Wikipedia claims this but only cites an offline proof:

$$\lim_{n\to\infty} 2^n \sqrt{2-\sqrt{2+\cdots+ \sqrt 2}} = \pi$$

for $n$ square roots and one minus sign. The formula is not the "usual" one, like Taylor series or something like that, so I can't easily prove it. I wrote a little script to calculate it and it's clearly visible, but that's not a proof.


3 Answers 3


It's the sequence of approximations obtained when you approximate the perimeter of the circle of diameter $1$ with inscribed regular $n$-gons for $n$ a power of $2$.

As I happen to have this TeXed' up, I'll offer:

Suppose regular $2^n$-gons are inscribed in a circle of radius $r$.

Suppose the side length(the length of one "face") $a_n$ of a the inscribed $2^{n}$-gon is known (so, $a_2$ is the side length of the square). To find the side length of the $2^{n+1}$-gon, one may apply the Pythagorean Theorem twice to obtain $$ \tag{1}a_{n+1} = r\sqrt{2-\sqrt{4-{a_n^2\over r^2}}} $$

Now, starting with a square, $$a_2=\sqrt 2 r.$$ Using the recursion formula (1) repeatedly gives: $$ a_3%= r\sqrt{2-\sqrt{4-{2r^2\over r^2}}} = r\sqrt{2-\sqrt2}, $$ $$ a_4%= r\sqrt{2-\sqrt{4-{ ( r\sqrt{2-\sqrt2})^2 \over r^2}}} = r\sqrt{2-\sqrt{4-{ ({2-\sqrt2} ) }}} = r\sqrt{2-\sqrt{{ {2+\sqrt2} }}}, $$ and $$ a_5%= r\sqrt{2-\sqrt{4-{ ( r\sqrt{2-\sqrt{{ {2+\sqrt2} }}} )^2\over r^2}}} = r\sqrt{2-\sqrt{ 2+\sqrt{{ {2+\sqrt2} }}} }. $$ $$\vdots$$

Let $b_n=2^n a_n$. Let $P_n=r\cdot b_n$ be the perimeter of the $2^n$-gon. Let $P$ be the perimeter of the circle. Then $$ \lim_{ n\rightarrow \infty} P_n = P. $$ Note that from the above identity, it follows that the ratio of the perimeter of a circle to its diameter must be a constant, namely $\lim\limits_{n \rightarrow \infty} b_n$. We call this number $\pi$.

Below are some particular calculations when the radius of the circle is $1/2$:

$$\eqalign{ P_2&=2^1\cdot\sqrt 2 \approx 2.82842712\cr P_3&=2^2\cdot\sqrt{2-\sqrt2}\approx 3.06146746\cr P_4&=2^3\cdot\sqrt{2-\sqrt{2+\sqrt2}}\approx3.12144515 \cr P_5&=2^4\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt2}}}\approx 3.13654849\cr P_6&=2^5\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt2}}}}\approx 3.14033116\cr P_7&=2^6\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt2}}}}}\approx 3.14127725\cr P_8&=2^7\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt2}}}}}}\approx 3.1415138 \cr P_9&=2^8\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt2}}}}}}}\approx 3.14157294 \cr } $$

For completedness:

Remark 1: Here is the proof that the recursion formula (1) holds:

Let $a_n$ be the side length of the $2^n$-gon.

To obtain the $2^{n+1}$-gon: take the "outer end point" of the radii of the circle that bisect the faces of the $2^n$-gon to form the new vertices of the $2^{n+1}$-gon.

