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:

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.

Answer

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.

Attribution
Source : Link , Question Author : MJD , Answer Author : user1551

Leave a Comment