I have taken the logarithm of this expression and computed the Taylor expansion of the $\log(1+\epsilon)$ term but by doing this we're required to calculate powers of this series when using the definition of the exponential function, but this gives a combinatorical mess.
eg. $(1+\epsilon)^{s/\epsilon}=\exp\left(s\right)\exp\left(s\delta\right),$ where $\delta=\sum_{i=1}^\infty \frac{(-1)^n \epsilon^n}{n+1}.$
I was wondering if anyone knows of a general formula for the coefficient of $\epsilon^n$ of $(1+\epsilon)^{s/\epsilon}$ as $\epsilon \rightarrow 0$. Any known papers on this topic too?
Thanks
I am stuck with the general formula.
For the first terms, it is quite simple : starting with $$A=(1+\epsilon )^{s/\epsilon }$$ $$\log(A)=\frac{s }{\epsilon }\log (1+\epsilon)=\frac{s }{\epsilon }\sum_{n=1}^\infty(-1)^{n+1}\frac{\epsilon^n} n=s \sum_{n=1}^\infty(-1)^{n+1}\frac{\epsilon^{n-1}} n$$ Now, using $A=e^{\log(A)}$ and using Taylor series again leads to $$A=e^s\left(1-\frac{s }{2}\epsilon+\frac{s (3 s+8)}{24} \epsilon ^2-\frac{s(s+2)(s+6) }{48} \epsilon ^3+O\left(\epsilon ^4\right) \right)$$ So, $$A=e^s \sum_{n=0}^\infty (-1)^n P_n(s) \epsilon^n$$ where $P_n(x)$ is a polynomial of degree $n$ in $s$.