# e⩽π

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

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.