How can I prove the following question

Question

Let $(A,+,.,*,\|.\|)$ denotes complex Banach algebra such that $\|.\|$ norm on $A$ satisfies $$\|f*g\|| \leq \| f\|.\|g\|$$ and $e$ is the identity element. How can I prove that if $\| x\|<1$ then $$(e-x)^{-1}=e+x+x^2+x^3+\cdots?$$

Answer 1

Define $S_n= e+x+...+x^n$. Since $A$ is a Banach algebra, $||x^n||\leq ||x||^n$, hence $\{S_n\}$ forms a Cauchy sequence. Thus, it is convergent to, say, $S$.

Since $S \cdot(e-x)= \lim_{n\to \infty} S_n \cdot (e-x)= \lim_{n\to\infty} e -x^{n+1}= e$ and analogously $(e-x) \cdot S= e$, $(e-x)^{-1}=S$.