Proof, [Theorem 1.8, Fol16]. Let $r = \limsup_{n \to \infty}\normn{x^n}_{A}^{1/n}$, then for any $\lambda \in \complex$ with $|\lambda| > r$, the series $\sum_{n = 0}^{\infty} \lambda^{-n-1}x^{n}$ converges absolutely, and to the inverse of $(x - \lambda)$. Therefore $r \ge [\cdot]_{sp}$.

Let $D = \bracs{\lambda \in \complex|\ |\lambda| > [x]_{sp}}$, then since the series $\sum_{n = 0}^{\infty} \lambda^{-n-1}x^{n}$ is the expansion of $R_{x}$ at infinity, which is defined on $D$, it must converge on $D$. Let $\phi \in A^{*}$, then by the preceding discussion, the series $\sum_{n = 0}^{\infty} \lambda^{-n-1}\dpn{x^n, \phi}{A}$ converges on $D$. Thus for any $\lambda \in D$,

By the Uniform Boundedness Principle,

Therefore $\limsup_{n \to \infty}\norm{x^n}_{A}^{1/n}\le [x]_{sp}$.$\square$