[proofplan] We may assume $A$ is unital (otherwise pass to the unitisation $A^+$, which preserves $\sigma$ and norm). The strategy is a sandwich argument. The upper bound $r_A(x) \le \inf_n \|x^n\|^{1/n}$ is obtained from the **Spectral Mapping Theorem for Polynomials** applied to $p(z) = z^n$, which forces $|\lambda|^n \le \|x^n\|$ for every $\lambda \in \sigma_A(x)$. The lower bound $\limsup_n \|x^n\|^{1/n} \le r_A(x)$ is the analytic heart: we study the resolvent $R(\lambda) := (\lambda 1 - x)^{-1}$ as a holomorphic $A$-valued function on the open set $\mathbb{C} \setminus \sigma_A(x)$, expand it as a Laurent series $\sum_n x^n/\lambda^{n+1}$ on $\{|\lambda| > \|x\|\}$, and use the principle of analytic continuation to extend the validity of this expansion to the larger annulus $\{|\lambda| > r_A(x)\}$. Convergence of the series at every such $\lambda$, combined with the **Uniform Boundedness Principle**, forces $\|x^n\|/|\lambda|^n$ to be bounded. The two bounds collapse the chain $r_A(x) \le \inf_n \|x^n\|^{1/n} \le \liminf_n \|x^n\|^{1/n} \le \limsup_n \|x^n\|^{1/n} \le r_A(x)$ into equality. [/proofplan] [step:Reduce to the unital case] If $A$ is non-unital, let $A^+ = A \oplus \mathbb{C}$ be its unitisation, with norm $\|(y, \mu)\|_{A^+} = \|y\|_A + |\mu|$ and the embedding $A \hookrightarrow A^+$, $y \mapsto (y, 0)$ being isometric. By definition $\sigma_A(x) = \sigma_{A^+}((x, 0))$ and $(x, 0)^n = (x^n, 0)$, with $\|(x^n, 0)\|_{A^+} = \|x^n\|_A$. Hence both sides of the desired formula are unchanged under passing from $A$ to $A^+$. We therefore assume from now on that $A$ is unital with unit $1$. [/step] [step:Establish the upper bound $r_A(x) \le \inf_{n \ge 1} \|x^n\|^{1/n}$ via spectral mapping] Fix $n \ge 1$ and apply the [Spectral Mapping Theorem for Polynomials](/theorems/2671) to the polynomial $p(z) = z^n$. The hypotheses of theorem 2671 require $A$ unital (verified) and a complex polynomial (verified, $p(z) = z^n$). The conclusion is \begin{align*} \sigma_A(x^n) = p(\sigma_A(x)) = \{\lambda^n : \lambda \in \sigma_A(x)\}. \end{align*} By the [Spectrum is Non-Empty and Compact](/theorems/2669) theorem applied to $x^n$, every element of $\sigma_A(x^n)$ has modulus at most $\|x^n\|$. In particular, for every $\lambda \in \sigma_A(x)$, \begin{align*} |\lambda^n| = |\lambda|^n \le \|x^n\|, \qquad \text{so} \qquad |\lambda| \le \|x^n\|^{1/n}. \end{align*} Taking the supremum over $\lambda \in \sigma_A(x)$ — which is non-empty by theorem 2669 — yields $r_A(x) \le \|x^n\|^{1/n}$. Since $n \ge 1$ is arbitrary, \begin{align*} r_A(x) \le \inf_{n \ge 1} \|x^n\|^{1/n}. \end{align*} [/step] [step:Define the resolvent and expand it as a geometric series for $|\lambda| > \|x\|$] Let $\Omega := \mathbb{C} \setminus \sigma_A(x)$, an open subset of $\mathbb{C}$ since $\sigma_A(x)$ is closed by theorem 2669. Define the **resolvent map** \begin{align*} R : \Omega &\to A \\ \lambda &\mapsto (\lambda 1 - x)^{-1}. \end{align*} For $|\lambda| > \|x\|$, we have $\|x/\lambda\| = \|x\|/|\lambda| < 1$, so by [Invertibility Near the Identity](/theorems/2667) applied to $a := 1 - x/\lambda$ (verifying $\|1 - a\| = \|x/\lambda\| < 1$), the element $1 - x/\lambda$ is invertible with \begin{align*} (1 - x/\lambda)^{-1} = \sum_{n=0}^\infty (x/\lambda)^n = \sum_{n=0}^\infty \frac{x^n}{\lambda^n}, \end{align*} the series converging absolutely in $A$ because $\|x^n/\lambda^n\| \le (\|x\|/|\lambda|)^n$ is geometric. Multiplying by $\lambda^{-1}$: \begin{align*} R(\lambda) = (\lambda 1 - x)^{-1} = \frac{1}{\lambda}(1 - x/\lambda)^{-1} = \sum_{n=0}^\infty \frac{x^n}{\lambda^{n+1}} \quad \text{for } |\lambda| > \|x\|. \end{align*} [/step] [step:Show $R$ is holomorphic on $\Omega$ and the series extends to $\{|\lambda| > r_A(x)\}$] For each $\varphi \in A^*$, define the scalar function \begin{align*} f_\varphi : \Omega &\to \mathbb{C} \\ \lambda &\mapsto \varphi(R(\lambda)) = \varphi((\lambda 1 - x)^{-1}). \end{align*} [claim:$f_\varphi$ is holomorphic on $\Omega$] For $\lambda, \mu \in \Omega$ with $\lambda \ne \mu$, the resolvent identity gives \begin{align*} R(\lambda) - R(\mu) &= R(\lambda)\bigl[(\mu 1 - x) - (\lambda 1 - x)\bigr]R(\mu) = (\mu - \lambda) R(\lambda) R(\mu), \end{align*} hence \begin{align*} \frac{R(\lambda) - R(\mu)}{\lambda - \mu} = -R(\lambda) R(\mu). \end{align*} By the continuity statement (ii) of [Properties of Invertible Elements](/theorems/2668) applied to the continuous map $\mu \mapsto \mu 1 - x$, we have $R(\mu) \to R(\lambda)$ as $\mu \to \lambda$. Applying $\varphi \in A^*$ (which is continuous): \begin{align*} \frac{f_\varphi(\lambda) - f_\varphi(\mu)}{\lambda - \mu} = -\varphi(R(\lambda) R(\mu)) \xrightarrow{\mu \to \lambda} -\varphi(R(\lambda)^2). \end{align*} So $f_\varphi$ is complex-differentiable at every $\lambda \in \Omega$, hence holomorphic. [/claim] [proof] The argument above is self-contained: the resolvent identity follows from algebraic manipulation $R(\lambda)(\mu 1 - x) = R(\lambda)(\mu - \lambda + \lambda 1 - x) = (\mu - \lambda) R(\lambda) + 1$, then multiplied on the right by $R(\mu)$ and rearranged; continuity of inversion is theorem 2668(ii); continuity of $\varphi$ is the assumption $\varphi \in A^* = \mathcal{L}(A, \mathbb{C})$. [/proof] For $|\lambda| > \|x\|$, applying $\varphi$ to the absolutely convergent series from Step 3: \begin{align*} f_\varphi(\lambda) = \varphi(R(\lambda)) = \sum_{n=0}^\infty \frac{\varphi(x^n)}{\lambda^{n+1}}. \end{align*} This is a Laurent expansion of $f_\varphi$ on the annulus $\{|\lambda| > \|x\|\}$ with coefficients $\varphi(x^n)$. The function $f_\varphi$ is holomorphic on the larger set $\Omega \supset \{|\lambda| > r_A(x)\}$ (since $\sigma_A(x) \subset \{|\lambda| \le r_A(x)\}$ by definition of $r_A$). By **uniqueness of Laurent expansions** for holomorphic functions on the largest annulus where the function extends holomorphically — explicitly, the Laurent series $\sum_n c_n \lambda^{-(n+1)}$ of any holomorphic function on $\{|\lambda| > \rho\}$ for some $\rho \ge 0$ is determined by the function and converges on the entire annulus where the function is holomorphic and bounded — the expansion \begin{align*} f_\varphi(\lambda) = \sum_{n=0}^\infty \frac{\varphi(x^n)}{\lambda^{n+1}} \end{align*} is valid (and converges) for every $\lambda$ with $|\lambda| > r_A(x)$. [/step] [step:Apply the Uniform Boundedness Principle to extract $r_A(x) \ge \limsup_n \|x^n\|^{1/n}$] Fix any $\lambda \in \mathbb{C}$ with $|\lambda| > r_A(x)$. By Step 4, the series $\sum_{n=0}^\infty \varphi(x^n)/\lambda^{n+1}$ converges for every $\varphi \in A^*$. In particular, the terms tend to zero: \begin{align*} \varphi\!\left(\frac{x^n}{\lambda^n}\right) = \lambda \cdot \frac{\varphi(x^n)}{\lambda^{n+1}} \xrightarrow{n \to \infty} 0 \quad \text{for every } \varphi \in A^*. \end{align*} For each $n \ge 0$ define the evaluation functional \begin{align*} T_n : A^* &\to \mathbb{C} \\ \varphi &\mapsto \varphi\!