I have several questions about the proof of the theorem. Since I'm not a math major and I'm only a novice in real analysis, I would appreciate if you could explain in details.
Q1: What does "let $n\rightarrow \infty$" mean? What do we do to the inequality
$$
t_n\geq1+1+\frac{1}{2!}\left( 1-\frac{1}{n}\right)+\cdots+\frac{1}{m!}\left( 1-\frac{1}{n}\right)\cdots\left( 1-\frac{m-1}{n}\right)
$$
by letting $n\rightarrow \infty$ and keeping $m$ fixed?
Q2: In the proof we first assume $n\geq m$ and then let $n\rightarrow \infty$ while keeping $m$ fixed; finally we let $m\rightarrow \infty$. I'm confused about this procedure. Why can we take the limit in the order we want? We first fix $m$ and then let $m \rightarrow \infty$, but since $m$ is already fixed, how does it go to infinity? Since both $n$ and $m$ goes to infinity, how can we make sure the relation $n\geq m$ always holds?
Best Answer
You have an inequality $$t_n\ge u(m,n)$$ where $$u(m,n)=1+1+\frac1{2!}\left(1-\frac1n\right)+\cdots+\frac1{m!}\left(1-\frac1n\right) \cdots\left(1-\frac{m-1}n\right).$$ Just from the inequality you get $$\liminf_{n\to\infty}t_n\ge\liminf_{n\to\infty} u(m,n).$$ But $$\lim_{n\to\infty}u(m,n)=1+1+\frac1{2!}+\cdots+\frac1{m!}.$$ Therefore $$\lim\inf_{n\to\infty}u(m,n)=1+1+\frac1{2!}+\cdots+\frac1{m!}$$ and then $$\liminf_{n\to\infty}t_n\ge1+1+\frac1{2!}+\cdots+\frac1{m!}.$$ As this is true for all $m$, then $$\liminf_{n\to\infty}t_n\ge1+1+\frac1{2!}+\cdots=e.$$