 Solution

The same Jordan calculation gives convergence of $\sum A^k$ exactly for $\rho (A)<1$. Finite sums satisfy $(I-A){\sum }_{k=0}^m{A}^k=I-A^{m+1}$; taking limits gives $(I-A{)}^{-1}$. Conversely convergence forces $A^k\to 0$, so part (a) applies.