Series expansion of $(1-cx)^{1/x}$ I am trying to understand the series expansion of $$(1-cx)^{1/x}$$ The wolframalpha seems to solve the problem by using taylor series for $ x\rightarrow 0$ and Puiseux series for $ x\rightarrow \infty$. Any ideas how can I calculate them ?
https://www.wolframalpha.com/input?i=%281-cx%29%5E%281%2Fx%29+series
Here is the link of the problem
 A: As already said in comments
$$y=(1-c x)^{\frac{1}{x}}\quad \implies \quad \log(y)=\frac{1}{x}\log(1-cx)$$ For small values of $cx$
$$\log(y)=-\sum_{n=1}^\infty \frac {c^n} n\,x^{n-1}$$ Now, using $y=e^{\log(y)}$
$$\color{blue}{y=1 +c\,e^{-c}\sum_{n=1}^\infty(-1)^n \frac{ c^{n}}{\alpha_n}\,P_n(c)\,x^n}$$ The $\alpha_n$ form the sequence $A053657$ in $OEIS$ (have a look here) and the first $P_n(c)$ are listed below
$$\left(
\begin{array}{cc}
n & P_n(c) \\
 1 & 1 \\
 2 & 3 c-8 \\
 3 & (c-6) (c-2) \\
 4 & 15 c^3-240 c^2+1040 c-1152 \\
 5 & (c-4) \left(3 c^3-68 c^2+408 c-480\right) \\
 6 & 63 c^5-2520 c^4+35280 c^3-211456 c^2+526176 c-414720 \\
 7 & (c-6) \left(9 c^5-450 c^4+7800 c^3-55792 c^2+151104 c-120960\right)
\end{array}
\right)$$
Looking separately at $P_{2n}(c)$ and  $P_{2n+1}(c)$, it seems that there are some interesting patterns (worth to be explored ?).
For large values of $x$, using $L=\log(-cx)$
$$\color{blue}{y=1+  c \sum_{n=1}^\infty \frac 1{n!\,c^n} Q_n(c,L) \frac 1 {x^n}}$$ where the first $Q_n(c,L)$ are
$$\left(
\begin{array}{cc}
n & Q_n(c,L) \\
 1 & L \\
 2 & c L^2-2 \\
 3 & c^2 L^3-6 c L-3 \\
 4 & c^3 L^4-12 c^2 L^2-12 c L+12 c-8 \\
 5 & c^4 L^5-20 c^3 L^3-30 c^2 L^2+60 c^2 L-40 c L+60 c-30 
\end{array}
\right)$$
Again, possible interesing patterns.
