#### Answer

The Maclaurin series for $f(x)=\frac{1}{1-x^4}$ is convergent if $|x^4|\lt 1$, that is $|x|\lt 1$.

#### 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 the comparison with the function $f(x)=\frac{1}{1- x^4}$, we have the Maclaurin series as follows
$$f(x)=\frac{x}{1-x^4}=x(1+x^4+(x^4)^2+(x^4)^3+(x^4)^4+...)\\
=x+ x^5+ x^9+x^{15}+x^{17}+...$$
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 $f(x)=\frac{1}{1-x^4}$ is convergent if $|x^4|\lt 1$, that is $|x|\lt 1$.