Convergent operatorial series 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 $|z|=1$? 

You solution is correct and solution in your book is wrong, because of the simple case $z=1$. For this case we have infinite sum of identical non-zero operators, more preciesly we have
$$
\sum\limits_{n=1}^\infty 1^nA^n=\sum\limits_{n=1}^\infty A= ???
$$
Since $A^n=A$ our series is of the form
$$
\sum\limits_{n=0}^\infty z^n A
$$
which is absolutely convergent for $|z|<1$, because convergence depends only on the series 
$$
\sum\limits_{n=0}^\infty z^n
$$
