Prove $\lim_{n\to\infty}(1+1/n)^n=e$

Question

I would like to show that $\lim_{n\to\infty}(1+1/n)^n=e$, using the binomial theorem and the power series expansion of $e$. So to be clear: I do not want to use that $e^x:=\lim_{n\to\infty}(1+x/n)^n$, because that's not how I've learned the definition of $e^x$.

Basically I would like to show the following: $$ \lim_{n\to\infty}\sum_{k=0}^n\begin{pmatrix}n\\k\end{pmatrix}\left(\frac{1}{n}\right)^k=\sum_{k=0}^\infty\frac{1}{k!} $$ I'm guessing this should be possible. Maybe some rewriting could work: $$ \sum_{k=0}^n\begin{pmatrix}n\\k\end{pmatrix}\left(\frac{1}{n}\right)^k=\sum_{k=0}^n\frac{1}{k!}\cdot\frac{n!}{(n-k)!}\cdot\left(\frac{1}{n}\right)^k $$

EDIT (I was shown a mistake in the comments)

So I have to show the following: $$ \lim_{n\to\infty}\sum_{k=0}^n\frac{1}{k!}\frac{n!}{(n-k)!}\cdot\left(\frac{1}{n}\right)^k=\lim_{n\to\infty}\sum_{k=0}^n\frac{1}{k!}. $$

This seems doable, given that I see the term $\begin{align}\frac{1}{k!}\end{align}$ at both sides. So what can I do with $\begin{align}\frac{n!}{(n-k)!}\cdot\left(\frac{1}{n}\right)^k\end{align}$?

Answer 1

Notice that, if $k \geq 1$ then

\begin{align*} \frac{n!}{(n-k)!} \cdot \frac{1}{n^k} &= \frac{n(n-1)\cdots(n-k+1)}{n^k} \\ &= \left(1 - \frac{0}{n}\right)\left(1 - \frac{1}{n}\right)\cdots \left(1 - \frac{k-1}{n}\right) \tag{1} \end{align*}

and thus the LHS converges to $1$ as $n \to \infty$. Then, loosely speaking, the result follows from

$$ \lim_{n\to\infty} \sum_{k=0}^{n} \frac{1}{k!} \cdot \frac{n!}{(n-k)! n^k} = \sum_{k=0}^{n}\frac{1}{k!} \left( \lim_{n\to\infty} \frac{n!}{(n-k)! n^k} \right) = \sum_{k=0}^{n} \frac{1}{k!} = e. \tag{2}$$

Of course, interchaning limit and infinite summation is a huge leap of faith (and often yields a simply wrong answer), so it should be accompanied by a rigorous justification.

---

Assuming that you have not been exposed to real analysis, let me give an elementary solution. (But I am not claiming that this is either the most intuitive nor the easiest-to-swallow one.)

We begin by proving the following lemma:

Lemma. If $a_1, \cdots, a_n \in [0, 1]$, then $$ (1 - a_1)\cdots(1-a_n) \geq 1 - (a_1 + \cdots + a_n). \tag{*}$$

This is easily proved by invoking mathematical induction. Indeed, the claim is trivial if $n = 1$. Next, assume that that $\text{(*)}$ is true for $n$. If $s_{n+1} := a_1 + \cdots + a_{n+1} > 1$ then

$$ (1 - a_1)\cdots(1-a_{n+1}) \geq 0 > 1 - s_{n+1} $$

and the claim is trivial. So we assume that $s_{n+1} \leq 1$. Then $s_n := a_1 + \cdots + a_n \leq 1$ as well and by the induction hypothesis,

\begin{align*} (1 - a_1)\cdots(1-a_{n+1}) & \geq (1 - s_n)(1 - a_{n+1}) \\ &= 1 - (s_n + a_n) + s_n a_n \\ &\geq 1 - s_{n+1}. \end{align*}

Therefore $\text{(*)}$ follows by mathematical induction. ////

Now returning to our problem, applying the lemma to $\text{(1)}$ gives

$$ 1 \geq \frac{n!}{(n-k)!} \cdot \frac{1}{n^k} \geq 1 - \sum_{j=0}^{k-1} \frac{j}{n} = 1 - \frac{k(k-1)}{2n}. $$

This inequality is also true when $k = 0$. Summing this over $k = 0, \cdots, n$, we have

$$ \sum_{k=0}^{n} \frac{1}{k!} - \frac{1}{2n}\sum_{k=2}^{n} \frac{1}{(k-2)!} \leq \left(1 + \frac{1}{n}\right)^n \leq \sum_{k=0}^{n} \frac{1}{k!}. $$

Taking $n \to \infty$ gives the desired proof.