Series expansion of $\frac{1}{\sqrt{x^3-1}}$ near $x \to 1^{+}$ How can I arrive at a series expansion for $$\frac{1}{\sqrt{x^3-1}}$$ at $x \to 1^{+}$? Experimentation with WolframAlpha shows that all expansions of things like $$\frac{1}{\sqrt{x^y - 1}}$$ have $$\frac{1}{\sqrt{y}\sqrt{x-1}}$$ as the first term, which I don’t know how to obtain.

Set $x:=1+\epsilon$, with $\epsilon \to 0^+.$ Then, by the binomial theorem,
$$
x^3=(1+\epsilon)^3=1+3\epsilon+3\epsilon^2+\epsilon^3
$$ giving
$$
\sqrt{x^3-1}=\sqrt{3\epsilon+3\epsilon^2+\epsilon^3}=\sqrt{3}\:\sqrt{\epsilon}\:\sqrt{1+\epsilon+O(\epsilon^2)} \tag1
$$ Observe that, by the Taylor expansion, as $\epsilon \to 0^+$,
$$
\sqrt{1+\epsilon+O(\epsilon^2)}=1+O(\epsilon). \tag2
$$ From $(1)$ and $(2)$, one gets
$$
\frac1{\sqrt{x^3-1}}=\frac1{\sqrt{3}\:\sqrt{\epsilon}}\frac1{\sqrt{1+\epsilon+O(\epsilon^2)}}=\frac1{\sqrt{3}\:\sqrt{\epsilon}}\frac1{\left(1+O(\epsilon)\right)}=\frac1{\sqrt{3}\:\sqrt{\epsilon}}\left(1+O(\epsilon) \right)
$$ or, using $\epsilon=x-1$,

$$
\frac1{\sqrt{x^3-1}}=\frac1{\sqrt{3}\:\sqrt{x-1}}+O(\sqrt{x-1}).
$$ 

Similarly, one obtains, for $y>0$, as $x \to 1^+$,

$$
\frac1{\sqrt{x^y-1}}=\frac1{\sqrt{y}\:\sqrt{x-1}}+O(\sqrt{x-1}).
$$

