TITLE: An elementary verification of the equivalence between two expressions for $e^x$
QUESTION [5 upvotes]: I would appreciate some constructive comments on the following argument for
\begin{equation*}
\sum_{n=0}^{\infty} \frac{x^{n}}{n!}
= \lim_{n\to\infty} \left(1 + \frac{x}{n}\right)^{n} .
\end{equation*}
I understand that there are several different arguments for it.  I like that this does not involve the natural logarithm function.  I looked in many elementary textbooks on real analysis for such an argument.  I only found one, but it was in the special case $x = 1$, and it was flawed.  The only analysis techniques used are the convergence of the sequence defined by
\begin{equation*}
\left(1 + \frac{x}{n}\right)^{n} ,
\end{equation*}
and the absolute convergence of
\begin{equation*}
\sum_{n=0}^{\infty} \frac{x^{n}}{n!} .
\end{equation*}
Here it is.
Demonstration
According to the Binomial Theorem, for every positive integer $n$,
\begin{align*}
&\sum_{i=0}^{n} \frac{x^{i}}{i!} - \left(1 + \frac{x}{n}\right)^{n} \\
&\qquad = \sum_{i=0}^{n} \frac{x^{i}}{i!} - \sum_{i=0}^{n} \binom{n}{i} \frac{x^{i}}{n^{i}} \\
&\qquad = \sum_{i=0}^{n} \left[\frac{x^{i}}{i!} - \binom{n}{i} \frac{x^{i}}{n^{i}} \right] \\
&\qquad = \sum_{i=0}^{n} \left[\frac{1}{i!} - \frac{1}{n^{i}} \binom{n}{i} \right] x^{i} \\
&\qquad = \sum_{i=2}^{n} \left[\frac{1}{i!} - \frac{1}{n^{i}} \binom{n}{i} \right] x^{i} .
\end{align*}
For each integer $2 \leq i \leq n$,
\begin{align*}
\frac{1}{n^{i}} \binom{n}{i} &= \frac{1}{n^{i}} \cdot \frac{n!}{i!(n-i)!} \\
&= \frac{1}{n^{i}} \cdot \frac{n(n - 1)(n - 2) \cdots (n - i + 1)}{i!} \\
&= \frac{1}{i!} \cdot \frac{n(n - 1) (n - 2) \cdots (n - (i - 1))}{n^{i}} \\
&= \frac{1}{i!} \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{i - 1}{n}\right) .
\end{align*}
So,
\begin{equation*}
\sum_{i=0}^{n} \frac{x^{i}}{i!} - \left(1 + \frac{x}{n}\right)^{n}
= \sum_{i=2}^{n} \left[1 - \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{i - 1}{n}\right) \right] \frac{x^{i}}{i!}
\end{equation*}
According to the Triangle Inequality, for each pair of positive integers $2 \leq k < n$,
\begin{align*}
&\left\vert \sum_{i=0}^{n} \frac{x^{i}}{i!} - \left(1 + \frac{x}{n}\right)^{n} \right\vert \\
&\qquad \leq \left\vert \sum_{i=2}^{k} \left[1 - \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{i - 1}{n}\right) \right] \frac{x^{i}}{i!} \right\vert \\
&\qquad\qquad + \left\vert \sum_{i=k+1}^{n} \frac{x^{i}}{i!} \right\vert
+ \left\vert \sum_{i=k+1}^{n} \frac{1}{n^{i}} \binom{n}{i} x^{i} \right\vert \\
&\qquad \leq \sum_{i=2}^{k} \left[1 - \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{i - 1}{n}\right) \right] \frac{\vert x \vert^{i}}{i!} \\
&\qquad\qquad + \sum_{i=k+1}^{n} \frac{\vert x \vert^{i}}{i!}
+ \sum_{i=k+1}^{n} \frac{1}{n^{i}}\binom{n}{i} \vert x \vert^{i} .
\end{align*}
\begin{align*}
&\sum_{i=k+1}^{n} \frac{1}{n^{i}} \binom{n}{i} \vert x \vert^{i} \\
&\qquad = \sum_{i=k+1}^{n} \frac{1}{i!} \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{i - 1}{n}\right)
\vert x \vert^{i} \\
&\qquad = \frac{1}{(k+1)!} \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{k}{n}\right) \vert x \vert^{k+1} \\
&\qquad\qquad \!\begin{aligned}[t]
&+ \frac{1}{(k+2)!} \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{k+1}{n}\right) \vert x \vert^{k+2} \\
&+ \frac{1}{(k+3)!} \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{k+2}{n}\right) \vert x \vert^{k+3} \\
&+\ldots
+ \frac{1}{n!} \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{n-1}{n}\right) \vert x \vert^{n} .
