TITLE: Q: $\lim_{n\to \infty}\left(1 + \frac{1}{n}\right)^{n} = e$
QUESTION [5 upvotes]: I am having difficulty with the proof $$\lim_{n\to \infty}\left(1 + \frac{1}{n}\right)^{n} = e$$ in Rudin's Principles of Mathematics.
In particular, the last few steps.  The proof is as follows:

Theorem. $\lim_{n\to\infty} \left(1 + \frac1n \right)^n = e.$
Proof. Let $$s_n = \sum_{k=0}^{n} \frac{1}{k!}, \qquad t_n = \left(1 + \frac1n \right)^n.$$
By the binomial theorem, $$t_n = 1 + 1 + \frac1{2!}\left( 1 - \frac1n \right) + \frac1{3!}\left(1 - \frac1n \right)\left(1 - \frac2n \right) + \dots \\ + \frac1{n!}\left(1 - \frac1n \right)\left(1 - \frac2n \right)\dots\left(1-\frac{n-1}{n}\right).$$
Hence $t_n \leq s_n$, so that $$\limsup_{n\to\infty} t_n \leq e, 
\tag{14}$$ by Theorem 3.19. Next, if $n \geq m$, $$t_n \geq 1 + 1 + \frac1{2!}\left(1-\frac1n\right) + \dots + \frac1{m!}\left(1-\frac1n\right)\dots\left(1-\frac{m-1}{n}\right).$$
Let $n\to\infty$, keeping $m$ fixed. We get $$\liminf_{n\to\infty} t_n \geq 1 + 1 + \frac1{2!} + \dots + \frac1{m!},$$ so that $$s_m \leq \liminf_{n\to\infty} t_n.$$
Letting $m\to\infty$, we finally get $$e \leq \liminf_{n\to\infty} t_n. \tag{15}$$
The theorem follows from (14) and (15).

(original image)
My question:

What allows us to say that $$\lim_{n\to \infty}\text{inf } t_{n} \geq s_{m}?$$

I see that $\lim_{n\to \infty}t_{n} \geq s_{m}$ (if it exists, I suppose otherwise it's vacuously true...), but why the $\liminf$?  The text does similar steps in other proofs without explanation, so I am not sure if I am misunderstanding something obvious...

REPLY [3 votes]: If we have a sequence $t_n$, we don't necessarily have a limit $\lim t_n$. This means that, until we show the limit exists, we can't write things like $\lim t_n \geq s_m$. However, a handy replacement is $\liminf$ and $\limsup$. These are nice because they always exist for any sequence of numbers, and so writing things like $\liminf t_n \geq e$ and $\limsup t_n \leq e$ are fine.
What makes these two things so handy is that $\lim t_n$ exists precisely when $\liminf t_n = \limsup t_n$, and is equal to the common value. Hence, by showing $e \leq \liminf t_n \leq \limsup t_n \leq e$, we have not only shown $\lim t_n$ exists, but computed its value.