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.
 A: By linear approximations.
Since the sequence is increasing, it converges iff it doesn't go to infinity. Therefore we need only to construct an upper bound.
Notice that all tangents to $\sqrt{x}$ are always above $\sqrt{x}$. The slope of each tangent is $\frac12\frac1{\sqrt x}$, which is at most $\frac 1 2$ when $x\geq 1$.
So $$\sqrt{1 + \sqrt{2 + \sqrt{3 + \dotsc}}} \leq 1 + \frac12(\sqrt 2 + \frac12(\sqrt{3} + \frac12(\dotsc))) = \sum_{i=1}^\infty \frac{\sqrt i}{2^{i-1}}$$ and it's easier to see that that converges.
A: Take a positive sequence $\{a_n\}$ and a constant $c>0$ such that $\sqrt{a_{n+1}}<ca_n$.
Set $b_n=\sqrt{a_1+\sqrt{a_2+\cdots \sqrt{a_n}}}$. By induction, $$\log_{c+1} \left(\frac{b_n}{\sqrt{a_1}}\right)<\sum_{i=1}^{n-1}2^{-i}<1$$
So $b_n<(c+1)\sqrt{a_1}$ and $b_n$ is monotonic increasing; by the Monotone Convergence Theorem, $\lim_{n\to\infty}b_n$ exists. Taking $a_n=n$ and $c>\sqrt{2}$ answers this question.
A: 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.
A: For any $n\ge4$, we have $\sqrt{2n} \le n-1$. Therefore
\begin{align*}
a_n
&\le \sqrt{1+\sqrt{2+\sqrt{\ldots+\sqrt{(n-2)+\sqrt{(n-1) + \sqrt{2n}}}}}}\\
&\le \sqrt{1+\sqrt{2+\sqrt{\ldots+\sqrt{(n-2)+\sqrt{2(n-1)}}}}}\\
&\le\ldots\\
&\le \sqrt{1+\sqrt{2+\sqrt{3+\sqrt{2(4)}}}}.
\end{align*}
Hence $\{a_n\}$ is a monotonic increasing sequence that is bounded above.
A: 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 $\le \sqrt{1+4}$.
A: An easy sloppy way to see it:
You know that $\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}=\phi$. Now multiply by $2$ to get $2\phi=2\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}=\sqrt{4+4\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}=\sqrt{4+\sqrt{16+\sqrt{256+\sqrt{256^2+\dots}}}}>\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\dots}}}}$. Of course this should be done more rigorous.
A: If $( a_k )_{k\in\mathbb{N}}$ is any sequences of positive numbers such that:
$$0 \le a_k \le \alpha \lambda^{2^k}\quad\text{ for some }\quad \alpha, \lambda \in \mathbb{R}_{+}$$
Using same convention $\;[r_1,r_2\ldots] = \sqrt{r_1 + \sqrt{r_2 + \ldots}}\;$ as in the question,
we have:
$$\begin{align}
[a_n] & \le \sqrt{\alpha \lambda^{2^n}} =   \sqrt{\alpha}\lambda^{2^{n-1}} = [\alpha] \lambda^{2^{n-1}}\\
\implies [a_{n-1},a_n ] &\le \sqrt{\alpha\lambda^{2^{n-1}}+\sqrt{\alpha}\lambda^{2^{n-1}}}
=\sqrt{\alpha+\sqrt{\alpha}}\lambda^{2^{n-2}} = [\alpha,\alpha]\lambda^{2^{n-2}}\\
\implies [a_{n-2},a_{n-1},a_n]&\le \sqrt{\alpha\lambda^{2^{n-2}} + \sqrt{\alpha+\sqrt{\alpha}}\lambda^{2^{n-2}}} = [\alpha,\alpha,\alpha]\lambda^{2^{n-3}}\\
&\;\vdots\\
\implies [a_1,\ldots,a_n] & \le \underbrace{ [ \alpha,\ldots,\alpha ] }_{n\text{ terms}}  \lambda\\
\implies [a_1, \ldots,a_n] & \le [ \alpha, \alpha, \ldots ]\lambda = \frac{1 + \sqrt{1+4\alpha}}{2}\lambda
\end{align}$$
Since $n \le \sqrt{2}^{2^n-2}$, we can take $\alpha = \frac12$ and $\lambda = \sqrt{2}$ to get:
$$[1,2,\ldots,n] \le \underbrace{ [ \frac12,\ldots,\frac12 ]}_{n\text{ terms}} \lambda \le \frac{1+\sqrt{3}}{\sqrt{2}} \sim 1.931851 $$
To get a better bound, observe for any $m, k \in \mathbb{Z}_{+}$, we have:
$$ m + k - 1 \le \frac{m^2}{m+1}\left(\sqrt{\frac{m+1}{m}}\right)^{2^k}$$
Using the same approach as above, we get:
$$[m,m+1,m+2,\ldots] \le \frac{\sqrt{m+1}+\sqrt{4m^2+m+1}}{2\sqrt{m}}$$
Take $m = 3$, we already get a bound accurate up to $O(10^{-2})$.
$$[3,4,\ldots] \le \frac{1+\sqrt{10}}{\sqrt{3}}
\implies [1,2,3,\ldots] \le \sqrt{1+\sqrt{2+\frac{1+\sqrt{10}}{\sqrt{3}}}} \sim 1.760214368
$$
