Find series expansion of $e^{-\frac{c}{1-x}}$ I would like to know what the series expansion of
 $$f(x)= e^{-\frac{c}{1-x}},  \quad  \mbox{for} \, 0<x<1 \, \mbox{and}\, c>0.$$
Thank you in advance
 A: It is known that $e^y=\sum_{n=0}^{+\infty}\frac1{n!}y^n$ for any $y\in \mathbb R$ hence, letting $y:=-\frac{c}{1-x}$ you get $$e^{-\frac{c}{1-x}}=\sum_{n=0}^{+\infty}\frac{1}{n!}\left(-\frac{c}{1-x}\right)^n=\sum_{n=0}^{+\infty}\frac{(-c)^n}{n!}\left(\frac{1}{1-x}\right)^n$$ Now $\frac1{1-x}$ is the sum of the geometric series for $|x|<1$ $$\frac{1}{1-x}=\sum_{m=0}^{\infty}x^m$$ Substituting in the previous equation this gives $$e^{-\frac{c}{1-x}}=\sum_{n=0}^{\infty}\frac{(-c)^n}{n!}\left(\sum_{m=0}^\infty x^m\right)^n$$ but this is not in the desired form $\sum a_n x^n$. 

The n-th derivative of $\frac1{1-x}$ is $$\frac{\partial}{\partial x}\left(\frac1{1-x}\right)=\frac{(-1)^{n+1}n!}{(1-x)^{n+1}}$$ Hence the Taylor series expansion of $f(x)$ around $x=0$ is \begin{align}f(x)&=f(0)+\sum_{n=1}^{+\infty}\frac{f^{(n)}(0)}{n!}x^n=e^{-c}+\sum_{n=1}^{+\infty}\left.ce^{-\frac{c}{1-x}}\frac{n!(-1)^{n+1}}{n!(1-x)^{n+1}}\right|_{x=0}x^n\\[0.2cm]&=e^{-c}+\sum_{n=1}^{+\infty}ce^{-c}(-1)^{n+1}x^n\end{align} This gives $a_0=e^{-c}$ and $$a_n=ce^{-c}(-1)^{n+1}, \quad \text{ for } n\ge1$$
A: If you need to expand $\frac{c}{1-x}$, then $$e^{-\frac{c}{1-x}}=\sum_{n=0}^{+\infty}\frac{1}{n!}\left(-c\sum_{m=0}^{+\infty}x^m\right)^n$$
A: Hint. One may use the Laguerre polynomials $L_n(\cdot)$.

Proposition.
  $$
e^{\Large-\frac{c}{1-x}}=e^{-c}+e^{-c}\sum_{n=1}^\infty  (L_n(c) -L_{n-1}(c)) \: x^n, \qquad c \in \mathbb{C},\, |x|<1,
$$ 

where
$$
\begin{align}
L_0(c)&=1
\\L_1(c)&=-c+1
\\L_2(c)&=\frac{c^2}{2}-2 c+1
\\L_3(c)&=-\frac{c^3}{6}+\frac{3 c^2}{2}-3c+1
\\\cdots&=\cdots 
\\L_n(c)&=\frac{e^c}{n!}\frac{d^n}{dc^n}\left(e^{-c} c^n\right) =\frac{1}{n!}\left(\frac{d}{dc}-1\right)^n c^n.
\end{align}
$$
Proof. One may recall the generating function of the Laguerre polynomials (see generating function):
$$
\frac{1}{1-x} e^{\Large-\frac{cx}{1-x}}=\sum_{n=0}^\infty  L_n(c)  \: x^n
$$ then write $\displaystyle -\frac{cx}{1-x}=-\frac{c}{1-x}+c$, multiply by $1-x$ and make a change of index. 
