A remarkable sequence

Theorem 1.

Let $x\in\mathbf{R}$, $x\gt 0$ is given. Then \[ \left( 1+\frac{x}{n}\right)^n\lt\left( 1+\frac{x}{n+1}\right)^{n+1}
\] for any positive integer $n$

By arithmetic-geometric mean inequality
\[
\sqrt [n+1] {\left( 1+\dfrac {x} {n}\right) ^{n}\cdot 1} < \dfrac {n\left( 1+\dfrac {x} {n}\right) +1} {n+1}=1+\dfrac {x} {n+1}. \,\,\blacksquare
\]

Theorem 2.

For any positive integer $n$ we have
\[
\left( 1+\frac{1}{n}\right)^n<3.
\]

\begin{align*}
\left( 1+\dfrac {1} {n}\right)^{n} &=1+\sum_{k=1}^{n}\binom{n}{k}\dfrac{1}{n^{k}}\\
&=1+\sum_{k=1}^{n}\dfrac {1} {k!}\dfrac {n\left( n-1\right) \cdots \left( n-k+1\right) } {n^{k}}\\
&\leq 1+\sum_{k=1}^{n}\dfrac {1} {k!}\\
&\leq 1+\sum_{k=1}^{n}\dfrac {1} {2^{k-1}}\\
&=1+\sum_{k=0}^{n-1}\dfrac {1} {2^{k}}\\
&< 3. \,\,\blacksquare
\end{align*}

We prove by induction that for $i\leq n$ we have
\[
\left( 1+\frac{1}{n}\right)^i < 1+\frac{i}{n}+\frac{i^2}{n^2}.
\]
(i)  $i=1$ $\checkmark$
(ii) $i$ $\checkmark$
(iii) $i+1$.
\[
\begin{align*}
\left( 1+\frac{1}{n}\right)^{i+1} &=\left( 1+\frac{1}{n}\right)^i \left( 1+\frac{1}{n}\right)\\
&\leq \left( 1+\frac{i}{n}+\frac{i^2}{n^2} \right) \left( 1+\frac{1}{n}\right)\\
&=1+\frac{i+1}{n}+\frac{n(i^2+i)+i^2}{n^3}\\
&\leq 1+\frac{i+1}{n}+\frac{n(i^2+i)+(i+1)n}{n^3}\\
&= 1+\frac{i+1}{n}+\frac{(i+1)^2}{n^2}. \,\,\blacksquare
\end{align*}
\]

Theorem 3.

The sequence  $\{\left(1+\frac{1}{n}\right)^n\}_{n=1,\infty}$ has limit.

Notation.

\[
\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=:e. \]

It follows from Theorem 1 and Theorem 2. $\blacksquare$

Theorem 4.

Let $k$ be positive integer. Then
\[
\lim_{n\to\infty}\left(1+\frac{k}{n}\right)^n=e^k.
\]

We have seen in Theorem 1 that the sequence
\[
b_n:=\left(1+\frac{k}{n}\right)^n
\]
is strictly increasing. Since
\[
b_{km}=\left(1+\frac{1}{m}\right)^{km}=\left[\left(1+\frac{1}{m}\right)^m\right]^k
\]
we have $\lim_{m\to\infty}b_{km}=e^k$, which implies
\[
\lim_{n\to\infty}b_n=e^k. \quad \blacksquare \]

Theorem 5.

Let $r=p/q$ be positive rational number $p,q>0$, $(p,q)=1$. Then
\[
\lim_{n\to\infty}\left(1+\frac{r}{n}\right)^n=e^r.
\]

\[
\left(1+\frac{r}{n}\right)^n =\left[\left(1+\frac{p}{qn}\right)^{qn}\right]^{1/q} .
\]
By Theorem 4 we obtain
\[
\lim_{n\to\infty}\left(1+\frac{r}{n}\right)^n=\lim_{n\to\infty}\left[\left(1+\frac{p}{qn}\right)^{qn}\right]^{1/q}=\left(e^p\right)^{1/q}=e^r.  \quad\blacksquare
\]

Theorem 6.

