I have a question on the second part of the proof where $n \geq m$.
I understand that when $n \rightarrow \infty$, $s_m \leq \lim \inf t_n$.
However, when $m \rightarrow \infty$, it states that $e \leq \lim \inf t_n$.
This should not be true since $$t_n = 1 + 1 + \frac{1}{2!}(1 – \frac{1}{n}) + \frac{1}{3!}(1 – \frac{1}{n})(1 – \frac{2}{n}) + … + \frac{1}{n!}(1 – \frac{1}{n})(1 – \frac{2}{n}) … (1 – \frac{n – 1}{n})$$ and $$s_m = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + … + \frac{1}{m!}$$.
So when both n and m converges to infinity, shouldn't $\lim \inf t_n \leq \lim \inf s_m = e$?
Best Answer
Note that the second expression for $t_n$ contains terms with $m$. The estimate with the $\liminf$ follows readily. Also, it seems like you missed the order in which $m,n \to \infty$.
To see that the second expression for $t_n$ is correct, look at the expression you wrote, and instead of writing all the $n$ terms, instead write only $m\leq n$ terms. Clearly $\leq$ will hold since all these terms are non-negative.