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.
 A: Maybe works, $$3=\sqrt{3^{2}}=\sqrt{5+4}=\sqrt{5+\sqrt{16}}=\sqrt{5+\sqrt{11+5}}$$
$$=\sqrt{5+\sqrt{11+\sqrt{25}}}=\sqrt{5+\sqrt{11+\sqrt{19+6}}}=\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{36}}}}$$
$$=\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+7}}}}=\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\sqrt{49}}}}}=\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\sqrt{41+8}}}}}=\ldots$$
This is just a beautiful way of writing 3.
A: 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, $$4 = \sqrt{11 + \sqrt{19 + \sqrt{29 + \sqrt{41 + 8}}}}.$$
A: Heuristic Idea. We may adopt the technique for Ramanujan's infinite radical.
Let $p(x) = x^2 + 3x + 1$ and define $F : [0, \infty) \to [0, \infty)$ by
$$ F(x) = \sqrt{p(x) + \sqrt{p(x+1) + \sqrt{p(x+2) + \cdots }}} $$
Then $F$ solves the functional equation
$$ F(x)^2 = p(x) + F(x+1). $$
Now we make an ansatz that $F(x)$ takes the form $F(x) = ax + b$. Plugging this into the functional equation and comparing the coefficients shows that
$$ F(x) = x + 2. $$
Finally, since $(p(1), p(2), p(3), \ldots) = (5, 11, 19, \ldots) $, the infinite radical in question corresponds to the case $x = 1$. Therefore we conclude
$$ \sqrt{5 + \sqrt{11 + \sqrt{19 + \cdots}}} = F(1) = 3. $$

Justification. Fix a constant $a \in (1, 2) $. Also, let $\mathcal{C}$ denote the set of all continuous functions $f : [0, \infty) \to \mathbb{R}$ such that $f(x) = \mathcal{O}(a^x)$ as $x \to \infty$. Then $\mathcal{C}$ is a Banach space with respect to the norm $\|\cdot\|$ defined by
$$ \| f\| := \sup_{x \geq 0} \left( a^{-x} |f(x)| \right). $$
Now, write $p(x) = x^2 + 3x + 1$ and define the subset $\mathcal{A}$ of $\mathcal{C}$ by
$$\mathcal{A} = \{ f \in \mathcal{C} : f(x) \geq 0 \text{ for all } x \geq 0 \}. $$
This is a closed subset of $\mathcal{C}$. Now define the operator $\Phi : \mathcal{A} \to \mathcal{A}$ by
$$ \Phi[f](x) = \sqrt{p(x) + f(x+1)}. $$
$\Phi$ is indeed a well-defined operator on $\mathcal{A}$, for if $f \in \mathcal{A}$, then we have
$$ \limsup_{x\to\infty} a^{-x} |\Phi[f](x)|
\leq \lim_{x\to\infty} a^{-x} \sqrt{p(x) + a^{x+1}\|f\|}
= 0 $$
and hence $\| \Phi[f] \| < \infty$. Moreover, if $f, g \in \mathcal{A}$, then
\begin{align*}
a^{-x} \left| \Phi[f](x) - \Phi[g](x) \right|
&= a^{-x} \cdot \frac{\left| f(x+1) - g(x+1) \right|}{\sqrt{p(x) + f(x+1)} + \sqrt{p(x) + g(x+1)}} \\
&\leq a^{-x} \cdot \frac{a^{x+1} \| f - g \|}{2} \\
&= \frac{a}{2} \| f - g \|.
\end{align*}
By the choice of $a$, $\Phi$ is a contraction mapping on $\mathcal{A}$. So, by the contraction mapping theorem, we conclude:

*

*There exists a unique $F \in \mathcal{A}$ for which $\Phi[F] = F$, and

*Such $F$ is realized as the limit $\Phi^{\circ n}[f]$ as $n\to\infty$ for arbitrary initial choice $f \in \mathcal{A}$.

Finally, we already know that $F(x) = x+2$ is an element of $\mathcal{A}$ that solves $\Phi[F] = F$. Therefore, for any $ f \in \mathcal{A}$, we have
$$ \lim_{n\to\infty} \Phi^{\circ n}[f](x) = x+2. $$
A: 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 $-1 + \sqrt{1 + \sqrt 5} > 0$.