Let $x\in\mathbf{R}$, $x> 0$ is given. Then
\[
\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=e^x .
\]

Choose $0<r_1<x<r_2$  ($r_1,r_2\in\mathbf{Q}$). Then
\[
\left(1+\frac{r_1}{n}\right)^n <\left(1+\frac{x}{n}\right)^n <\left(1+\frac{r_2}{n}\right)^n.
\]
\[
\lim_{n\to\infty} \left(1+\frac{r_1}{n}\right)^n \leq \liminf_{n\to\infty} \left(1+\frac{x}{n}\right)^n
\leq \limsup_{n\to\infty} \left(1+\frac{x}{n}\right)^n \leq \lim_{n\to\infty} \left(1+\frac{r_2}{n}\right)^n,
\]
that is,
\[
e^{r_1}\leq \liminf_{n\to\infty} \left(1+\frac{x}{n}\right)^n
\leq \limsup_{n\to\infty} \left(1+\frac{x}{n}\right)^n \leq e^{r_2}.
\]
Since
\[
\sup_{r_1\leq x,\, r_1\in\mathbf{Q}}e^{r_1}=e^x=\inf_{x\leq r_2,\, r_2\in\mathbf{Q}} e^{r_2}
\]
it follows that
\[
\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=e^x .\qquad\blacksquare
\]

In the next Theorem we show that the restriction $x>0$ can be dropped.

Theorem 7.

Let $x\in\mathbf{R}$, $x$ is given. Then
\[
\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=e^x .
\]

It is enough to prove for $x<0$. Let $x=:-y$. Then $y>0$.
Now we can write
\[
\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n =\lim_{n\to\infty}\left(1-\frac{y}{n}\right)^n
=\lim_{n\to\infty} \frac{\left(1-\frac{y}{n^2}\right)^n}{\left(1+\frac{y}{n}\right)^n}.
\]
Here by Bernoulli's inequality
\[
1\geq \left(1-\frac{y}{n^2}\right)^n \geq 1-\frac{y}{n}
\]
which yields
\[
\lim_{n\to\infty} \left(1-\frac{y}{n^2}\right)^n=1.
\]
Using Theorem 6 we obtain
\[
\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=\frac{1}{e^y}=e^{-y}=e^x.\qquad\blacksquare
\]

Theorem 8.

Let $x\in\mathbf{R}$, $x\gt 0$ is given. Then
\[ \lim_{n\to\infty}\sum_{k=0}^{n}\,\frac{x^k}{k!}=e^x. \]

\begin{align} \left(1+\frac{x}{n} \right)^n &=\sum_{k=0}^n \binom{n}{k}\frac{x^k}{n^k}\nonumber\\ &=1+x+\sum_{k=2}^n \frac{x^k}{k!}\cdot\frac{(n-1)\cdots(n-k+1)}{n^{k-1}}\nonumber\\ &\leq \sum_{k=0}^n \frac{x^k}{k!} \label{eq:lowerineq}. \end{align} On the other hand, let $n$ be a fixed positive integer and $m\gt n$ an arbitrary positive integer. Then \[ \begin{align*} e^x\gt\left(1+\frac{x}{m} \right)^m &=\sum_{k=0}^m \binom{m}{k}\frac{x^k}{m^k}\\ &=1+x+\sum_{k=2}^m \frac{x^k}{k!}\cdot\frac{(m-1)\cdots(m-k+1)}{m^{k-1}}\\ &\geq 1+x+\sum_{k=2}^n \frac{x^k}{k!}\cdot\left( 1-\frac{1}{m}\right)\cdots\left(1-\frac{k-1}{m} \right). \end{align*} \] Taking the limit $m\to\infty$ it follows \begin{align*} e^x\geq 1+x+\sum_{k=2}^n \frac{x^k}{k!}, \end{align*} that is, \begin{equation} \label{eq:upperineq} e^x\geq \sum_{k=0}^n \frac{x^k}{k!}. \end{equation} From \eqref{eq:lowerineq} and \eqref{eq:upperineq} we obtain \[ \left(1+\frac{x}{n} \right)^n\leq \sum_{k=0}^n \frac{x^k}{k!}\leq e^x. \] Now taking the limit $n\to\infty$ and taking into account of Theorem 7 we obtain statement in Theorem 8.$\quad\blacksquare$

In the next Theorem we show that the restriction $x>0$ can be dropped.

Theorem 9.

Let $x\in\mathbf{R}$ be given. Then
\[ \lim_{n\to\infty}\sum_{k=0}^{n}\,\frac{x^k}{k!}=e^x. \]

Lemma.

