# 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.) – George V. Williams Jul 6 '13 at 3:23
• possible duplicate of Sum and Product of Infinite Radicals – MJD Jul 6 '13 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 '13 at 3:28
• Related: Nested radicals – MJD Jul 6 '13 at 3:31
• This may help. – Maazul Jul 6 '13 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 $n\ge 4$. – Tom Collinge Jun 6 '14 at 14:17
• Very elegant estimate! +1 – ZFR Nov 15 '15 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 $O(10^{-2})$.

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

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.

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 $2 \text{Li}_{-\frac{1}{2}}\left(\frac{1}{2}\right)\approx 2.69451$ – Claude Leibovici Oct 17 '17 at 6:44

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.



• 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 – TwoStones Jul 6 at 17:20
• Sorry again! It was really just a mistake of mine – TwoStones Jul 6 at 17:22
• Thanks, this is really nice. – MJD Jul 7 at 21:52