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.

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.
