Evaluation of $\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots+\sqrt{\frac{1}{2^n}}}}}$

Question

I was just playing around with a calculator, and came to the conclusion that:

$$\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots+\sqrt{\frac{1}{2^n}}}}} \approx 1.29$$

Now I'm curious. Is it possible to evaluate the exact value of the following?

Answer 1

This question is related to at least five others:

• Problem 6 - IMO 1985
• How to find this limit: $A=\lim_{n\to \infty}\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+\cdots+\sqrt{\frac{1}{n}}}}}$
• Find the limit $L=\lim_{n\to \infty} \sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\cdots+\sqrt[n]{\frac{1}{n}}}}$
• Evaluating the sum $\lim_{n\to \infty}\sqrt[2]{2+\sqrt[3]{2+\sqrt[4]{2+\cdots+\sqrt[n]{2}}}}$
• Limit of $\lim_{x\rightarrow 1}\sqrt{x-\sqrt[3]{x-\sqrt[4]{x-\sqrt[5]{x-…}}}}$

This makes a short answer possible (and desirable).
For a numerical calculation backward recursion is proposed (again): $$ a_{n-1} = \sqrt{1/2^n+a_n} \qquad \mbox{with} \quad \lim_{n\to\infty} a_n = 0 $$ Here comes the Pascal program snippet that is supposed to do the job:

program apart;

procedure again(n : integer);
var
  a,two : double;
  k : integer;
begin
  two := 1;
  for k := n downto 2 do
    two := two/2;
  a := 0;
  for k := n downto 2 do
  begin
    a := sqrt(two+a);
    two := two*2;
  end;
  Writeln(a);
end;

begin
  again(52);
end.

Note that an error analysis is not implemented in the program. This has not much sense because the accuracy is determined by the smallest $1/2^n$ that can be represented with some significance; that is for $n\approx 52$ in double precision Pascal. The outcome is, of course, in concordance with the value already found by Lucian:

 1.28573676335699E+0000

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.

Answer 2

$$y=\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots+\sqrt{\frac{1}{2^n}}}}}\equiv\sqrt{\frac12+\frac{1}{\sqrt{2}}\left(\sqrt{\frac12+\sqrt{\frac12+\cdots+\sqrt{\frac{1}{2}}}}\right)}$$

$$y=\sqrt{\frac12+\frac{1}{\sqrt{2}}\left(\sqrt{\frac12+\sqrt{\frac12+\cdots+\sqrt{\frac{1}{2}}}}\right)}$$ But the Let the term in bracket be x therefore:$$x=\sqrt{\frac12+\sqrt{\frac12+\cdots+\sqrt{\frac{1}{2}}}}\equiv\sqrt{\frac12+x}$$ squaring both sides $$x^2=\frac12+x$$ $$x^2-x-\frac12=0$$solving the equation gives: $$x=\frac{1+\sqrt{3}}{2}$$ but we have $$y=\sqrt{\frac12+\frac{1}{\sqrt{2}}\left(\sqrt{\frac12+\sqrt{\frac12+\cdots+\sqrt{\frac{1}{2}}}}\right)}=\sqrt{\frac12+\frac{1}{\sqrt{2}}x}$$ Putting the value of x into y gives$$y=\sqrt{\frac12+\frac{1}{\sqrt{2}}\frac{1+\sqrt{3}}{2}}=1.211$$ Therefore $$\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots+\sqrt{\frac{1}{2^n}}}}}=1.211$$ This is an exact value.

Answer 3

It is easy to show, assuming the limit exists:

$$\sqrt{a+\sqrt{a+\sqrt{a+...}}}=\frac{1+\sqrt{1+4a}}{2}$$

It is also easy to see the following set of inequalities:

$$\sqrt{\frac{1}{2}+1} < \sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots}}}<\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+...}}}$$

$$\sqrt{\frac{1}{2}+\sqrt{\frac{1}{4}+1}} < \sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots}}}<\sqrt{\frac{1}{2}+\sqrt{\frac{1}{4}+\sqrt{\frac{1}{4}+...}}}$$

$$\sqrt{\frac{1}{2}+\sqrt{\frac{1}{4}+\sqrt{\frac{1}{8}+1}}} < \sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots}}}<\sqrt{\frac{1}{2}+\sqrt{\frac{1}{4}+\sqrt{\frac{1}{8}+\sqrt{\frac{1}{8}+...}}}}$$

And so on, getting better and better approximations.

Calculating the nested radicals:

$$\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+...}}}=\frac{1+\sqrt{3}}{2}$$

$$\sqrt{\frac{1}{4}+\sqrt{\frac{1}{4}+\sqrt{\frac{1}{4}+...}}}=\frac{1+\sqrt{2}}{2}$$

$$\sqrt{\frac{1}{8}+\sqrt{\frac{1}{8}+\sqrt{\frac{1}{8}+...}}}=\frac{1+\sqrt{1.5}}{2}$$

We get following set of boundaries for the value we need:

$$R=\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots}}}=1.285737\dots$$

$$1.22474<R<1.36603$$

$$1.27202<R<1.30656$$

$$1.28251<R<1.29120$$

And so on. There is no closed form, but it's not hard to evaluate this nested radical with good precision.