Why radius of convergence is $\frac{1}{r}=\liminf_{n\to \infty }\left|\frac{a_{-n-1}}{a_{-n}}\right| ?$ Let consider the series $$\sum_{n\in\mathbb Z}a_nz^n.$$
We denote $R$ the radius of $\sum_{n=0}^\infty a_nz^n$ and $r$ the radius of $\sum_{n=-\infty }^{-1}a_nz^n$, i.e.e the series converge absolutely if $r<|z|<R$. The thing I don't understand is why $$\frac{1}{r}=\liminf_{n\to \infty }\left|\frac{a_{-n-1}}{a_{-n}}\right|.$$
Indeed, $$\sum_{n=-\infty }^{-1}a_nz^n=\sum_{n=1}^\infty a_{-n}z^{-n},$$
and thus, by d'Alembert, it converge if $$\limsup_{n\to \infty }\left|\frac{a_{-n-1}}{a_{-n}}\right|\frac{1}{|z|}<1\implies |z|>\limsup_{n\to \infty }\left|\frac{a_{-n-1}}{a_{-n}}\right|:=r.$$
Therefore $$r=\limsup_{n\to \infty }\left|\frac{a_{-n-1}}{a_{-n}}\right|.$$
What's wrong here ? Why this $\liminf$ ? I can't get it.

This is wrong:
$$\frac{1}{r}=\liminf_{n\to \infty }\left|\frac{a_{-n-1}}{a_{-n}}\right|$$
A better formula is
$$\frac{1}{r}=\liminf_{n\to \infty }\left|\frac{a_{-n}}{a_{-n-1}}\right|$$
But as Daniel Fischer points out in a comment, this is not generally correct. It is correct if $|a_{-n}|$ is monotonic.
