Find the sum $\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\sqrt{41+\cdots}}}}}$
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
$$ F(x)^2 = p(x) + F(x+1). $$
Now we make an ansatz that $F(x)$ takes the form $F(x) = ax + b$. Plugging this and comparing coefficients shows that
$$ F(x) = x + 2 $$
solves the functional equation. Finally, since $(p(1), p(2), p(3), \cdots) = (5, 11, 19, \cdots) $, the infinite radical in question corresponds to the case $x = 1$, giving
$$ \sqrt{5 + \sqrt{11 + \sqrt{19 + \cdots}}} = F(1) = 3. $$
Rigorous justification. Let $\mathcal{C}$ be the set of all continuous functions $f : [0, \infty) \to \mathbb{R}$ such that
$$ \| f\| := \sup_{x\to\infty} \left( 2^{-x/2} |f(x)| \right) $$
is finite. Notice that $\mathcal{C}$ is a complete normed space with respect to $\|\cdot\|$. Write $p(x) = x^2 + 3x + 1$ and define
$$\mathcal{A} = \{ f \in \mathcal{C} : f(x) \geq 0 \text{ for all } x \geq 0 \}. $$
This is a closed subset of $\mathcal{C}$. Now define $\Phi : \mathcal{A} \to \mathcal{A}$ by
$$ \Phi[f](x) = \sqrt{p(x) + f(x+1)}. $$
If $f \in \mathcal{A}$, then $ 2^{-x/2}|\Phi[f](x)| \leq 2^{-x/2}\sqrt{p(x) + 2^{(x+1)/2}\|f\|} $ shows that $\|\Phi[f]\| < \infty$, hence $\Phi$ is well-defined. Moreover, if $f, g \in \mathcal{A}$, then
\begin{align*} 2^{-x/2} \left| \Phi[f](x) - \Phi[g](x) \right| &= 2^{-x/2} \cdot \frac{\left| f(x+1) - g(x+1) \right|}{\sqrt{p(x) + f(x+1)} + \sqrt{p(x) + g(x+1)}} \\ &\leq 2^{-x/2} \cdot \frac{2^{(x+1)/2} \| f - g \|}{2} \\ &= \frac{1}{\sqrt{2}} \| f - g \|. \end{align*}
So $\Phi$ is a contraction mapping on $\mathcal{A}$, and hence, by the contraction mapping theorem,
- 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 $\Phi[F] = F$, and therefore, $$ \forall f \in \mathcal{A} \ : \quad \lim_{n\to\infty} \Phi^{\circ n}[f](x) = x+2 $$
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.
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.
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, $$4 = \sqrt{11 + \sqrt{19 + \sqrt{29 + \sqrt{41 + 8}}}}.$$