<s>[INST] Convergence/Divergence of a particular infinite nested radical Is it known if the following infinite nested radical converges or diverges (for $n \in \mathbb N$)?:
$$R(n) = \sqrt{n+\sqrt{(n+1)+\sqrt{(n+2)+ \cdots}}}$$
I recently became interested in these problems when reading about Ramanujan's (famous?) formula: $$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\cdots}}}} = 3$$ Which he showed to be a special case of: $$\sqrt{ax+(n+a)^2+x\sqrt{a(x+n)+(n+a)^2+(x+n)\sqrt{\cdots}}} = x+n+a$$
My formula does not fit into that of Ramanujan and the farthest I am able to get is:
$$R^2(n) = n + R(n+1)$$ which I prefer to rewrite as $$R(n+1) = R^2(n) - n$$
This is well...not far at all.  Knowing $R(0)$ would supply me with all values of $R(n)$ and so I end up needing to solve $$\sqrt{\sqrt{1+\sqrt{2+ \sqrt{3+\cdots}}}}$$
I feel that the general method for solving these problems is currently beyond me without any instruction.  Would anyone be able to offer hints and get me a little bit further down the path?
 [/INST] If you accept
$$
\phi=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{\cdots}}}}=\sqrt{1+\phi}=\frac{1+\sqrt{5}}{2}
$$
then
$$
\begin{align}
\phi < R(1) &< \sqrt{1+\sqrt{2+\sqrt{2^2+\sqrt{2^4+\sqrt{2^8+\cdots}}}}}\\
&=\sqrt{1+\sqrt{2}\phi}
\end{align}
$$
so
$1.615<R(1)<1.815$ and as you say you can get the other $R(n)$ from there.
Edit: You can use this technique for any $n$:
$$
\phi_n=\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{\cdots}}}}=\sqrt{n+\phi_n}=\frac{1+\sqrt{4n+1}}{2} \\
\phi_n < R(n) < \sqrt{n+\sqrt{\frac{n+1}{n}}\phi_n} \\
n+\frac{1}{2}+\sqrt{n+\frac{1}{4}} < R^2(n) < n+\frac{1}{2}\sqrt{1+\frac{1}{n}}+\sqrt{n+\frac{5}{4}+\frac{1}{4n}}
$$
</s>