How can I show that $\sqrt{1+\sqrt{2+\sqrt{3+\sqrt\ldots}}}$ exists? I would like to investigate the convergence of
$$\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\sqrt\ldots}}}}$$
Or more precisely, let $$\begin{align}
a_1 & = \sqrt 1\\
a_2 & = \sqrt{1+\sqrt2}\\
a_3 & = \sqrt{1+\sqrt{2+\sqrt 3}}\\
a_4 & = \sqrt{1+\sqrt{2+\sqrt{3+\sqrt 4}}}\\
&\vdots
\end{align}$$
Easy computer calculations suggest that this sequence converges rapidly to the value 1.75793275661800453265, so I handed this number to the all-seeing Google, which produced:


*

*OEIS A072449

* "Nested Radical Constant" from MathWorld
Henceforth let us write  $\sqrt{r_1 + \sqrt{r_2 + \sqrt{\cdots + \sqrt{r_n}}}}$ as $[r_1, r_2, \ldots r_n]$ for short, in the manner of continued fractions.
Obviously we have $$a_n= [1,2,\ldots n] \le \underbrace{[n, n,\ldots, n]}_n$$ 
but as the right-hand side grows without bound (It's $O(\sqrt n)$) this is unhelpful.  I thought maybe to do something like:
$$a_{n^2}\le [1, \underbrace{4, 4, 4}_3, \underbrace{9, 9, 9, 9, 9}_5, \ldots,
\underbrace{n^2,n^2,\ldots,n^2}_{2n-1}] $$
but I haven't been able to make it work.

I would like a proof that the limit $$\lim_{n\to\infty} a_n$$
  exists. The methods I know are not getting me anywhere.

I originally planned to ask "and what the limit is", but OEIS says "No closed-form expression is known for this constant".
The references it cites are unavailable to me at present.

The proofs previous to this one have all been non-constructive. They all state an upper bound, observe that the sequence is monotonically increasing, and then appeal to the Monotone Convergence Theorem. This type of argument cannot provide a computable error bound. The following does.
Call the expression $\sqrt{1 + \sqrt{2 + \sqrt{3+\dotsc}}}$ by the name $R$. Observe that if you partially expand $R$, and leave an $x$ for the unexpanded bit, you get something like $\sqrt{1 + \sqrt{2 + \sqrt{3 + x}}}$. Now consider replacing $x$ with some upper bound $U$, and some lower bower bound $L$. You get:
$$\sqrt{1 + \sqrt{2 + \sqrt{3 + L}}} \leq R \leq \sqrt{1 + \sqrt{2 + \sqrt{3 + U}}}.$$
The top answer for this question suggests to set $U = \sqrt 4 \phi$, where $\phi$ is the golden ratio. Why does this work? Because $\sqrt4\phi = \sqrt4\sqrt{1 + \sqrt{1 + \sqrt{1+\dotsc}}} = \sqrt{4^{2^0}+\sqrt{4^{2^1}+\sqrt{4^{2^2}+\dotsc}}} \geq \sqrt{4 + \sqrt{5 + \sqrt{6 + \dotsc}}}$.
We can set $L = \sqrt 4$.
We now estimate the difference between the upper bound and lower bound. We get:
$$\begin{align}
&\sqrt{1 + \sqrt{2 + \sqrt{3 + U}}} - \sqrt{1 + \sqrt{2 + \sqrt{3 + L}}}
\\&\leq \sqrt{0 + \sqrt{0 + \sqrt{0 + U}}} - \sqrt{0 + \sqrt{0 + \sqrt{0 + L}}} \tag{drive 1,2,3 to 0}
\\&=U^{2^{-3}} - L^{2^{-3}}
\\&=(\sqrt{4}\phi)^{2^{-3}} - (\sqrt{4})^{2^{-3}}
\\&=(\sqrt{4})^{2^{-3}}(\phi^{2^{-3}} - 1)
\\&\approx 0.06760864818
\end{align}$$
In general, you have that when $R$ is partially expanded to $\sqrt{1+\sqrt{2+\dotsc\sqrt{n+\sqrt{n+1}}}}$, the error is at most $(\sqrt{n+1})^{2^{-n}}(\phi^{2^{-n}} - 1)$. We demonstrated this in the special case where $n=3$. The error clearly converges to zero. The technique generalises to all similar problems.
