For every $n\in\mathbb{N}$, show that $0<e-\left(1+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}\right)<\frac{1}{n!n}$.

Question

I'm stucked in this problem: For every $n\in\mathbb{N}$, show that

$$0<e-\left(1+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}\right)<\frac{1}{n!n}$$ From this, prove $e$ is irrational.

I don't know how to prove $e-\left(1+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}\right)<\frac{1}{n!n}$. I already used induction, but it didn't work. I don't really mind about $e$ being irrational, just the inequality. Thanks for your help.

Answer 1

We need to use the Taylor expansion of $e$, namely, $$e = \sum_{k=0}^{\infty} \frac{1}{k!}.$$ From this, one can deduce $$e - \sum_{k=0}^{n}\frac{1}{k!} = \sum_{k=n+1}^{\infty}\frac{1}{k!} = \sum_{k=0}^{\infty}\frac{1}{(n+1+k)!}, \quad n=0,1,\dots.$$ This automatically provides the lower bound. To get the upper bound, realize that, for each $k\geq 1$, $$(n+1+k)! = (n+1+k)\cdots (n+2)(n+1)! > (n+1)!(n+1)^k,$$ hence $$\sum_{k=n+1}^{\infty}\frac{1}{k!} < \frac{1}{(n+1)!}\sum_{k=0}^{\infty}(n+1)^{-k},$$ which is a geometric series with a known formula. Simplifying this geometric series will yield the desired upper bound.

Source: Rudin's Principles of Mathematical Analysis, 3d ed, pp 65.