We then have, for $a_{n+1}$, the scenario shown in the following diagram (not to scale):

enter image description here

Now $$ b^2=r^2-{a_n^2\over4}; $$ whence $$\eqalign{ a_{n+1}^2={a_n^2\over4} + \Biggl((r-\sqrt{ r^2-{a_n^2\over4}}\ \Biggr)^2 &={a_n^2\over4}+ r^2-2r\sqrt{r^2-{a_n^2\over4}}+r^2 -{a_n^2\over4}\cr &= 2r^2-2r\sqrt{r^2-{a_n^2\over4}}\cr &= 2r^2-r^2\sqrt{4-{a_n^2\over r^2}}\cr &= r^2 \Biggl(2-\sqrt{4-{a_n^2\over r^2}}\ \Biggr).}$$ And, thus $$ a_{n+1}= r \sqrt{2-\sqrt{4-{a_n^2\over r^2}}}. $$

Remark 2: To explain why limit $\lim\limits_{n\rightarrow\infty} P_n=P\ $ holds, I can do no better than refer you to Eric Naslund's comment in his answer.

See also, here.

  • $\begingroup$ Weird how mathjax (I'm presuming) decides to use slant-y square root signs vs. rectangular square roots signs. $\endgroup$ Dec 14, 2011 at 17:43
  • $\begingroup$ Think I might rob this for my discrete dynamical systems class... $\endgroup$ Jan 15, 2013 at 17:05

Here is a slightly different way to see why it is the area of a $2^k$-gon. (It is really the same, I just also want to point out that the nested radical expression is the sin of $\frac{\pi}{2^k}$. This type of argument gives Vietas product formula for $\frac{2}{\pi}$)

Recall: If $a,b$ are sides of a triangle, and $\theta$ is the angle between them, then the area of this triangle is $\frac{1}{2}ab\sin(\theta)$.

For any regular $n$-gon inscribed in the unit circle, this means that by splitting it into the $n$ identical isosceles triangles with two sides equal to $1$, we have that $$\text{Area of regular n-gon}=\frac{n}{2}\cdot \sin\left(\frac{2\pi}{n}\right).$$

In the limit, this must approach $\pi$. Now, the nested radical expression can be rewritten as sin of angle as follows:

We have that $$\sin\left(\frac{\pi}{2^k}\right)=\frac{1}{2}\sqrt{2-\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}.$$

Proof: Notice that $$\cos(x)=\sqrt{\frac{1}{2}\left(1+\cos(2x)\right)}.$$ Using this iteratively allows us to find $\cos\left(\frac{\pi}{2^k}\right)$. For example, since $\cos\left(\frac{\pi}{2}\right)=0$ we see that $$\cos\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}=\frac{1}{2}\cdot \sqrt{2}$$ and similarly

$$\cos\left(\frac{\pi}{8}\right)=\sqrt{\frac{1}{2}\left(1+\frac{1}{\sqrt{2}}\right)}=\frac{1}{2}\cdot \sqrt{2+\sqrt{2}}.$$ Iterating again we have $$\cos\left(\frac{\pi}{16}\right)=\frac{1}{2}\cdot \sqrt{2+\sqrt{2+\sqrt{2}}},$$ and by induction we'll arrive at our result.

Since $\cos^2(x)+\sin^2(x)=1$, one final manipulation will allow us to conclude our result.

  • $\begingroup$ If you'd add that we know $sin x\over x$ is 1 as x goes to 0 then it'd be a complete proof. $\endgroup$
    – chx
    Nov 24, 2011 at 14:04
  • $\begingroup$ Yeah -- my comment was for an earlier edit of the answer. $\endgroup$
    – chx
    Nov 24, 2011 at 14:07
  • $\begingroup$ @chx: Oops sorry. I edit many times. Originally I had read the other answer, and then explained why the fractions come up. When I posted I saw that he had elaborated and my comment wasn't really needed! So I just added more as it seemed different enough. (can't let all that good typing go to waste!) $\endgroup$ Nov 24, 2011 at 14:12
  • 1
    $\begingroup$ One small problem -- how do you know the area of the series regular n-gon actually has the limit of the area of the circle? $\endgroup$
    – chx
    Nov 24, 2011 at 18:05
  • 2
    $\begingroup$ @chx: For this, it actually helps to look at $2^k$-gons, because the $2^{k+1}$-gon always contains the $2^k$-gon. Then we have a sequence of measurable compact sets, each one contained in the next. We know that the union contains the open unit disk, since for any point strictly inside, I can find a $2^k$-gon which contains it. (it is also contained in the closed unit disk) Then the result follows from the fact that measures are continuous from below, and specifically the Lebesgue measure is continuous from below. $\endgroup$ Nov 24, 2011 at 18:15

This is more of an extended comment than an answer, but I figure that since the OP is interested in the computational aspect ("...I wrote a little script to calculate it..."), this might be of some interest. The following is adapted from Ole Østerby's unpublished manuscript.

