Prove that $e$ is irrational using this result

Question

I have to prove that $e$ is irrational using this result

$1/e = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!}$

and getting the estimation for the partial sums $0 < 1/e - s_{2k-1} < \frac{1}{(2k)!}$

I tried, but I obtain $0 < 1/e + \frac{1}{(2k-1)!} = \frac{1}{(2k)!} < 1$, and I don't think it's good

Answer 1

Once you know that $0<1/e-s_{2k-1}<\frac{1}{(2k)!}$,

suppose that $e=\frac{p}{q}$ for integers $p,q$ and $p\neq 0$ ($e\neq 0$ obviously). From above

$$ 0<\frac{(2k)!}{e}-m_k<1 $$

where $m_k=(2k)!s_{2k-1}$ is an integer. Since this is true for all $k$, we can pick some $k$ such that $2k\ge |p|$ so that $\frac{(2k)!}{e}=\frac{q(2k)!}{p}$ is also an integer.

But this is a contradiction, since then $\frac{(2k)!}{e}-m_k$ is an integer between $0$ and $1$.

Answer 2

Assume that $e=\frac{p}{q}$ with $p,q\in\mathbb{Z}^+$ and $\gcd(p,q)=1$. We have $$ \frac{q}{p}=\sum_{k\geq 0}\frac{(-1)^k}{k!} = \sum_{k=0}^{p}\frac{(-1)^k}{k!}+\sum_{k>p}\frac{(-1)^k}{k!} $$ from which $$ q(p-1)! = \underbrace{\sum_{k=0}^{p}(-1)^k\frac{p!}{k!}}_{\in\mathbb{Z}}+\underbrace{\sum_{k>p}\frac{(-1)^k p!}{k!}}_{\text{absolute value }\in(0,1)} $$ leading to a contradiction.