Consider the power series expansion $$ \sqrt{1-z} = 1+\sum\limits_{k=1}^\infty c_k z^k, $$ converging absolutely in the ball $|z| \leq 1$. Let $H$ be a Hilbert space and $A \in \mathcal L(H)$ a bounded linear operator with $\|\mathrm{Id}-A\| \leq 1$. Then the series $$ 1 + \sum\limits_{k=1}^\infty c_k (I-A)^k $$ converges in norm to some $B \in \mathcal L(H)$. How to show that $B^2 = A$?
Clearly, we have $$ \biggl| \bigl(1-\|I-A\|\bigr) - \bigl(1+\sum_{k=1}^N c_k\|I-A\|^k \bigr)^2 \biggr| \to 0, \quad N \to +\infty, $$ but I need to show somehow that $$ \biggl\| (I-(I-A)) - \bigl(I+\sum\limits_{k=1}^Nc_k(I-A)^k\bigr)^2 \biggr\| \to 0, \quad N \to + \infty. $$ If it is necessary, $A = A^*$ and $A \geq 0$.