# Evaluating the limit of $\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt[n]{2}}}}$ when $n\to\infty$

The following nested radical $$\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$$ is known to converge to $2$.

We can consider a similar nested radical where the degree of the radicals increases:

$$\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt[n]{2}}}}$$ which converges to a constant $C=1.8695973...$. Is there a closed-form expression for this limit?

• Is there any reason to think so? – GEdgar Jun 17 '14 at 12:44
• Can you even give a recursive formula for the $n^{\text{th}}$ term of your sequence? – Ted Shifrin Jun 17 '14 at 13:06
• In my opinion, there are no particular reasons to think that a closed-form expression exists. – Anatoly Jun 17 '14 at 15:34
• The reason that the first nested radical possesses a closed form is because its exponents form a cycle, which ultimately leads to solving a polynomial of finite degree, in this case, $x=\sqrt{2+x}=>x^2=x$ $+2$. – Lucian Jun 17 '14 at 20:29
• This may be linked to a related question – gar Jun 18 '14 at 4:05

## 4 Answers

This is only a partial answer , or a too long comment. Please read the Disclaimer at the end.
From a computational point of view , this question is related to at least three others:

The common denominator is that, in order for a numerical calculation to be stable, backward recursion is to be preferred. For the problem at hand, this backward recursion goes as follows: $$a_{n-1} = \sqrt[n]{2+a_n} \qquad \mbox{with} \quad \lim_{n\to\infty} a_n = 1$$ The latter condition means in practice that for sufficient large values of $\;n\;$ the starting (or rather ending) value for $a_n$ may be set equal to $1 \approx \sqrt[n]{3}$ .
An estimate for the errors (update) may be obtained by differentiation: $$d a_{n-1} = d \sqrt[n]{2+a_n} = d a_n \frac{1}{n} \left(2+a_n\right)^{1/n-1} = \frac{a_{n-1}}{n(2+a_n)}da_n$$ starting with $\:da_n = \sqrt[n]{3} - 1\:$ for some sufficiently large value of $n$ , in our case $n = 13$ . It is seen that the error in $\;a_1\;$ can be calculated (more or less) by backward recursion as well. And it may be conjectured that some reasonable upper bound must go like $\,1/(2^n.n!)$ , meaning that convergence is really fast !
The accompanying Delphi Pascal program is a logical consequence of the above considerations.

program apart;
function power(x,r : double) : double;
begin
power := exp(r*ln(x));
end;
procedure anatoly(x : double; n : integer);
var
a,d,b : double;
k : integer;
begin
a := power(x+1,1/n);
d := a-1; { error }
a := 1;
for k := n downto 2 do
begin
b := a;
a := power(x+a,1/k);
d := d*a/(x+b)/k;
end;
Writeln(x,' ',a,' +/-',d);
end;
begin
anatoly(2,13); { at double precision }
end.

The outcome is, of course, in concordance with the value already found by the OP: $$1.86959730667536$$ It is noted that forward recursion can be employed in principle as well, but only in a quite different manner, because, as mentioned by others too, it's highly unstable.
Disclaimer: I certainly would have tried the closed form - whatever that means in modern times - if I only could believe that such a thing does indeed exist here. If this answer nevertheless clarifies some issues, that would be nice.

BONUS. But the "closed form" phrase keeps ringing in everyone's head ..
As suggested in a comment by Semiclassical, making a graph of the following more general function $f(x)$ might be interesting. $$f(x) = \lim_{n\to \infty}\sqrt{x+\sqrt{x+\sqrt{x+\cdots+\sqrt[n]{x}}}}$$ Because, if the graph of that function could be recognized as belonging to some "elementary" function, say $g(x)$, then we could say that the outcome of the OP's question is simply $g(2)$ .
So here goes, for $\;0 < x < 4\;$ and $\;0 < y < 4\;$ : It is conjectured that $f(0) = 0$ and that there is a singularity ( jump $0\to 1$ ) at that place.

• I think a closed form is pretty hopeless myself. But what I'd be curious to see is to see a plot of $C$ when 2 is changed to a range of different values (say, 1 to 3). – Semiclassical Jul 25 '14 at 19:27
• @Semiclassical: If you would formulate that as a question at MSE, then maybe someone would think of trying to answer that question. Please notify me if you do :-) – Han de Bruijn Jul 25 '14 at 19:58
• Fair enough! I may do that in a bit. – Semiclassical Jul 25 '14 at 20:00
• @Semiclassical: Couldn't resist. Please take a look at the BONUS part. – Han de Bruijn Jul 27 '14 at 11:52
• I guess it's some kind of singular perturbation near 0? I suppose it somehow reflects the fact that the order of limits ($x_0\to 0$ or $n\to \infty$) matters a lot. (In the computation, it's definitely the latter; I'd think that the $f(x)\to1$ limit reflects the former.) – Semiclassical Jul 28 '14 at 19:21

As already mentioned by others, there is no closed form known for this expression (like the majority of infinite nested radicals).

