Let $A$ be an unital Banach algebra over a vector space $X$. Let us consider the power series:
\begin{equation*}
\sum_{k=0}^{+\infty} c_k x^k
\end{equation*}
with coefficients in $\mathbb{K}$ and the $x$ in $X$.
Suppose the series converges for some $\bar{x}\neq 0$. then the sequence $c_k \bar{x}^k$ converges to zero, which means that it is also bounded, so that there exists an $L \in \mathbb{R}_+$ such that $|c_k \bar{x}^k |\leq L , \forall k \in \mathbb{N}$. Now, let $x$ be such that $|x|<|\bar{x}|$. We have the following:
$$|c_k x^k| \leq |c_k| |x|^k = |c_k| |x|^k \left(\frac{|\bar{x}|}{|\bar{x}|}\right)^k = |c_k||\bar{x}|^k q^k$$
Having defined $q = \frac{|x|}{|\bar{x}|} $. I then want to say that the last term of the inequalities is less or equal to $L q^k$, with $|q| < 1$, so that the series has normal convergence by comparison with the geometric series. But that inequality doesn't follow from any property of the norm!! Am i missing some other way one could go about it?
Best Answer
The problem you found is essential. What you are trying to do works with numbers, but it doesn't necessarily work in a Banach algebra.
For example let $X=M_2(\mathbb C)$ with the operator norm, and consider the series $$ \sum_{k=0}^\infty x^k. $$ Let $$ \bar x=\begin{bmatrix} 0&3\\0&0\end{bmatrix}. $$ Since $\bar x^2=0$, the series converges at $\bar x$ (and is equal to $1+\bar x$). If now you take $$ x=\begin{bmatrix} 1&1\\0&1\end{bmatrix}, $$ then $$ \|x\|=\sqrt{\frac{3+\sqrt5}2}\leq 3=\|\bar x\|. $$ But $$ \sum_{k=0}^\infty x^k=\sum_{k=0}^\infty \begin{bmatrix} 1&k\\0&1\end{bmatrix} $$ does not converge.