Convergence range of Binomial Expansions with square unknowns?

Question

So the binomial expansion of $\left(\displaystyle\frac{1}{a^2+x^2}\right)^{5/2}$ would converge for $\left|\displaystyle\frac{x^2}{a^2}\right|<1.$

Would this not mean that

$-1<\displaystyle\frac{x^2}{a^2}<1$?

But isn't $0\leq\displaystyle\frac{x^2}{a^2}$, and so shouldn't the convergence range just be $0\leq\displaystyle\frac{x^2}{a^2}<1$? Also wouldn't the inequality $-1<\displaystyle\frac{x^2}{a^2}<1$'s left hand side produce a square-rooting problem?

Answer 1

Regarding your first question: ... shouldn't the convergence range just be $0\leq\displaystyle\frac{x^2}{a^2}<1$?

Note the following sets are equal:

\begin{align*} \left\{x\in\mathbb{R}:\left|\frac{x^2}{a^2}\right|<1\right\} &=\left\{x\in\mathbb{R}:-1<\frac{x^2}{a^2}<1\right\}\\ &=\left\{x\in\mathbb{R}:0\leq \frac{x^2}{a^2}<1\right\}\\ &=\left(-|a|,|a|\right) \end{align*}

This means whenever an element $x$ is in one of the sets it is also in all the other sets. So, you are free to choose the representation which seems to be the most convenient for your needs.

Regarding your second question: ... wouldn't the inequality $-1<\displaystyle\frac{x^2}{a^2}<1$'s left hand side produce a square-rooting problem?

No, since it belongs to you to do correct manipulations with this double inequality in order to determine the range of validity for $x$.