Convergence/Divergence of a particular infinite nested radical

Question

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?

Answer 1

Define $$ S(n,m)=\sqrt{n+\sqrt{n+1+\sqrt{n+2+\dots+\sqrt{m-1+\sqrt{m}}}}}\tag{1} $$ Note that for $m\ge n\ge1$, we have $S(n,m)\ge1$ and $$ \sqrt{n+x}-\sqrt{n}=\frac{x}{\sqrt{n+x}+\sqrt{n}}\tag{2} $$ Suppose that $S(n+1,m)\le\sqrt{n+1}+1$. Then by $(2)$, $$ \begin{align} S(n,m) &=\sqrt{n+S(n+1,m)}\\[6pt] &=\sqrt{n}+\frac{S(n+1,m)}{\sqrt{n+S(n+1,m)}+\sqrt{n}}\\ &\le\sqrt{n}+\frac{\sqrt{n+1}+1}{\sqrt{n+1}+1}\\[9pt] &=\sqrt{n}+1\tag{3} \end{align} $$ Since $S(m,m)=\sqrt{m}\le\sqrt{m}+1$ for all $m$, we have shown that for $m\ge n\ge1$, $$ S(n,m)\le\sqrt{n}+1\tag{4} $$ For a given $n$, $S(n,m)$ is an increasing sequence in $m$ that is bounded above by $\sqrt{n}+1$, therefore its limit exists and $$ \lim_{m\to\infty}S(n,m)\le\sqrt{n}+1\tag{5} $$

Answer 2

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 $n$: $$ \phi_n=\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{\cdots}}}}=\sqrt{n+\phi_n}=\frac{1+\sqrt{4n+1}}{2} \\ \phi_n < R(n) < \sqrt{n+\sqrt{\frac{n+1}{n}}\phi_n} \\ n+\frac{1}{2}+\sqrt{n+\frac{1}{4}} < R^2(n) < n+\frac{1}{2}\sqrt{1+\frac{1}{n}}+\sqrt{n+\frac{5}{4}+\frac{1}{4n}} $$