How can I prove $\lim \limits_{n \to \infty}\left(1+\frac{1}{a_n}\right)^{a_n}=e$ without involving function limit?

Question

If I already know that $$\lim \limits_{n \to \infty} a_n=+\infty$$ Then how can I prove $$\lim \limits_{n \to \infty}\left(1+\frac{1}{a_n}\right)^{a_n}=e$$ without involving function limit?

This question comes because you may find some books on calculus or analysis (maybe they are badly written) require you to prove something like $$\lim \limits_{n \to \infty}\left(1+\frac{2}{n}\right)^{n}=e^2$$ or something more complex even before they formally introduce the definition of limit of a function, They are hard to prove because you can't simply take something like $\frac{n}{2}$ as a subsequence of $n$.

The definition of limit of a function (at infinity) here mean:

For a real function $f$ which is well-defined on $[a, +\infty)$, if for any $\epsilon >0$, there is a positive number $M \geq a$ such that when $x>M$ we can say $|f(x)-A|<\epsilon$, then $$\lim \limits_{x \to \infty}f(x)=A.$$

While the definition of limit of a sequence here mean:

For a sequence $\{a_n\}$, if for any $\epsilon>0$, there is a positive integer $N$ such that when $n>N$ we can say $|a_n-A|<\epsilon$, then $$\lim \limits_{n \to \infty}a_n=A.$$

I know sequence is a "special" kind of function whose domain is $\mathbb{N}$ and thus sequence limit is but a special case of function limit. Here I say avoid involving the idea of function limit means not to use the idea above but only to prove it by the "special case" below. After all, $(1+\frac{1}{a_n})^{a_n}$ is still a "special" function - a sequence.

p.s. $e$ is defined by $$\lim \limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n=e.$$

My try so far:

Since $$\lim \limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n=e,$$ for every $\epsilon>0$, there is $N \in \mathbb{N}$ s.t. for all $n>N$,$$|\left(1+\frac{1}{n}\right)^n-e|<\epsilon.$$

Meanwhile, since $$\lim \limits_{n \to \infty} a_n=+\infty,$$ for $N' \in \mathbb{N}$ and $N'>N$, there is $N'' \in \mathbb{N}$ s.t. for all $n>N''$, $a_n>N'>N.$

However, if $a_n$ become bigger then $1+\frac{1}{a_n}$ will be smaller, and vise versa, so I don't know how to deal with $\left(1+\frac{1}{a_n}\right)^{a_n}.$

Answer 1

Any subsequence of a Cauchy sequence is a Cauchy sequence with the same limit point, hence $$ \lim_{n\to +\infty}\left(1+\frac{1}{a_n}\right)^{a_n}=e $$ as soon as $\{a_n\}_{n\in\mathbb{N}}$ is a diverging sequence of natural numbers. The very last assumption can be dropped by noticing that $\frac{\log(1+x)}{x}$ is a positive, decreasing and convex function on $[0,1]$, hence if $a_m\in\mathbb{R}$ is between $n$ and $n+1$, $$\left(1+\frac{1}{a_m}\right)^{a_m}$$ is between $\left(1+\frac{1}{n}\right)^n$ and $\left(1+\frac{1}{n+1}\right)^{n+1}$, so the same conclusion as above follows by squeezing.

Answer 2

For $$\displaystyle\lim_{n\to\infty }\left(1+\frac{k}{n}\right)^n=e^k$$ with $k\in\mathbb N$, do it by induction. The case $k=1$ is the definition. For $k>1$, you have that $$\left(1+\frac{k+1}{n}\right)^n=\left(\frac{n+k+1}{n}\right)^n=\left(1+\frac{k}{n}\right)^n\left(1+\frac{1}{n+k}\right)^n=\underbrace{\left(1+\frac{k}{n}\right)^n}_{\to e^k\ (hyp\ induction)}\underbrace{\left(1+\frac{1}{n+k}\right)^{n+k}}_{\to e}\underbrace{\left(1+\frac{1}{n+k}\right)^{-k}}_{\to 1}\underset{n\to\infty }{\longrightarrow }e^{k+1}$$