Answer
$$f(x)=\frac{1-\cos x}{x}
= \frac{x}{2!}-\frac{x^3}{4!}+\frac{x^5}{6!}+..$$
Convergent for $x\ne 0$.
Work Step by Step
By making use of Table 2, we have the Maclaurin series for $f(x)=\cos x$ as follows
$$f(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...$$
Now by the comparison with the function $f(x)=\frac{1-\cos x}{x}$, we have the Maclaurin series as follows
$$f(x)=\frac{1-\cos x}{x}\\
=\frac{1-(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...)}{x}\\
= \frac{x}{2!}-\frac{x^3}{4!}+\frac{x^5}{6!}+..$$
Moreover, again from Table 2, the Maclaurin series for $f(x)=\cos x$ is convergent for any value of $x$. In our case, the series does not converge for $x=0$ (division by 0). Hence, the Maclaurin series for $f(x)=\frac{1-\cos x}{x}$ is convergent for $x\ne 0$.
