# 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.

• This is probably a good place to start: "It was discovered by T. Vijayaraghavan that the infinite radical $\sqrt{ a_1 + \sqrt{ a_2 + \sqrt{ a_3 + \sqrt{a_4 + \ldots }}}}$ where $a_n \ge 0$, will converge to a limit if and only if the limit of $\log a_n / 2^n$ exists" - Clawson, p. 229. (Taken from OEIS.) Jul 6, 2013 at 3:23
• possible duplicate of Sum and Product of Infinite Radicals
– MJD
Jul 6, 2013 at 3:26
• I misread the solution at Sum and Product of Infinite Radicals. It asks several questions, one of which is mine, but all the answers provided are for the other questions.
– MJD
Jul 6, 2013 at 3:28
– MJD
Jul 6, 2013 at 3:31
• This may help. Jul 6, 2013 at 4:44

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 part following from $n^2 -4n +1 >0$ for $n\ge 4$. Jun 6, 2014 at 14:17
• Very elegant estimate! +1
– RFZ
Nov 15, 2015 at 17:06

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

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

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.

• The last term is just $2 \text{Li}_{-\frac{1}{2}}\left(\frac{1}{2}\right)\approx 2.69451$ Oct 17, 2017 at 6:44

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.

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.

• I am verry sorry, I just clicked wrong. Btw I wouldnt consider myself a coward! I tried to undo my vote, but I cant until you edit your answer, sorry Jul 6, 2019 at 17:20
• Sorry again! It was really just a mistake of mine Jul 6, 2019 at 17:22
• Thanks, this is really nice.
– MJD
Jul 7, 2019 at 21:52