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 first number $1$ is a nuisance, so at first we disregard it.
We proceed by induction, and deal with finite nested radicals that start with $\sqrt{k+\sqrt{(k+1)+\cdots}}$, where $k\ge 2$. 
We will show that such a radical is $\lt 2k$, by induction on depth.  The result  is  certainly true for all nested radicals of depth $1$, since $\sqrt{q}\lt 2q$. 
For the induction step, a nested radical of depth $n$ that starts with $k$ is $\sqrt{k+R}$, where $R$ is a nested radical of depth $n-1$ that starts with $k+1$. By the induction assumption, we have $R\lt 2k+2$. But then $\sqrt{k+R}\lt \sqrt{3k+2}\lt 2k$ if $k \ge 2$.  
So (finite) nested radicals of any depth that start with $2$ are $\lt 4$. The sequence of nested radicals is clearly increasing, so it converges. It follows that the nested radical of the post is $\le \sqrt{1+4}$.
