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.
 A: 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 $x \to 1^+$,

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

A: I like to expand around zero.
So, in
$\frac{1}{\sqrt{x^3-1}}
$,
let $x = 1+y$.
Then
$\begin{array}\\
x^3-1
&=(1+y)^3-1\\
&=1+3y+3y^2+y^3-1\\
&=3y+3y^2+y^3\\
&=y(3+3y+y^2)\\
\end{array}
$
so
$\begin{array}\\
\frac{1}{\sqrt{x^3-1}}
&=\frac{1}{\sqrt{y(3+3y+y^2)}}\\
&=\frac1{\sqrt{y}}\frac{1}{\sqrt{3+3y+y^2}}\\
&=\frac1{\sqrt{3y}}\frac{1}{\sqrt{1+y+y^2/3}}\\
&=\frac1{\sqrt{3y}}(1+y+y^2/3)^{-1/2}\\
\end{array}
$
Now apply the generalized binomial theorem
in the form
$(1+a)^b
=\sum_{n=0}^{whatever}\binom{b}{n}a^n
$
with
$a=y+y^2/3$
and
$b = -\frac12$.
Remember that
$\binom{b}{n}
=\frac{\prod_{k=0}^{n-1} (b-k)}{n!}
$,
so that
$\binom{-\frac12}{0}
=1
$,
$\binom{-\frac12}{1}
=-\frac12
$,
and
$\binom{-\frac12}{1}
=\frac12(-\frac12)(-\frac12-1)
=\frac12(-\frac12)(-\frac32)
=\frac38
$.
This will give you the first few terms
in terms of $y$.
To get them in terms of $x$,
replace $y$ by
$x-1$ and expand.