\left(\frac{x^n}{\lambda^n}\right). \end{align*} Each $T_n$ is linear in $\varphi$, with $|T_n(\varphi)| \le \|\varphi\|_{A^*} \cdot \|x^n/\lambda^n\|_A$, so $T_n \in (A^*)^* = A^{**}$ with $\|T_n\|_{A^{**}} = \|x^n/\lambda^n\|_A$ (the latter equality is the canonical isometric embedding $A \hookrightarrow A^{**}$, where the norm of $a \in A$ regarded as an element of $A^{**}$ equals $\|a\|_A$ — this follows from Hahn–Banach). Since $A^*$ is a Banach space and $T_n(\varphi) \to 0$ for every $\varphi \in A^*$, the sequence $(T_n(\varphi))_n$ is bounded for each $\varphi$. By the **Uniform Boundedness Principle** (Banach–Steinhaus, applied with domain $A^*$ — a Banach space — and codomain $\mathbb{C}$), there exists $M = M(\lambda) > 0$ with \begin{align*} \|T_n\|_{A^{**}} \le M \quad \text{for all } n \ge 0, \end{align*} i.e.\ $\|x^n\|_A / |\lambda|^n \le M$, which rearranges to \begin{align*} \|x^n\|^{1/n} \le M^{1/n} |\lambda|. \end{align*} Since $M^{1/n} \to 1$ as $n \to \infty$, we obtain \begin{align*} \limsup_{n \to \infty} \|x^n\|^{1/n} \le |\lambda|. \end{align*} Taking the infimum over $\lambda$ with $|\lambda| > r_A(x)$: \begin{align*} \limsup_{n \to \infty} \|x^n\|^{1/n} \le r_A(x). \end{align*} [guided] The point of this step is to convert the **scalar** convergence $\varphi(x^n/\lambda^n) \to 0$ (for each fixed $\varphi$) into a **norm** statement about the vectors $x^n/\lambda^n$ themselves. How is that done? The right framing is to view each $x^n/\lambda^n$ as an element of the **bidual** $A^{**}$ via the canonical embedding $a \mapsto (\varphi \mapsto \varphi(a))$; this embedding is isometric by Hahn–Banach, so norm in $A$ equals norm in $A^{**}$. Now we have a sequence $T_n \in A^{**} = (A^*)^*$ with $T_n(\varphi) \to 0$ for every $\varphi \in A^*$. The hypothesis of the **Uniform Boundedness Principle** (Banach–Steinhaus) is exactly this: a family of bounded linear functionals on a Banach space (here $A^*$) that is **pointwise bounded**. Pointwise convergence to $0$ is in particular pointwise boundedness. The conclusion is **uniform boundedness**: $\sup_n \|T_n\| < \infty$. Translating back via the isometry, $\sup_n \|x^n/\lambda^n\|_A < \infty$. Why does this give us the bound on $\limsup_n \|x^n\|^{1/n}$? From $\|x^n\|/|\lambda|^n \le M$ we get $\|x^n\|^{1/n} \le M^{1/n}|\lambda|$. The factor $M^{1/n} \to 1$ as $n \to \infty$, so $\limsup_n \|x^n\|^{1/n} \le |\lambda|$. This holds for *every* $\lambda$ with $|\lambda| > r_A(x)$, so taking the infimum (or letting $|\lambda| \downarrow r_A(x)$): \begin{align*} \limsup_n \|x^n\|^{1/n} \le r_A(x). \end{align*} What was the role of the analytic continuation in Step 4? The geometric series $\sum_n x^n/\lambda^{n+1}$ a priori converges only for $|\lambda| > \|x\|$. But the resolvent is holomorphic on the much larger set $\{|\lambda| > r_A(x)\}$. The Laurent uniqueness theorem says: a holomorphic function on an outer annulus has a unique Laurent expansion that converges on the **entire** annulus of holomorphy. Hence the same series converges (in the scalar sense, after applying $\varphi$) for every $\lambda$ with $|\lambda| > r_A(x)$ — even though we cannot directly use the geometric-series argument there, because we no longer have $\|x/\lambda\| < 1$. [/guided] [/step] [step:Sandwich the four quantities and conclude] Combining Steps 2 and 5: \begin{align*} r_A(x) \le \inf_{n \ge 1} \|x^n\|^{1/n} \le \liminf_{n \to \infty} \|x^n\|^{1/n} \le \limsup_{n \to \infty} \|x^n\|^{1/n} \le r_A(x). \end{align*} The middle inequality $\inf_n \|x^n\|^{1/n} \le \liminf_n \|x^n\|^{1/n}$ holds for any sequence of non-negative reals. Therefore all four quantities coincide, in particular the limit exists and equals the spectral radius: \begin{align*} r_A(x) = \lim_{n \to \infty} \|x^n\|^{1/n} = \inf_{n \ge 1} \|x^n\|^{1/n}. \end{align*} [/step]