TITLE: How can I show that $\sqrt{1+\sqrt{2+\sqrt{3+\sqrt\ldots}}}$ exists?
QUESTION [60 upvotes]: 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.

REPLY [4 votes]: 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.
Let $L_n = \sqrt{1 + \sqrt{2 + \dots\sqrt{n}}}$ and $U_n = \sqrt{1 + \sqrt{2 + \dots \sqrt{n}\phi}}$, where $\phi$ is the golden ratio. We assume $n > 1$.
Observe that since $\phi = \sqrt{1 + \sqrt{1 + \dots}}$, we have that $\sqrt{n}\phi = \sqrt{n^{2^0} + \sqrt{n^{2^1} + \sqrt{n^{2^2} + \dots}}}$. Comparing $n^{2^1}$ to $n+1$ and $n^{2^2}$ to $n+2$ and so on, we have that $L_i < U_j$ for all $i$ and $j$.
Observe that $L_n$ is monotonically increasing.
Define $f_n(x) = \sqrt{1 + \sqrt{2 + \dots\sqrt{n + x}}}$. It then follows that $L_{n} = f_n(0)$ and $U_{n+1} = f_n(\phi\sqrt{n+1})$.
So for $i \geq 0$, we have
$$\begin{align}
  &L_n &\leq L_{n + i} &\leq U_{n+1}
\\\iff  &0 &\leq L_{n + i} - L_n &\leq U_{n+1} - L_n
\\&&&= f_n(\phi\sqrt{n+1}) - f_n(0)
\\&&&\leq f_n'(0) \times \phi\sqrt{n+1} \tag{by Mean Value Theorem}
\end{align}$$
The last inequality follows from the Mean Value Theorem. Using routine calculus, we have that $$\begin{align}
f_n'(0) &= \frac{1}{2^n\sqrt{n} \cdot\sqrt{(n-1) + \sqrt{n}} \dots \sqrt{1 + \sqrt{2 + \dots{\sqrt{(n-1) + \sqrt{n}}}}}}
\\&\leq \frac{1}{2^n\sqrt{n!}}.
\end{align}$$
Therefore $$\begin{align}
  &0 &\leq L_{n + i} - L_n &\leq \frac{\phi\sqrt{n+1}}{2^n\sqrt{n!}}.
\end{align}$$
The rightmost side is exponentially decreasing. It follows then that the sequence $L_n$ is Cauchy, and must therefore converge.
PS. The following approximation $\ell_n = [1,2,\dotsc,n-1,n+1] = f_n(1)$ is a lower bound which is stronger than $L_n = [1, 2, \dotsc, n-1, n] = f_n(0)$. The error is then at most $\frac{1}{2^n\sqrt{n!}}$ (true for $n \geq 2$). In particular, $\ell_3 = \sqrt{3} \approx 1.7$ is accurate to 2 significant figures.
$$ $$