Q: $\lim_{n\to \infty}\left(1 + \frac{1}{n}\right)^{n} = e$

Question

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).

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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...

Answer 1

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.

Answer 2

The problem with talking about $\lim_{n \to \infty} t_n$ is that you don't know that the limit exists. However, the $\liminf$ and $\limsup$ always exist (if you include $\pm\infty$). To prove the limit exists, they show that the $\liminf$ equals the $\limsup$.

Answer 3

In situations where your don't know that limit a priori exists, it's a standard trick to first calculate both $\liminf$ and $\limsup$ (since they always exist) and then see that they are equal (and finite) which is necessary and sufficient condition for limit to exist (and in this case we have $\lim=\liminf=\limsup$).

Note also that the crucial thing that allowed you to do these manipulations was that if $a_n\leq b_n$ for all $n$, then $\liminf a_n\leq \liminf b_n$ and $\limsup a_n\leq \limsup b_n$ (a property that holds for limits as well).