I am asked by my teacher : Find the Maclaurin series generated by $f(x)=e^{3x}$ and show that the series converges to the function $\forall x\in\mathbb{R}$.
My teacher also gave me a hint: Prove that $|M(x)-f(x)|$ approaches $0$ by using Sandwich Theorem, i.e. $\lim_{\epsilon\to0}0\leq|M(x)-f(x)|\leq\epsilon$, where $M(x)$ is the Maclaurin Series for $f(x)=e^{3x}$.
I found the Maclaurin series, but I could not find a function that I can use to use Sandwich Theorem. Can you help me?
We have
$\forall x\in\mathbb R \;\forall n>0$
$$e^{3x}=\sum_{k=0}^n \frac{(3x)^k}{k!}+R(x,n,c)$$
with $c\in(0,3x)$ and
$$R(x,n,c)=\frac{(3x)^{n+1}}{(n+1)!}3^{n+1}e^{3c}$$
but $$0<|R(x,n,c)|<U(x,n)$$
with $$U(x,n)=\frac{(9|x|)^{n+1}}{(n+1)!}e^{3|x|}$$
and by ratio test,
$$\lim_{n\to+\infty}\frac{U(x,n+1)}{U(x,n)}=0$$
$$\implies \lim_{n\to+\infty}U(x,n)=0$$
qed.