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

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...

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.

• I think I understand the overall argument and the way the text uses the $\liminf$. Though, I think somehow I am confused with the definition of $\liminf$ and $\limsup$. My understanding is that $\liminf$ of a sequence, in this case $\{t_{n}\}$, is the greatest lower bound of all possible sub-sequences of $\{t_{n}\}$. My thinking right now is, say we consider $t_{4}$. Then choose $m = 2$ so that $$t_{4} \geq 1+1+\frac{1}{2!}\left(1-1/2\right)$$ Can we consider a finite sub-sequence of $\{t_{n}\}$ consisting of only $t_{1}$? This is seems to apparently converge by definition, to $1+1 = 2$. Jun 21, 2014 at 19:09
• I meant $g.l.b$ of all possible sub-sequential limit points. Though, I do follow after that step. But I don't see exactly what allows us to say that the g.l.b of all sub-sequential limit points of $\{t_{n}\}$ is greater than or equal to $s_{m}$. Jun 21, 2014 at 19:17
• I find it somewhat confusing that we call them sub-sequential limit points after studying what were limit points in metric spaces (chapter $2$). Jun 21, 2014 at 19:27
• @DavidJhoo 1) By definitions sequences (and subsequences) are not finite. 2) Here, "sub-sequential limit points" are the limits of convergent subsequences. Yes "limit point" was used to mean something else in the context of metric spaces and topology, which is slightly different. You shouldn't dwell too much on the fact that they share the same name, although you should see why they are similar. Jun 21, 2014 at 22:17
• @DavidJhoo 3) In the proof above, we are given that $s_n$ is a convergent sequence, whereas we are not sure if $t_n$ is convergent or not. By comparing terms of then $t_n$ sequence with terms of the $s_n$ sequence, you can then take the limit as $n \to \infty$ to compare the limit of the $s_n$ sequence ($e$ in this case) with the $\limsup$ or $\liminf$ of the $t_n$ sequence (if you're not convinced, use the subsequence definition to take a subsequence of $t_n$ and compare it with the corresponding $s_n$). Jun 21, 2014 at 22:20

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 $\limsup$.

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).