TITLE: The series expansion of $\frac{1}{\sqrt{e^{x}-1}}$ at $x=0$
QUESTION [6 upvotes]: The function $ \displaystyle\frac{1}{\sqrt{e^{x}-1}}$ doesn't have a Laurent expansion at $x=0$.
But according to Wolfram Alpha, it does have a series expansion that includes terms raised to noninteger powers.  Specifically, $\displaystyle\frac{1}{\sqrt{x}}- \frac{\sqrt{x}}{4} + O(x^{\frac{3}{2}})$.
How is that series derived?
My initial thought was to use general binomial theorem.  But I don't seem to get anywhere with that.

REPLY [3 votes]: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
&{1 \over \root{\expo{x} - 1}}
=
{1 \over \root{x}} + \pars{{1 \over \root{\expo{x} - 1}} - {1 \over \root{x}}}
=
{1 \over \root{x}}
- {\root{\expo{x} - 1} - \root{x} \over \root{x}\root{\expo{x} - 1}}
\\[3mm]&=
{1 \over \root{x}}
-
{\pars{\expo{x} - 1} - x
\over \root{x}\root{\expo{x} - 1}\pars{\root{\expo{x} - 1} + \root{x}}}
\\[3mm]&=
{1 \over \root{x}}
-
{\expo{x} - 1 - x \over \root{x}\pars{\expo{x} - 1} + x\root{\expo{x} - 1}}
=
{1 \over \root{x}}
-
\root{x}\bracks{%
{\expo{x} - 1 - x \over x\pars{\expo{x} - 1} + x^{3/2}\root{\expo{x} - 1}}}
\end{align}

Also,
$$
\lim_{x \to 0^{+}}
{\expo{x} - 1 - x \over x\pars{\expo{x} - 1} + x^{3/2}\root{\expo{x} - 1}}
=
\lim_{x \to 0^{+}}
{x^{2}/2 \over x\pars{x} + x^{3/2}\pars{x^{1/2}}} = {1 \over 4}
$$

Then,
\begin{align}
&{1 \over \root{\expo{x} - 1}}
=
{1 \over \root{x}} - {1 \over 4}\,\root{x}
+
\root{x}\bracks{%
{1 \over 4}
-
{\expo{x} - 1 - x \over x\pars{\expo{x} - 1} + x^{3/2}\root{\expo{x} - 1}}}
\\[3mm]&=
{1 \over \root{x}} - {1 \over 4}\,\root{x}
+
x^{3/2}\
\underbrace{\bracks{%
{1 \over 4x}
-
{1 \over x^{2}}\,{\pars{\expo{x} - 1}/x - 1 \over \pars{\expo{x} - 1}/x + \root{\pars{\expo{x} - 1}/x}}}}
_{\ds{\to -\,{1 \over 12}\ \mbox{when}\ x \to 0^{+}}}
\end{align}

$$
\color{#0000ff}{\large%
{1 \over \root{\expo{x} - 1}}
\sim {1 \over \root{x}} - {1 \over 4}\,\root{x} - {1 \over 12}\,x^{3/2}}
$$