vrotterigzl

2022-06-20

Let $\left\{{ϵ}_{n}\right\}$ be a sequence where ${ϵ}_{n}$ is either $1$ or $-1$. How can We Showthat the sum of the series
$\sum _{n=0}^{\mathrm{\infty }}\frac{{ϵ}_{n}}{n!}$
is an irrational number?

Aaron Everett

If there exist $p\in \mathbb{Z}$ and $q\in \mathbb{N}$, such that $\frac{p}{q}=\sum _{n=0}^{\mathrm{\infty }}\frac{{ϵ}_{n}}{n!}$, then $q!\cdot \sum _{n=q+1}^{\mathrm{\infty }}\frac{{ϵ}_{n}}{n!}$ must be an integer. However,
$|q!\cdot \sum _{n=q+1}^{\mathrm{\infty }}\frac{{ϵ}_{n}}{n!}-\frac{{ϵ}_{q+1}}{q+1}|\le \sum _{n=q+2}\frac{q!}{n!}<\frac{1}{\left(q+1\right)\left(q+2\right)}\cdot \sum _{m=0}\frac{1}{{2}^{m}}=\frac{2}{\left(q+1\right)\left(q+2\right)},$
which implies that
$0<\frac{1}{q+1}-\frac{2}{\left(q+1\right)\left(q+2\right)}\le |q!\cdot \sum _{n=q+1}^{\mathrm{\infty }}\frac{{ϵ}_{n}}{n!}|\le \frac{1}{q+1}+\frac{2}{\left(q+1\right)\left(q+2\right)}<1,$