I would consider some bounds for this number, which give quite good approximations.

First, instead of truncating the number, it's much better to replace the remainder by $1$ at any step, for example:

$$\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}>\sqrt{2+\sqrt{2+\sqrt{2+1}}}=1.8684804$$

For the upper bound we can use the approximation:

$$\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}<\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$$

To find $\sqrt{2+\sqrt{2+\cdots}}$, we use the equation:

$$x^3-x-2=0$$

$$x=\frac{1}{3} \left(\sqrt{27-3\sqrt{78}}+\sqrt{27+3\sqrt{78}} \right)=1.5213797$$

$$\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}<\sqrt{2+1.5213797}=1.876534$$

For a better upper bound we find:

$$\sqrt{2+\sqrt{2+\cdots}}=\frac{1}{3} \left(1-2\sqrt{\frac{2}{47+3 \sqrt{249}}}+\sqrt{\frac{47+3 \sqrt{249}}{2}} \right)=1.3532099$$

$$\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}<\sqrt{2+\sqrt{2+1.3532099}}=1.8699639$$

So, we get, at this stage:

$$1.8685<\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}<1.8700$$

We get the value with uncertainty less than $0.1 \%$ in just a couple of steps.

We won't get a closed form this way, but this is much more accurate, than just truncating the nested radical.

One of the well known formula by Indian mathematician Ramanujan is this one: $$f(x)=x+1=\sqrt { \left( x+1 \right) ^{2}}=\sqrt {1+{x}^{2}+2\,x}=\sqrt {1+x\sqrt {(x+2)^2}}$$ $$=\sqrt {1+x\sqrt {1+{x}^{2}+4\,x+3}}=\sqrt {1+x\sqrt {1+ \left( x+1 \right) \sqrt {(x+3)^2}}}$$ $$=\sqrt {1+x\sqrt {1+ \left( x+1 \right) \sqrt {1+ \left( x+2 \right) \sqrt {1+ \left( x+3 \right) \sqrt {1+ \left( x+4 \right) \sqrt {\cdots}}}}}}$$ So it is interesting to show that why your mentioned limit tends to $f(0.86959730667536)$?

• Very nice answer for possibly a different question. – Simply Beautiful Art Jul 19 '16 at 20:52

Just a hint.

We have the recursion $$b_n^n=2+b_{n+1} \qquad \mbox{with} \quad \lim_{n\to\infty} b_n = 1 \tag{1}$$ We are interested in $b_2 \approx 1.8695973$ (my $b_n=a_{n-1}$ in Han de Bruijn's answer).

Let's try $$b_n=1 + \frac{d_1}{n}+ \frac{d_2}{2! \,n^2}+\cdots \tag{2}$$

Consider that if $g(x)=1 + d_1 x + d_2 \, x^2/2 +\cdots$ around $x=0$ then $$h(x)=g(x)^{1/x}=e^{d_1}+e^{d_1}\frac{(d_2 -d_1^2)}{2}x+\cdots \tag{3}$$

Further, replacing $b_n = g(1/x)$ in $(1)$ we get

$$g(x)^{1/x}=2+g\left(\frac{x}{1+x}\right) \tag{4}$$

Replacing, we can obtain the first coefficients:

$$d_1=\log(3)=1.09861228866811$$ $$d_2=\frac{2}{3}\log(3)+\log^2(3)=1.939357153257989$$ And

$$b_n=1 + \log(3)\frac{1}{n}+ \log(3) \left(\frac{1}{3}+\frac{\log(3)}{2}\right) \frac{1}{n^2}+O\left(\frac{1}{n^3}\right)$$ Plugin this in $(2)$ is still not great for computing $b_2$ directly ($\approx 4\%$ relative error) but it's a good start point for the backward recursion $(1)$. For example, starting with $n=5$ the relative error for $b_2$ (ony three iterations) is below $10^{-5}$ ; starting from $n=9$ the error is around $2 \cdot 10^{-11}$

Obviously, higher order terms could be obtained for the Taylor expansion in $(3)$, but I don't think we can get a simple closed form for the general term.