Numerically, the series is
1 + 3x + \frac{15}{2}x^2 + \cdots. Equating this to
1 + abx + \frac{1}{2}b(b-1)a^2 x^2 + \cdots we get
3=ab and
b(b-1)a^2 = 15. This gives
\displaystyle{\frac{b-1}{b} = \frac{15}{9} = \frac{5}{3}}, so that
b = -\frac{3}{2} and
a=-2. Now check that the series actually is
(1 - 2x)^{-\frac{3}{2}}. At this point, you can deduce that the radius of convergence is
\frac{1}{2} since you know the radius of the general binomial expansion. To check the endpoints
\pm \frac{1}{2} there are two approaches. One is to use Stirling's formula (conveniently missing from most calculus textbooks). It says that
\lim_{n \rightarrow \infty} \frac{n!}{\sqrt{2\pi} n^{n+\frac{1}{2}} e^{-n}} = 1.
This gives
\displaystyle{\frac{(2n+1)!}{2^n (n!)^2} |x^n| } | \displaystyle{\sim} | \displaystyle{ \frac{(2n+1)^{2n+\frac{3}{2}} e^{-2n-1}}{\sqrt{2\pi} 2^n n^{2n+1} e^{-2n}} |x^n| = \frac{(n+\frac{1}{2})^{2n} 2^{2n+\frac{3}{2}}(n + \frac{1}{2})^{\frac{3}{2}}}{e\sqrt{2\pi} n^{2n} n 2^n}|x^n|} | |
\displaystyle{ } | \displaystyle{=} | \displaystyle{ \frac{2\sqrt{2}}{e\sqrt{2\pi}} \left( 1 + \frac{1}{2n}\right)^{2n} n^{-1} \left(n+\frac{1}{2}\right)^\frac{3}{2} |2x|^n} | |
\displaystyle{ } | \displaystyle{\sim} | \displaystyle{ \frac{2}{\surd \pi} n^{-1} \left(n+\frac{1}{2}\right)^\frac{3}{2} |2x|^n} | |
and we see that the terms do not tend to zero as
n \rightarrow \infty when
x = \pm \frac{1}{2}.
The second approach is to use the ratio test. Let
a_n = \frac{(2n+1)!}{2^n (n!)^2} x^n. then we find
\displaystyle{\left|\frac{a_{n+1}}{a_n}\right| } | \displaystyle{=} | \displaystyle{ \frac{(2n+3)! 2^n (n!)^2}{2^{n+1} ((n+1)!)^2 (2n+1)!} |x| = \frac{(2n+3)(2n+2)}{2 (n+1)^2} |x|} | |
\displaystyle{ } | \displaystyle{=} | \displaystyle{ \frac{2n+3}{n+1} |x| \longrightarrow 2|x|} | |
as
n \rightarrow \infty. This shows that the radius of convergence is
\frac{1}{2} which we already knew. But the key point is that for
x = \pm \frac{1}{2} we have
\left|\frac{a_{n+1}}{a_n}\right| = \frac{2n+3}{2n+2} \geq 1.
The terms of the series are increasing in absolute value and hence not tending to zero. The series cannot converge at
x = \pm \frac{1}{2}. The interval of convergence is
-\frac{1}{2} < x < \frac{1}{2}.