An exercise I was doing asks (among other things) for the values of $z\in\mathbb{C}$ for which the following (operatorial) series converges absolutely: $\sum_{n=0}^{\infty}z^nA^n$ where $A$ is an operator in the Hilbert space $L^2(0,2\pi)$ such that $(Af)(x)=\frac{1}{\pi}\int_0^{2\pi}[\cos(x)\cos(y)+\sin(x)\sin(y)]f(y)dy$ I understand that $A$ is basically a projection operator in the form $Af = c\langle c,f\rangle+s\langle s,f\rangle$, where $s=\frac{1}{\sqrt{\pi}}\sin (x)$ and $c=\frac{1}{\sqrt{\pi}}\cos (x)$, so $||A||=1$ and $A^n = A$.
I also understand that, if I interpreted well, you should apply the Cauchy-Hadamard theorem to $\sum_{n=0}^{\infty}z^nA^n$ and search if it converges absolutely in the Banach space of all bounded operators between $L^2(0,2\pi)$ and itself. But with the Cauchy-Hadamard theorem you can conclude that the radius of convergence is $(\limsup||A^n||^{1/n})=1$. The answer to the exercise, however, is different and more cryptic:
"The series converges in norm when $||zA||\le 1$ ($|z|\le1$)"
How could you include the case $|z|=1$?