The series expansion of $\frac{1}{\sqrt{e^{x}-1}}$ at $x=0$ 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.
 A: Related problems: (I), (II). Just find the Taylor series of the function 
$$ \frac{\sqrt{x}}{\sqrt{e^{x}-1}}. $$
Added: Here is a formula for the $n$th derivative of the function 

$$\left(\frac{\sqrt{x}}{\sqrt{e^{x}-1}}\right)^{(n)}= \frac{\pi}{2}\sum _{k=0}^{n}  \sum _{i=0}^{k}  \sum _{m=0}^{i}{
\frac { \binom {k}{i}\left[\matrix{i\\m}\right] \left\{\matrix{n\\k}\right\} 
x^{\frac{1}{2}-m}\, {\rm e}^{(k-i) x}  \left( {\rm e}^x - 1
\right)^{i-k-\frac{1}{2}}    }{\Gamma  \left( \frac{1}{
2}-k+i \right) \Gamma  \left( \frac{3}{2}-m \right) }} .$$

Note: I'll appreciate it if someone can verify this formula with Maple or Mathematica. 
A: Binomial expansion version... as $x \to 0^+$,
$$
e^x-1 = \left(1 + x +\frac{x^2}{2} + \frac{x^3}{6} +\dots\right) - 1 = x \left(1+\frac{x}{2}+\frac{x^2}{6}+\dots\right) = x(1+S),
$$
where $S \to 0$.  So
$$
\left(e^x-1\right)^{-1/2} = x^{-1/2}\left(1+S\right)^{-1/2}
= x^{-1/2}\sum_{k=0}^\infty \binom{-1/2}{k} S^k
$$
Then put in what $S$ is (as many terms as needed...)
A: $\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}

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

