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Integral and series proofs that $\frac{22}{7}>\pi$

We can prove that $\frac{22}{7}$ exceeds $\pi$ by using Dalzell integral $$\int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx=\frac{22}{7}-\pi$$

or its equivalent series $$\sum_{k=1}^{\infty} \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\frac{22}{7}-\pi$$

(see Series and integrals for inequalities and approximations to $\pi$)

Equivalent expressions

This series may be written in terms of factorials, binomial coefficients or Beta integrals as $$\begin{align} \frac{22}{7}-\pi &= 3840\sum_{k=1}^\infty \frac{(k+2)!(4k)!}{(4k+8)!k!} \\ &= \frac{4}{21} \sum_{k=1}^\infty \frac{{k+2 \choose 2}}{{4k+8\choose 8}} \\ &= \frac{16}{21} \sum_{k=1}^\infty \frac{B(4k+1,8)}{B(k+1,2)} \end{align} $$ (see A series to prove $\frac{22}{7}-\pi>0$)

Can $\frac{22}{7}-\pi$ be given a combinatorial or probabilistic interpretation?

Some situations where $\pi$ appears are Buffon's needle or the probability that two random integers are relatively prime. See also $\pi$ in random phenomena by Boris Gourévitch and Occorrenze in calcolo delle probabilità e statistica by Mauro Fiorentini.

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    $\begingroup$ A sketch: the numerator in the integral looks like the $p^n(1-p)^m$ in the binomial distribution, while the denominator $1+x^2$and the missing binomial coefficient may be related to some a priori distribution of $p$. So, maybe something like "$\frac{22}{7}- \pi$ is the probability to get 4 heads and 4 tails when flipping a coin 8 times given that $p$ follows a priori distribution... " could make sense. $\endgroup$
    – Jaume Oliver Lafont
    Feb 19, 2016 at 7:57
  • $\begingroup$ $\pi$ appears in the limit of the probability of getting the same number of heads and tails as the number of tosses grows. pi314.net/eng/aleatoire.php $\endgroup$
    – Jaume Oliver Lafont
    Feb 19, 2016 at 8:26
  • $\begingroup$ Related: math.stackexchange.com/a/1646548/134791 $\endgroup$
    – Jaume Oliver Lafont
    Feb 22, 2016 at 8:20
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    $\begingroup$ $$\int_0^1 \frac{x^2}{1+x^2}dx = 1-\frac{\pi}{4}$$ is the probability that a point inside a square is outside the inscribed circle. Maybe $\frac{11}{14}-\frac{\pi}{4}$ should be considered. $\endgroup$
    – Jaume Oliver Lafont
    Mar 16, 2016 at 22:22

1 Answer 1

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Here's one in the spirit of Boufon's needle and Jaume's comment that follows more or less directly from the Dalzell integral.

Put down a tin with square cross-section and walk far away in a random direction. Shoot at it until you've hit it eight times. We call a hit "proper" if it exits the tin through the face that's opposite the one it entered through. What's the probability that exactly half of the eight hits were proper?

Let's say the tin has diagonal length 1. Parametrize the possible orientations of the tin with an angle $\theta$ ranging from $0$ to $\pi/4$, where $0$ corresponds to looking at the tin edge-on and $\pi/4$ corresponds to looking at it face-on. Note that $\theta$ is distributed uniformly. Then the apparent width of the tin is $\cos \theta$ and the apparent width of the region you need for a proper hit is $\sin \theta$. So the probability of a proper hit is $\sin \theta \, / \cos\theta = \tan \theta$.

Hence, the probability of exactly four proper hits out of eight is

$$ \binom{8}{4} \frac1{\pi/4} \int_0^{\pi/4} (\tan \theta)^4(1-\tan\theta)^4\,d\theta \\ = \binom{8}{4} \frac1{\pi/4} \int_0^1 \frac{x^4(1-x)^4}{1+x^2}\,dx \\= \binom{8}{4} \frac1{\pi/4} \left(\frac{22}7 - \pi\right).$$

I can't think of a nice reason to expect a priori that this particular setup would give you a good rational approximation to $\pi$, but there you have it.

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