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.

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.

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 $\le \sqrt{1+4}$.

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.

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 $O(10^{-2})$.

$$[3,4,\ldots] \le \frac{1+\sqrt{10}}{\sqrt{3}} \implies [1,2,3,\ldots] \le \sqrt{1+\sqrt{2+\frac{1+\sqrt{10}}{\sqrt{3}}}} \sim 1.760214368 $$