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?

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    $\begingroup$ In general, expressions like this that are easy to evaluate usually have some sort of symmetry such that we can relate the value to its self via e.g. a square like for example $x^2 = ax + b$. One such example is $x = \sqrt{\frac{1}{2^2} + \sqrt{\frac{1}{2^4} + \ldots + \frac{1}{2^{2^n}}}}$ since then $x^2 = \frac{1}{2^2} + \frac{1}{2}x \to x = \frac{1+\sqrt{5}}{4}$. This one does not have such a symmetry - which does not rule out an analytic solution, but it makes less likely / much harder to find it if it exist. $\endgroup$
    – Winther
    Dec 28, 2014 at 13:23
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    $\begingroup$ Well, proving convergence is at least not too difficult. $\endgroup$ Dec 28, 2014 at 13:27
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    $\begingroup$ $1.2857367633569968672336\ldots$ does not possess a known closed form. $\endgroup$
    – Lucian
    Dec 28, 2014 at 15:14
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    $\begingroup$ The two-letter answer "No" is not allowed. Had it been allowed, it would have got the bounty! $\endgroup$
    – GEdgar
    Dec 31, 2014 at 16:14
  • $\begingroup$ How are people numerically evaluating this? It seems like it also suffers from some bad cancellation when using the obvious way (for loop from n = big to n = 0). $\endgroup$
    – Y. S.
    Jan 4, 2015 at 23:35

3 Answers 3


This question is related to at least five others:

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:
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.


It is easy to show, assuming the limit exists:


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:




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





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



$$y=\sqrt{\frac12+\frac{1}{\sqrt{2}}\left(\sqrt{\frac12+\sqrt{\frac12+\cdots+\sqrt{\frac{1}{2}}}}\right)}$$ 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.

  • $\begingroup$ @Adriandmen I just posted a solution for $\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots+\sqrt{\frac{1}{2^n}}}}}$ $\endgroup$ Jan 1, 2015 at 19:53
  • $\begingroup$ Please note that $$ \sqrt{\frac 1 2 + \frac{1}{\sqrt{2}}*\sqrt{\frac 1 2+\sqrt{\frac 1 4+\sqrt{...}}}} = \sqrt{\frac 1 2 + \sqrt{\frac 1 4+\frac 1 2\sqrt{\frac 1 4+\sqrt{...}}}} \neq \sqrt{\frac 1 2 + \sqrt{\frac 1 4+\sqrt{\frac 1 8+\sqrt{...}}}}$$ $\endgroup$
    – cirpis
    Jan 1, 2015 at 20:02
  • $\begingroup$ I see the mistake. thanks $\endgroup$ Jan 1, 2015 at 20:16
  • $\begingroup$ will correct it now $\endgroup$ Jan 1, 2015 at 20:20
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    $\begingroup$ @sholaOtitoju Note that $\sqrt{1/2 + \sqrt{1/4+\sqrt{1/8+\sqrt{1/16+\ldots}}}} = \sqrt{1/2 + 1/\sqrt{2}\sqrt{1/2 + \sqrt{1/2+\sqrt{1+\sqrt{8+\ldots}}}}}$ so your formula is not correct. This is also easily checked numerically as the solution is $x = 1.2857\ldots$ while you get $x= 1.211\ldots$. $\endgroup$
    – Winther
    Jan 6, 2015 at 5:10

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