Mathematicians do like giving proofs of known results from some specific point of view.

Here goes the combinatorial proof of the inequality

e\leqslant \pi. \quad \quad (1)

To be precise, I define our heroes. \pi is a length of a half-circle of the unit radius, and e is the common point of the segments [(1+1/n)^n,(1+1/n)^{n+1}], n=1,2,\ldots

The idea is that for a positive integer n and a complex q, |q|\leqslant 1, we have

{\displaystyle A:=\left|1\cdot (1+q)(1+q+q^2)\cdot \ldots  \cdot (1+q+\ldots+q^{n-1})\right| \leqslant n!\leqslant n^2(n/e)^{n-1}.}

The latter inequality is obtained by multiplying the definitive inequalities e\leqslant (1+1/k)^{k+1} for k=1,2,\ldots,n-1.

We apply it for q=e^{i\pi/n}, the primitive root of unity of degree 2n.

Note that |1-q|< \pi/n, since the chord of the unit circle joining 1 and q is shorter than the arc between these two points (which equals \pi/n, since the whole circle is a union of 2n congruent arcs.)

We have {\displaystyle B:=(1-q)(1-q^2)\cdot \ldots \cdot(1-q^{2n-1})=\prod_{w^{2n}-1=0, w\ne 1} (1-w)=(z^{2n}-1)'|_{z=1}=2n.}

On the other hand, since q^n=-1 and 1-q^{n+k} is the complex conjugate to 1-q^{n-k} for k=1,\ldots,n-1, we get

2n=B=2\left|(1-q)(1-q^2)\ldots (1-q^{n-1})\right|^2 =2|1-q|^{2n-2}A^2\leqslant 2(\pi/n)^{2n-2}n^4(n/e)^{2n-2}.

Thus (e/\pi)^{2n-2}\leqslant n^{3} that yields (1) as (e/\pi)^{2n-2} would grow exponentially fast if e>\pi.

Well, if you really prefer combinatorial proofs, here is a variant of the above argument for you. Although I do not provide any combinatorial quantity counted by \pi-e (that what we expect from a combinatorial proof of inequality (1)), I instead do this for something like (n/e)^n-(n/\pi)^n.

Count the number N of permutations of 1,\ldots,n for which the number of inversions is divisible by 2n. Since

\sum_{\pi\in S_n} x^{inv(\pi)}=1\cdot (1+x)(1+x+x^2)\cdot \ldots  \cdot (1+x+\ldots+x^{n-1})

is a generating polynomial for the number of permutations with the given number of inversions, by the polarization formula

[x^a]f+[x^{a+m}]f+[x^{a+2m}]f+\ldots=m^{-1}\sum\limits_{w^m=1} w^{-a}f(w), 0\leqslant a<m, f(x)\in\mathbb{C}[x],

for a=0,m=2n we get

N=(2n)^{-1}\sum_{w:w^{2n}=1} 1\cdot (1+w)(1+w+w^2)\cdot \ldots  \cdot (1+w+\ldots+w^{n-1}).

If e>\pi, we see that the summands corresponding to w=e^{\pm i\pi/n} are negative reals provided that n=8k+5 (this follows from evaluating the argument of 1+q+\ldots+q^{\ell}, which equals \frac{\ell\pi}{2n}, and summing over \ell=1,2,\ldots,n-1: we get \frac{(n-1)n\pi}{4n} that is \pi modulo 2\pi for n=8k+5). By absolute value they are exponentially greater both than n! (which corresponds to w=1, see the explanation above) and all other summands (since say |1-w|>(3/2)|1-q| for other roots, and |(1-w)(1-w^2)\ldots (1-w^n)| is either 0 or the same as |(1-q)(1-q^2)\ldots(1-q^n)|. So we get N<0 that is absurd.


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