As already noted, the OP's limit is equivalent to saying that

$$\lim_{h\to 0}\frac{\sin\,\pi h}{h}=\pi$$

where we identify $h$ with $2^{-n}$. That is, to borrow notation from David's answer (and taking $r=1$):

$$b_n=2^n \sin\left(\frac{\pi}{2^n}\right)$$

If we expand $\dfrac{\sin\,\pi h}{h}$ as a series, like so:

$$\frac{\sin\,\pi h}{h}=\pi-\frac{\pi^3 h^2}{6}+\frac{\pi^5 h^4}{120}+\cdots$$

we see that only even powers of $h$ occur in the expansion (as expected, since the function in question is even).

If we halve $h$ (equivalently, increase $n$), we have a slightly more accurate approximation of $\pi$. The upshot is that one can take an appropriate linear combination of $\dfrac{\sin\,\pi h}{h}$ and $\dfrac{\sin(\pi h/2)}{h/2}$ to yield an even better approximation to $\pi$:

$$\frac13\left(\frac{4\sin(\pi h/2)}{h/2}-\frac{\sin\,\pi h}{h}\right)=\pi -\frac{\pi^5 h^4}{480}+\frac{\pi^7 h^6}{16128}+\cdots$$

This game can be repeatedly played, by taking successive linear combinations of values corresponding to $h/2$, $h/4$, $h/8$... The method is known as Richardson extrapolation.

(I'll note that I have already brought up Richardson extrapolation in a number of my previous answers, like this one or this one.)

More explicitly, taking $T_n^{(0)}=b_n$, and performing the recursion


the "diagonal" sequence $T_n^{(n)}$ is a sequence that converges faster to $\pi$ than the sequence $b_n$. Christiaan Huygens used this approach (way before even Richardson considered his extrapolation method) to refine the Archimedean estimates from circumscribing and inscribing polygons.

Sundry Mathematica code:

Table[2^(n - 1)*Sqrt[2 - Nest[Sqrt[2 + #1] & , 0, n - 2]] ==
      FunctionExpand[2^n*Sin[Pi/2^n]], {n, 2, 15}]

{True, True, True, True, True, True, True, True, True, True,
 True, True, True, True}

This verifies the equivalence of the iterated square root and sine expression.

Here is an implementation of the application of Richardson extrapolation to the computation of $\pi$:

huygensPi[n_Integer, prec_: MachinePrecision] := 
 Module[{f = 1, m, res, s, ta},
  res = {ta[1] = s = N[2, prec]};
   If[k > 1, s = s/(2 + Sqrt[4 - s])];
   f *= 2; ta[k + 1] = f Sqrt[s];
   m = 1;
   Do[m *= 2;
    ta[j] = ta[j + 1] + (ta[j + 1] - ta[j])/(m - 1);, {j, k, 1, -1}];
   res = {res, ta[1]};, {k, n - 1}];

Note that I used a stabilized version of the recursion for generating the $b_n$ (f Sqrt[s] in the code) to minimize errors from subtractive cancellation.

Here's a sample run, where I generate 10 successive approximations to 25 significant digits:

huygensPi[10, 25] - Pi

{2.000000000000000000000000, 3.656854249492380195206755,
 3.173725640962868268333725, 3.13944246625722809089242,
 3.14157581875151427853903, 3.14159291451874033422144,
 3.14159265388327967181647, 3.14159265358866759077617,
 3.14159265358979303435864, 3.14159265358979323865872}

where the $10$-th approximant is good to 19 digits. By comparison, $b_{10}=2^{10}\sin\left(\dfrac{\pi}{2^{10}}\right)= 3.1415877\dots$ is only good to five digits.


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