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.