\end{aligned} \\
&\qquad = \frac{1}{k!} \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{k}{n}\right) \vert x \vert^{k} \\
&\qquad\qquad \!\begin{aligned}[t]
&\biggl[\frac{1}{k+1} \left(1 - \frac{k}{n}\right) \vert x \vert \\
&\hphantom{\biggl[\vphantom{\frac{1}{k+1}}}+ \frac{1}{(k+1)(k+2)} \left(1 - \frac{k}{n}\right) \left(1 - \frac{k+1}{n}\right) \vert x \vert^{2} \\
&\hphantom{\biggl[\vphantom{\frac{1}{k+1}}}+ \frac{1}{(k+1)(k+2)(k+3)}  \left(1 - \frac{k}{n}\right) \left(1 - \frac{k+1}{n}\right) \left(1 - \frac{k+2}{n}\right) \vert x \vert^{3} \\
&\hphantom{\biggl[\vphantom{\frac{1}{k+1}}}+ \ldots \\
&\hphantom{\biggl[\vphantom{\frac{1}{k+1}}}
+ \frac{1}{(k+1)(k+2) \cdots n} \left(1 - \frac{k}{n}\right) \left(1 - \frac{k+1}{n}\right) \cdots \left(1 - \frac{n-1}{n}\right)
\vert x \vert^{n-k}
\biggr]
\end{aligned} \\
&\qquad < \frac{\vert x \vert^{k}}{k!} \left[
\frac{\vert x \vert}{k+1} + \left(\frac{\vert x \vert}{k+1}\right)^{2}
+ \ldots + \left(\frac{\vert x \vert}{k+1}\right)^{n-k}
\right] . \\
\text{So, if $k \geq \vert x \vert$,} \\
&\sum_{i=k+1}^{n} \frac{1}{n^{i}} \binom{n}{i} \vert x \vert^{i} \\
&\qquad < \frac{\vert x \vert^{k}}{k!} \sum_{i=1}^{\infty} \left(\frac{\vert x \vert}{k+1} \right)^{i} \\
&\qquad = \frac{\vert x \vert^{k}}{k!} \cdot \frac{\dfrac{\vert x \vert}{k + 1}}{1 - \dfrac{\vert x \vert}{k+1}} \\
&\qquad = \frac{\vert x \vert^{k}}{k!} \cdot \frac{\vert x \vert}{k + 1 - \vert x \vert} , \\
\text{and if $k \geq 2\vert x \vert$,} \\
&\sum_{i=k+1}^{n} \frac{1}{n^{i}} \binom{n}{i} \vert x \vert^{i}
< \frac{\vert x \vert^{k}}{k!} .
\end{align*}
By the absolute convergence of
\begin{equation*}
\sum_{n=0}^{\infty} \frac{x^{n}}{n!} ,
\end{equation*}
for every $\epsilon > 0$, there is a big enough positive integer $K$ such that for every integer $k \geq K$,
\begin{equation*}
\sum_{i=k}^{\infty} \frac{\vert x \vert ^{i}}{i!}
< \frac{\epsilon}{3} ,
\end{equation*}
and so,
\begin{equation*}
\frac{\vert x \vert^{k}}{k!}
< \sum_{i=k}^{\infty} \frac{\vert x \vert ^{i}}{i!}
< \frac{\epsilon}{3}
\qquad \text{and} \qquad
\sum_{i=k+1}^{\infty} \frac{\vert x \vert ^{i}}{i!}
< \sum_{i=k}^{\infty} \frac{\vert x \vert ^{i}}{i!}
< \frac{\epsilon}{3} .
\end{equation*}
So, if $k \geq \max\{2\vert x \vert, \, K\}$, and if $n > k$,
\begin{equation*}
\sum_{i=k+1}^{n} \frac{1}{n^{i}} \binom{n}{i} \vert x \vert^{i}
< \frac{\vert x \vert^{k}}{k!}
< \frac{\epsilon}{3} .
\end{equation*}
Likewise, since for each integer $2 \leq i \leq k$,
\begin{equation*}
\lim_{n\to\infty} \left[1 - \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{i - 1}{n}\right) \right] = 0 ,
\end{equation*}
there is a big enough positive integer $N$ such that for every integer $n \geq N$,
\begin{equation*}
\sum_{i=2}^{k} \left[ 1 - \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{i - 1}{n}\right) \right]
< \frac{\epsilon}{3 \cdot \max\{\vert x \vert^{k}, \, 1\}} ,
\end{equation*}
and so,
\begin{equation*}
\sum_{i=2}^{k} \left[1 - \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{i - 1}{n}\right) \right] \frac{\vert x \vert^{i}}{i!}
< \frac{\epsilon}{3} .
