spectral radius formula - why does $\sum \xi^n \zeta^{-n}$ converge for $|\zeta| > r(\xi)$? Every proof of the spectral radius formula $r(\xi) = \limsup \|\xi^n\|^{1/n}$ for Banach algebras (that I have seen) uses the fact, that if $\zeta \in \mathbb{C}$ with $|\zeta| \geq s > r(\xi)$ (for $s>0$), then the series $$\sum \xi^n\zeta^{-n}$$ converges uniformly and absolutely on $\{\zeta \colon |\zeta| \geq s\}$. But isn't this already using the fact, that $r(\xi) \geq \limsup \|\xi^n\|^{1/n}$, which we essentially wanted to prove in the first place? 

I guess the question basically boils down to why holomorphicity of the function $\zeta \mapsto (\zeta-\xi)^{-1}$ on the resolvent $\rho(\xi)$ of $\xi$ implies, that $(\zeta-\xi)^{-1} = \sum \xi^n\zeta^{-n-1}$ for all $|\zeta| > r(\xi)$.

 A: I will use the symbol $a$ for a nonzero element in a Banach algebra and $\lambda$ for a complex number.
Suppose $|\lambda|\gt \|a\|\gt0$. Then $\frac{\|a\|}{|\lambda|}\lt 1$ and hence $(1-\lambda^{-1}a)^{-1}$ exists.  Also, $$(1-\lambda^{-1}a)^{-1}=\sum_{n\geq 0}\lambda^{-n}a^n.$$ Now, $(1-\lambda^{-1}a)^{-1}=\lambda(\lambda-a)^{-1}$. Thus, $$(\lambda-a)^{-1}=\sum_{n\geq 0}\lambda^{-n-1}a^n\tag1\label1.$$
In other words, if $\lambda^{-1}\in B(0,1/{\|a\|})\setminus\{0\}$(the deleted open ball of radius $\frac1{\|a\|}$ centred at $0$), then $(\lambda-a)^{-1}$ exists and the holomorphic function $\lambda^{-1}\mapsto (\lambda-a)^{-1}$ has a power series representation given as in $\eqref1.$
Notice that $r(a)\leq \|a\|$, so $B(0,1/{\|a\|})\subseteq B(0,1/{r(a)}).$ Now, if $0\lt|\lambda^{-1}| \lt \frac1{r(a)}$(i.e. $|\lambda|\gt r(a)\gt0$), then $(\lambda-a)^{-1}$ exists and the holomorphic function $\lambda^{-1}\mapsto (\lambda-a)^{-1}$ has a power series expansion on $B(0,1/{r(a)})\setminus\{0\}.$
Since we already have a power series on a smaller neighbourhood, the uniqueness implies that the same should hold on $B(0,1/{r(a)})\setminus\{0\}$ as well.
