The function $ \displaystyle\frac{1}{\sqrt{e^{x}-1}}$ doesn't have a Laurent series expansion at $x=0$.
But according to Wolfram Alpha, it does have a series expansion with terms raised to non-integer powers:
$$ \frac{1}{\sqrt{e^{x}-1}} = \frac{1}{\sqrt{x}}- \frac{\sqrt{x}}{4} + \cdots$$
How is that series derived?
My initial thought was to use the general binomial theorem, but I'm not sure how.
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...)
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.
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}
$$ \color{#0000ff}{\large% {1 \over \root{\expo{x} - 1}} \sim {1 \over \root{x}} - {1 \over 4}\,\root{x} - {1 \over 12}\,x^{3/2}} $$
At the site https://math.stackexchange.com/a/4657078, it was derived that \begin{equation*} \lim_{x\to0}\Biggl[\sqrt{\frac{\ln(1+x)}{x}}\,\Biggr]^{(n)} =-\sum_{k=0}^n\frac{(2k-3)!!}{(2k)!!} \sum_{m=0}^k(-1)^m\binom{k}{m}\frac{s(n+m,m)}{\binom{n+m}{m}},\quad n\ge0, \end{equation*} where $s(n,k)$ denotes the Stirling numbers of the first kind. Hence, we have \begin{align} \biggl(\sqrt{\frac{x}{\operatorname{e}^x-1}}\,\biggr)^{(n)} &=\Biggl[\sqrt{\frac{\ln(1+u(x))}{u(x)}}\,\Biggr]^{(n)}, \quad u=u(x)=\operatorname{e}^x-1\\ &=\sum_{k=0}^n\Biggl[\sqrt{\frac{\ln(1+u)}{u}}\,\Biggr]^{(k)} B_{n,k}(\operatorname{e}^x,\operatorname{e}^x,\dotsc,\operatorname{e}^x)\\ &=\sum_{k=0}^n\Biggl[\sqrt{\frac{\ln(1+u)}{u}}\,\Biggr]^{(k)} \operatorname{e}^{kx} B_{n,k}(1,1,\dotsc,1)\\ &\to \sum_{k=0}^n\lim_{u\to0}\Biggl[\sqrt{\frac{\ln(1+u)}{u}}\,\Biggr]^{(k)} S(n,k), \quad x\to0\\ &=-\sum_{k=0}^n\Biggl[\sum_{\ell=0}^k\frac{(2\ell-3)!!}{(2\ell)!!} \sum_{m=0}^\ell(-1)^m\binom{\ell}{m}\frac{s(k+m,m)}{\binom{k+m}{m}}\Biggr] S(n,k) \end{align} and \begin{equation} \sqrt{\frac{x}{\operatorname{e}^x-1}}\, =-\sum_{n=0}^\infty\Biggl[\sum_{k=0}^nS(n,k)\sum_{\ell=0}^k\frac{(2\ell-3)!!}{(2\ell)!!} \sum_{m=0}^\ell(-1)^m\binom{\ell}{m}\frac{s(k+m,m)}{\binom{k+m}{m}}\Biggr]\frac{x^n}{n!}, \quad |x|<\pi, \end{equation} where $S(n,k)$ denotes the Stirling numbers of the second kind. Consequently, we arrive at \begin{equation} \frac{1}{\sqrt{|\operatorname{e}^x-1|}}\, =-\frac1{\sqrt{|x|}\,}\sum_{n=0}^\infty\Biggl[\sum_{k=0}^nS(n,k)\sum_{\ell=0}^k\frac{(2\ell-3)!!}{(2\ell)!!} \sum_{m=0}^\ell(-1)^m\binom{\ell}{m}\frac{s(k+m,m)}{\binom{k+m}{m}}\Biggr]\frac{x^n}{n!} \end{equation} for $0\ne|x|<\pi$.
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