\end{equation*}
Consequently, for any positive integers $k \geq \max\{2\vert x \vert, \, K\}$ and $n > \max\{k, \, N\}$,
\begin{equation*}
\left\vert \sum_{i=0}^{n} \frac{x^{i}}{i!} - \left(1 + \frac{x}{n}\right)^{n} \right\vert < \epsilon .
\end{equation*}
Equivalently,
\begin{equation*}
\sum_{i=0}^{\infty} \frac{x^{i}}{i!} = \lim_{n\to\infty} \left(1 + \frac{x}{n}\right)^{n} .
\end{equation*}

REPLY [2 votes]: Here is another simpler approach which does not use logarithms (so that this is not exactly an answer, but it is rather too long for a comment). Let $$E_{n}(x) = 1 + x + \frac{x^{2}}{2!} + \cdots + \frac{x^{n}}{n!}\tag{1}$$ then we know that $$E(x) = \lim_{n \to \infty}E_{n}(x)$$ exists for all $x \in \mathbb{R}$. Using binomial theorem it is easy to show that $$F_{n}(x) = \left(1 + \frac{x}{n}\right)^{n} \leq E_{n}(x) \leq \left(1 - \frac{x}{n}\right)^{-n} = G_{n}(x)\tag{2}$$ for $x > 0$ and $n > x$. (Clearly by binomial theorem each of the expressions $F_{n}(x), E_{n}(x)$ is a finite series of $(n + 1)$ terms and each term of $F_{n}(x)$ is less than or equal to the corresponding term of $E_{n}(x)$. For $0 < x < n$ the function $G_{n}(x)$ can be expressed as an infinite series via the binomial theorem for general index. And again each term of $E_{n}(x)$ is less than or equal to the corresponding term of $G_{n}(x)$. The restriction $0 < x < n$ is needed for the convergence of infinite series representation of $G_{n}(x)$.) 
Further it can be shown that both $(1 + x/n)^{n}$ and $(1 - x/n)^{-n}$ tend to same limit as $n \to \infty$ (you show that the ratio of these two expressions tends to $1$). It thus follows that for $x > 0$ we have $$E(x) = \lim_{n \to \infty}E_{n}(x) = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n} = \lim_{n \to \infty}\left(1 - \frac{x}{n}\right)^{-n}\tag{3}$$ The relation obviously holds for $x = 0$. For negative $x$ it is easy to see that that $E(x)E(-x) = 1$ (by multiplication of infinite series) and the proof is easily extended to negative values of $x$. (The fact that $F_{n}(-x) = 1/G_{n}(x)$ will be of help here because it implies that the common limit of $F_{n}(x)$ and $G_{n}(x)$, say $F(x)$, will satisfy $F(x)F(-x) = 1$ similar to $E(x)E(-x) = 1$.)

The approach given by OP is correct but bit lengthy with lot of intermediate steps and needs some patience to follow. A better approach in the same direction is to appeal to the general theorem called "Monotone Convergence Theorem":
If for all natural numbers $j, k$, the dual indexed sequence $a_{j, k}$ is non-negative and $a_{j, k} \leq a_{j + 1, k}$ for all natural numbers $j, k$ then $$\lim_{j \to \infty}\sum_{k = 1}^{\infty}a_{j, k} = \sum_{k = 1}^{\infty}\lim_{j \to \infty}a_{j, k}$$
For the current case let's set $$a_{j, k} = \binom{j}{k}\left(\frac{x}{j}\right)^{k}$$ if $k \leq j$ and $a_{j, k} = 0$ otherwise. Then we can see that $$\left(1 + \frac{x}{j}\right)^{j} = \sum_{k = 0}^{\infty}a_{j, k}$$ It is easily verified that $a_{j, k} \leq a_{j + 1, k}$ for all $j, k$ if $x > 0$. Hence the monotone convergence theorem applies and clearly $$\lim_{j \to \infty}a_{j, k} = \frac{x^{k}}{k!}$$ hence we have the result $$\lim_{j \to \infty}\left(1 + \frac{x}{j}\right)^{j} = \sum_{k = 0}^{\infty}\frac{x^{k}}{k!}$$ For $x < 0$ we again need the multiplicative properties of the series for $e^{x}$ namely that $e^{x}e^{-x} = 1$.