Denote \[ f(x):=\sum_{k=0}^{\infty}\frac{x^k}{k!},\qquad x\in\mathbf{R}. \] Then \[ f(x_1)f(x_2)=f(x_1+x_2)\qquad\text{for all }x_1,x_2\in\mathbf{R}. \]

Since power series are absolutely convergent in the domain of convergence we can apply the Cauchy-product to determine $f(x_1)f(x_2)$. \begin{align*} f(x_1)f(x_2) &=\left(\sum_{k_1=0}^{\infty}\frac{x_1^{k_1}}{k_1!}\right)\left(\sum_{k_2=0}^{\infty}\frac{x_2^{k_2}}{k_2!}\right)\\ &=\sum_{k=0}^{\infty} \frac{1}{k!}\left(x_1^k+\ldots+\frac{k!}{m!(k-m)!}x_1^{k-m}x_2^m+\ldots+x_2^k\right) \\ &=\sum_{k=0}^{\infty}\frac{(x_1+x_2)^k}{k!}=f(x_1+x_2), \end{align*} where we used the binomial theorem in the last step. $\quad \blacksquare$

It is enough to prove for $x\lt 0$. Putting \[ x_1:=x\gt 0 \] and \[ x_2:=-x \] in Lemma we obtain \[ f(x)f(-x)=f(0)=1. \] By Theorem 8 $f(x)=e^x$, so it follows \[ f(-x)=\lim_{n\to\infty}\sum_{k=0}^{n}\,\frac{(-x)^k}{k!}=e^{-x}.\qquad\blacksquare \]

Theorem 10.

The function $e^z$ is defined uniquely by \[ e^z:=\sum_{k=0}^{\infty}\frac{z^k}{k!},\qquad z\in\mathbf{C}. \]

Since $f(z):=\displaystyle\sum_{k=0}^{\infty}\frac{z^k}{k!}$ is an analytical function, $f(x)=e^x$ for $x\in\mathbf{R}$, and analytical continuation is unique, we may write $f(z)=e^z$. $\blacksquare$

Theorem 11.

\[ \lim_{n\to\infty}\left( 1+\frac{z}{n}\right)^n=e^z, \qquad z\in\mathbf{C}. \] In fact, the convergence is uniform on every bounded subsets of $\mathbf{C}$.

We have seen in Theorem 10 that $e^z:=\displaystyle\lim_{n\to\infty}\sum_{k=0}^{n}\frac{z^k}{k!}$, $z\in\mathbf{C}$. By binomial theorem \begin{align*}
\left(1+\dfrac {z} {n}\right)^{n} &=\sum_{k=0}^{n}\binom{n}{k}\dfrac{z^k}{n^{k}}\\
&=\sum_{k=0}^{n}\dfrac {z^k} {k!}\dfrac {n\left( n-1\right) \cdots \left( n-k+1\right) } {n^{k}}\\
\end{align*}
So it is enough to prove \begin{equation} \label{eq:limitis0} \lim_{n\to\infty}\sum_{k=0}^{n}\frac{z^k}{k!}\left(1-\frac{n(n-1)\cdots(n-k+1)}{n^k} \right)=0. \end{equation} We note that \[ 1\gt 1-\frac{n(n-1)\cdots(n-k+1)}{n^k} \geq 0. \] Let $\varepsilon\gt 0$ be arbitrary. Then there exists $N$ such that \[ \sum_{k=N}^{\infty} \frac{|z|^k}{k!}\lt \varepsilon. \] For any fixed $k$ \[ \lim_{n\to\infty}\left(1-\frac{n(n-1)\cdots(n-k+1)}{n^k} \right)=0. \] So there exists $M\gt N$ such that \[ \left(1-\frac{n(n-1)\cdots(n-k+1)}{n^k} \right)\lt \varepsilon, \quad k=0,\ldots,N-1 \quad\text{for }n\gt M. \] Using the above estimations we have for $n\gt M$ \begin{align*} &\left|\sum_{k=0}^{n}\frac{z^k}{k!}\left(1-\frac{n(n-1)\cdots(n-k+1)}{n^k} \right)\right| \\ &\leq \sum_{k=0}^{n}\frac{|z|^k}{k!}\left(1-\frac{n(n-1)\cdots(n-k+1)}{n^k} \right) \\ &\leq \sum_{k=0}^{N-1}\frac{|z|^k}{k!}\varepsilon +\sum_{k=N}^{\infty}\frac{|z|^k}{k!}\\ &\leq \varepsilon e^{|z|} + \varepsilon. \end{align*} This estimation inplies \eqref{eq:limitis0}. $\quad\blacksquare$