Answer
The Maclaurin series for $f(x)=\frac{1}{1-2x}$ is convergent if $|2x|\lt 1$ -- that is, $|x|\lt \frac{1}{2}$.
Work Step by Step
By making use of Table 2, we have the Maclaurin series for $f(x)=\frac{1}{1-x}$ as follows
$$f(x)=1+x+x^2+x^3+x^4+...$$
Now by comparison with the function $f(x)=\frac{1}{1-2x}$, we have the Maclaurin series as follows
$$f(x)=\frac{1}{1-2x}=1+2x+(2x)^2+(2x)^3+(2x)^4+...\\
=1+2x+4x^2+8x^3+16x^4+...$$
Moreover, again from Table 2, the Maclaurin series for $f(x)=\frac{1}{1-x}$ is convergent if $|x|\lt 1$. Hence the Maclaurin series for $f(x)=\frac{1}{1-2x}$ is convergent if $|2x|\lt 1$; that is, if $|x|\lt \frac{1}{2}$.
