I am working through the following theorem and proof, but I struggle to understand the last part.
Theorem:
Let $A$ be a Banach algebra with unit element $e$, $x \in A$ and $||x||< 1$. Then $1_A-x$ is invertible and $$(1_A-x)^{-1}= \sum_{k=0}^\infty x^k. $$
Proof:
Let $||x||=r < 1$ and let $\displaystyle S_n = \sum_{k=0}^n x^k$. Now, for $n<m$ we have \begin{align*} ||S_n - S_m|| &= ||\sum_{k=0}^n x^k -\sum_{k=0}^m x^k|| \\ &\le \sum_{k=n+1}^m ||x||^k\le \frac{r^{n+1}}{1-r}. \end{align*} Therefore $(S_n)$ is a Cauchy sequence in $A$ converging to $\displaystyle a =\sum_{k=0}^\infty x^k$ .
Up onto this part I understand completely, however, the proof then finishes with the following
Because $xS_n = S_{n+1}-1_A$, we conclude that $a(1_A-x)=(1_A-x)a=1_A$.
I do not see how this is true? Can anyone please show me why?
Because $a=\lim_{n\to\infty}S_n$, and because multiplication is continuous: For all $n\in\Bbb{N}$ we have $$xS_n=S_{n+1}-1_A,$$ as you already noted. We may rewrite this to get $S_{n+1}-xS_n=1_A$, and hence $$\lim_{n\to\infty}(S_{n+1}-xS_n)=\lim_{n\to\infty}1_A=1_A.$$ Because addition and multiplication are continuous we have $$\lim_{n\to\infty}(S_{n+1}-xS_n)=\lim_{n\to\infty}S_{n+1}-x\lim_{n\to\infty}S_n=a-xa=a(1_A-x),$$ yielding the desired result $a(1_A-x)=1_A$.