# The transcendence of e (part 3)

This is the third part in a 3-part blog post in which we prove that ${e}$ is transcendental.

Three-step proof that $e$ is transcendental
Step 1
Step 2
Step 3

Recall that in step 1 and step 2 we proved that for any prime ${p}$ sufficiently large ${c_{0}F(0)+\cdots+c_{n}F(n)=c_{1}\beta_{1}+\cdots+c_{n}\beta_{n}}$ and that ${c_{0}F(0)+\cdots+c_{n}F(n)}$ is a nonzero integer. In this step we will prove that if ${p}$ is large enough, then ${|c_{1}\beta_{1}+\cdots+c_{n}\beta_{n}|<1}$ and hence $c_{1}\beta_{1}+\cdots+c_{n}\beta_{n}$ cannot be a nonzero integer. This will give a contradiction and we will conclude that ${e}$ is transcendental.

Step 3.

Recall that ${\beta_{k}=-ke^{k(1-\alpha_{k})}f(k\alpha_{k})}$. Expanding this using our polynomial $\displaystyle f(x)=\frac{x^{p-1}(1-x)^{p}(2-x)^{p}\cdots(n-x)^{p}}{(p-1)!}$

we obtain $\displaystyle \beta_{k}=\frac{-ke^{k(1-\alpha_{k})}(k\alpha_{k})^{p-1}(1-k\alpha_{k})^{p}(2-k\alpha_{k})^{p}\cdots(n-k\alpha_{k})^{p}}{(p-1)!}.$

Recall that ${0<\alpha_{k}<1}$ and ${k\le n}$. So ${e^{k(1-\alpha_{k})}\le e^{n}}$, ${k(k\alpha_{k})^{p-1}\le n^{p}}$, and ${(1-k\alpha_{k})^{p}(2-k\alpha_{k})^{p}\cdots(n-k\alpha_{k})^{p}\le (n!)^{p}}$. So we have $\displaystyle |\beta_{k}|\le \frac{e^{n}n^{p}(n!)^{p}}{(p-1)!},$

and (using the well-known fact that in the limit factorials grow faster than exponentials) $\displaystyle \lim_{p\rightarrow\infty}|\beta_{k}|\le\lim_{p\rightarrow\infty}\frac{e^{n}n^{p}(n!)^{p}}{(p-1)!}=e^{n}\lim_{p\rightarrow\infty}\frac{(n\cdot n!)^{p}}{(p-1)!}=0.$

In particular, we can choose the prime ${p}$ large enough so that for ${k=1,\ldots n}$, ${|c_{k}\beta_{k}|<1/n}$. So $\displaystyle |c_{1}\beta_{1}+\cdots+c_{n}\beta_{n}|\le|c_{1}\beta_{1}|+\cdots+|c_{n}\beta_{n}|<1,$

as promised. This yields a contradiction and we conclude that ${e}$ is transcendental.