Answer
We give steps required to show that the Maclaurin series for $f\left( x \right) = {{\rm{e}}^x}$ converges to $f\left( x \right)$ for all $x$.
Work Step by Step
The steps required:
Step 1. Expand the function
Let $x \in \left( { - R,R} \right)$. We expand $f\left( x \right) = {{\rm{e}}^x}$ using Maclaurin series:
$f\left( x \right) = {{\rm{e}}^x} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n}$
Since ${f^{\left( k \right)}}\left( 0 \right) = 1$ for all $k$, we obtain
$f\left( x \right) = {{\rm{e}}^x} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{x^n}}}{{n!}}$
Step 2. Apply Theorem 2
Because $\left| {{f^{\left( k \right)}}\left( 0 \right)} \right| \le 1$, we can set $K=1$ such that $\left| {{f^{\left( k \right)}}\left( 0 \right)} \right| \le K$, for all $k$, where $x \in \left( { - R,R} \right)$.
Since $R$ is arbitrary, we can have $x \in \left( { - \infty ,\infty } \right)$. By Theorem 2, $f\left( x \right) = {{\rm{e}}^x}$ is represented by
$f\left( x \right) = {{\rm{e}}^x} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{x^n}}}{{n!}}$ ${\ \ \ }$ for $x \in \left( { - \infty ,\infty } \right)$
This implies that the series is valid for all $x$.
Hence, the Maclaurin series for $f\left( x \right) = {{\rm{e}}^x}$ converges to $f\left( x \right)$ for all $x$.
