Let $\mathcal{A}$ be a unital Banach algebra and define $r^{\text{Gelf}}(A) = \lim_{n \rightarrow +\infty} \| A^n \|^{1/n}$.
It is possible to show that $r^{\text{Gelf}}(A) = \lim_{n \rightarrow +\infty} \| A^n \|^{1/n} = \inf_{n \rightarrow +\infty} \| A^n \|^{1/n}$.
I need to show that if $A$ is invertible and $B$ is such that $r^{\text{Gelf}}(BA^{-1}) < 1$, then $(A - B)$ is invertible.
What I did is the following: consider the element $C = A^{-1} + A^{-1}\sum_{n=1}^{+\infty}(BA^{-1})^n$. This is known as the Neumann series.
If this series converges, it follows from a direct computation that $(A - B)^{-1} = C$, but I am struggling to show that this series in fact converges.
I tried to show if $r^{\text{Gelf}}(BA^{-1}) < 1$, then $\| BA^{-1} \| < 1$, but I did not succeed (and maybe this isn't even true).
I would appreciate very much if someone could help me!
HINT:
Consider the Sequence $D_n=\sum_{k=1}^n(BA^{-1})^n$. Since the series $\sum_{k=1}^\infty\|(BA^{-1})^n\|$ converges by the root test, the sequence $\{D_n\}$ converges to some $D\in\mathcal A$. Then the $C$ you write in your question is $A^{-1}D$. Can you finish?
Side note: It is not true that $r^{\text{Gelf}}(A) < 1$ implies $\| A\| < 1$. For example, consider $\mathcal A=M_2(\mathbb C)$ and
$$A=\begin{pmatrix}0&\frac{3}{2}\\\frac{3}{8} &0\end{pmatrix}.$$
Then $r^{\text{Gelf}}(A)=\frac{3}{4}<1 $ while $\|A\|=\frac{3}{2}>1$.