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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  1. Dear Fedya: simpler than all the above is this: (1+(1/n))^{i}<1+(i/n)+ (i/n)^{2} by a simple induction on i=1,2,…n, which gives for i=n and then taking the limit for n→∞ that "e" ≤ 3. Divide the semicircle by 2 radiuses emanating from the midpoint of its diameter joining the base radiuses and one another at 60 degrees. Join the endpoints of the radiuses with the nearer endpoint of the diameter and with one another. We get 3 triangles which are regular and similar. Therefore the 3 new sides also have length 1 which are strictly smaller than the corresponding disjoint arc of the semicircle. All this shows that its length is strictly bigger than 3.


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