Is the following series expansion $$(1+x)^{\frac {1}{x}}=e\left(1-\frac{x}{2}+\frac{11}{24}x^2-...\right)$$ valid for all $x\in(-1,0)\cup(0,\infty)$
I derived the expansion in following manner:
$$(1+x)^{\frac {1}{x}}=e^{\frac{1}{x}\ln(1+x)}=e^{\frac{1}{x}(x-\frac{x^2}{2}+\frac{x^3}{3}+...)}=e.e^{-\frac{x}{2}+\frac{x^2}{3}...}=e\left(1+\frac{1}{1!}(-\frac{x}{2}+\frac{x^2}{3}-...)+\frac{1}{2!}(-\frac{x}{2}+\frac{x^2}{3}-...)^2+...\right)=e\left(1-\frac{x}{2}+\frac{11}{24}x^2-...\right)$$
But I realized that the expansion of $\ln(1+x)$ is valid only if $-1<x<1$.
So the above expansion should be valid only for $x\in(-1,1)$ or is there some other way to derive it so that there are no restrictions.
There is no way to derive it, as it's simply false for $(0, \infty)$.
If say we have $(1+x)^{1/x}=\sum_{i=0}^\infty a_ix^i$ for all $x\in (0, \infty)$, then the radius of convergence of the power seires must be $\infty$, hence the power series defines an entire function over $\mathbb C$, that is $(1+x)^{1/x}$ has holomorphic extension to the entire $\mathbb C$. But $f(x)=(1+x)^{1/x}$ has a singularity at $x=-1$ (it's easy to check that $f(x)$ is not differentiable at $x=-1$).