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?
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.