Let $\mathcal{A}$ be a Banach $^*$-algebra and $a=a^*\in \mathcal{A}$. Furthermore, suppose that $\sigma_{\mathcal{A}}(a)\subset \{0\}\cup [C,D]$ with $D>C>1$.
Question: Does the sequence $x_n=a(1-a^n)^{-1}$ converge to zero in $\mathcal{A}$?
Remark 1: Note that the sequence of polynomials $p_n(z)=\frac{z}{1-z^n}$ converges uniformly to zero on the spectrum of $a$ but not on any compact subset of an open neighborhood of the spectrum.
No.
Let $a$ be a nontrivial nilpotent element, i.e. $a^m=0$ for some $m\in\mathbb N$, then by spectral mapping theorem, we know $\sigma(a)=\{0\}$, and $x_n\rightarrow a\not=0$.
To show such an algebra exists, we may simply take $\mathcal A=\mathbb C[x]/(x^2)$ (and by abuse of notation, we shall use $x$ for the element $x+(x^2)$ in the algebra), and pick some arbitrary norm on $\mathcal A$, e.g. $\|\alpha+\beta x\|=|\alpha|+|\beta|$ and the star operation $(\alpha+\beta x)^*=\bar{\alpha}+\bar{\beta} x$. Now let $a=x$ in $\mathcal A$.
Take $\mathcal A\times\mathcal A$ with a suitalbe norm where $\mathcal A=\mathbb C[x]/(x^2)$ as above, then the element $a:=(2, x)$ has spectrum $\{0, 2\}$, and since as above, the second component of $a(1-a^n)^{-1}$ doesn't go to $0$, we know even if the spectrum of $a$ contains at least one number $>1$, it still doesn't work.
However, if $a$ is invertible, i.e. $\sigma_{\mathcal A}(a)\subset[C, D]$, then this is correct, which follows from the homolorphic functional calculus, as $p_n$ converges uniformly to $0$ on a an open neighborhood of $[C, D]$.