Definition 26.6.2 (Radius of Convergence).label Let $E$ be a normed space over $K \in \RC$, $F$ be a Banach space over $K$, and $\sum_{n = 0}^{\infty} T_{n}(x - a)^{(n)}$ be a power series about $a \in E$, then $R \in [0, \infty]$ be defined by[1]
\[\frac{1}{R}= \limsup_{n \to \infty}\norm{T_n}_{L^n(E; F)}^{1/n}\]
is the radius of convergence of $\sum_{n = 0}^{\infty} T_{n}(x - a)^{(n)}$. For each $0 < r < R$, the series converges uniformly and absolutely on $B_{E}(a, r)$.
Proof. For all $x \in B_{E}(a, r)$,
\[\sum_{n = 0}^{\infty} \normn{T_n(x - a)^{(n)}}_{F} \le \sum_{n \in \natz}\norm{T_n}_{L^n(E; F)}\norm{x - a}_{E}^{n} \le \sum_{n \in \natz}\norm{T_n}_{L^n(E; F)}r^{n}\]
For any $s \in (r, R)$, there exists $N \in \natp$ such that $\norm{T_n}_{L^n(E; F)}^{1/n}\le 1/s$ for all $n \ge N$. In which case,
\[\sum_{n = 0}^{\infty} \norm{T_n}_{L^n(E; F)}r^{n} \le \sum_{n = 0}^{N} \norm{T_n}_{L^n(E; F)}r^{n} + \sum_{n \ge N}\frac{r^{n}}{s^{n}}< \infty\]
As this estimate holds uniformly on $B_{E}(a, r)$, the series converges uniformly and absolutely on $B_{E}(a, r)$.$\square$
- Under the abuse that $1/\infty = 0$ and $1/0 =\infty$.keyboard_return