I do not understand an inequality used to prove that e is bigger/equal to exp(1)

Question

I have seen the following inequalies:

$$e \geq 1+1+\sum \limits_{k=2}^{n}\frac{1}{k!}=s_{n} \implies e\geq \lim_{n\to\infty}s_{n}=exp(1)$$

Based on this information, how can one explain the implication? Why would $$e\geq \lim_{n\to\infty}s_{n}$$ be true? I only know that $e \geq 1+1+\sum \limits_{k=2}^{n}\frac{1}{k!}=s_{n}$.

Wouldn't the implication also imply $\sum \limits_{k=0}^{n}\frac{1}{k!} = \sum \limits_{k=0}^{\infty}\frac{1}{k!} = \lim_{n\to\infty}s_{n}$?

Answer 1

Whenever you have a convergent sequence $(x_n)_{n\in\Bbb N}$ and a number $M$ such that $(\forall n\in\Bbb N):x_n\leqslant M$, you also have $\lim_{n\to\infty}x_n\leqslant M$. If you apply this to $(s_n)_{n\in\Bbb N}$ and to the number $e$, you get what you want.

Answer 2

This is a consecuence of the following statement:

Let be $\{x_n\}_{n\in\mathbb{N}}, \{y_n\}_{n\in\mathbb{N}}$ two convergent sequences, such that $x_k\ge y_k,~~~ \forall k\ge N_0, N_0\in\mathbb{N}$. If $x_n\to a$ and $y_n\to b$, then: $a\ge b$.

In this case, $x_n$ is a constant sequence with unique value is $e$.