The question asks me to find the Taylor series for $$f(z)=\frac{e^z}{1-3z}.$$

The radius of convergence is $|z|<1/3$ and I know the expansions for $e^z$ and $1/(1-3z)$ are \begin{align} e^z &=\sum_{n=0}^{\infty}\frac{z^n}{n!}=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots, \\ \frac{1}{1-3z} &= \sum_{n=0}^{\infty}(3z)^n=1+3z+9z^2+27z^3+\cdots.\\ \end{align}

Something tells me that since both series converge in the radius of convergence, I should be able to multiply them out and not worry about it. The answers provide the first three terms of the final answer which are $$f(z)=1+4z+\frac{25}{2}z^2+\cdots$$ which I can get by multipying the expanded expressions out for $e^z$ and $1/(1-3z)$. Why is this intuition correct? Or why does it seem to be so?

Furthermore, how would I go about finding the coefficient $a_n$ for $z^n$ in the Taylor series? Since the series converge can I just multiply their coefficients together to get the new one?