Answer
$$f(x)=(1-x)^{-1/2}=1+\frac{1}{2}x+\frac{3}{2^2}\frac{x^2}{2!}+\frac{3*5}{2^3}\frac{x^3}{3!}+...$$
Convergent for $|x|\lt 1$
Work Step by Step
By making use of Table 2, we have the Maclaurin series for $f(x)=(1+x)^a$ as follows
$$f(x)=(1+x)^a=1+ax+\frac{a(a-1)x^2}{2!}+\frac{a(a-1)(a-2)x^3}{3!}+...$$
Now by the comparison with the function $f(x)=(1-x)^{-1/2}$, we have the Maclaurin series as follows
$$f(x)=(1-x)^{-1/2}=1+\frac{1}{2}x+\frac{3}{2^2}\frac{x^2}{2!}+\frac{3*5}{2^3}\frac{x^3}{3!}+...$$ Moreover, again from Table 2, the Maclaurin series for $f(x)=(1+x)^a$ is convergent for the values $|x|\lt 1$. Hence, the Maclaurin series for $f(x)=(1-x)^{-1/2}$ is convergent for the values $|-x^2|=x^2\lt 1$ or $|x|\lt 1$.
