= r^2 - p(r + 1)\\
&= r^2 - ((r + 1 - 1)^2 - (r + 1))\\
&= r + 1\\
\end{array}$$
for $r\in\mathbb N$ #.
Note on $0 \le r \lt \frac{1 + \sqrt 5}{2}$. When $p(r + 1) < 0$, we have to resort to the complex square root to make everything work. In that case, the monotonous convergence theorem has to be applied separately to the real and imaginary part of $F$. In particular (after a few terms), the real part increases towards $r$ (as expected), while the imaginary part decreases towards $0$.
$$\begin{array}{rcl}
F(0,1,0) &=& i\\
F(0,2,0) &=& e^{3\pi i/8}\\
F(0,3,0) &=& i\\
F(0,4,0) &\approx& 0.3259i\\
F(0,5,0) &\approx& 0.1062i\\
F(0,6,0) &\approx& 0.04115i\\
F(1,2,0) &=& i\\
F(1,3,0) &=& 0\\
F(1,4,0) &\approx& 0.8938\\
F(1,5,0) &\approx& 0.9852\\
F(1,6,0) &\approx& 0.9983
\end{array}$$
A key observation here is that $-1 + \sqrt{1 + \sqrt 5} > 0$.
