It is easy to prove the existence of the limit

$$\lim_{n\to\infty} \left(\pi - 2^n \sqrt{2-\sqrt{2+\cdots+\sqrt 2}} \right) 4^n = C = 1.29\ldots$$

I am not confident about more digits but this number seems familiar as if it has a closed form or it has appeared elsewhere.

Is it possible to find a closed form for $C$ ? Can we prove that $C$ is irrational ?

$C$ seems close to $(5/4 + 4/3)/2 = 31/24$.

  • $\begingroup$ This answer may help. $\endgroup$
    – A.Γ.
    Aug 8, 2016 at 8:04

1 Answer 1


Well, let's try to make some sense out of it. $$\sqrt2=2\cos{\pi\over4}$$ $$\sqrt{2+\sqrt2}=\sqrt{2+2\cos{\pi\over4}}=2\cos{\pi\over8}$$ $$\sqrt{2+\dots\sqrt2}=2\cos{\pi\over2^n}$$ (I may be off by 1, since you never defined what $n$ is. Whatever.)

Now we put one minus on top of that chain of pluses. $$\sqrt{2-\sqrt{2+\dots\sqrt2}}=\sqrt{2-2\cos{\pi\over2^n}}=2\sin{\pi\over2^{n+1}}$$ Now the Taylor series kicks in. $$\sin{\pi\over2^{n+1}}={\pi\over2^{n+1}}-{1\over6}\cdot\left({\pi\over2^{n+1}}\right)^3+o\left({1\over2^{3n}}\right)$$ $$4^n\left(\pi-2^n\cdot\sqrt{2-\sqrt{2+\dots\sqrt2}}\right)=4^n\cdot{1\over6}\cdot{\pi^3\over2^{2n+2}}+o(1)={\pi^3\over24}+o(1)$$ So $\pi^3\over24$ is your limit. (BTW, yes, it's irrational and even transcendental.) Considering that $\pi^3\approx31.0063$, your approximation was indeed pretty close.

  • $\begingroup$ Could you say how you obtained the second equation? $\endgroup$
    – Bananach
    Aug 8, 2016 at 8:10
  • 1
    $\begingroup$ Why, that's the double angle formula applied in reverse. $\cos2x=\cos^2x-\sin^2x$, hence $\sqrt{2+2\cos2x}=\sqrt{2+2\cos^2x-2\sin^2x}=\sqrt{4\cos^2x}=2\cos x$. $\endgroup$ Aug 8, 2016 at 8:13
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    $\begingroup$ The power of Ivan and Taylor !! :) $\endgroup$
    – mick
    Aug 8, 2016 at 8:26
  • $\begingroup$ I have seen pi^3 / 24 before as Some Nice integral or zeta like thing... Thinking about it :) $\endgroup$
    – mick
    Aug 8, 2016 at 8:32

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