eigenvalues for an Infinte sum of Diagnolizable matrix

Question

Suppose that A is a diagonalizable matrix. Show that $\lim_{n\rightarrow\infty} \sum_{j=0}^n (A^j)$exists if the absolute values of all the eigenvalues of A are smaller than one. Can someone please give me a hint? I feel that A is diagonalizable and all of its eigenvalues lie in the half-open interval $(−1,1]$, and a necessary condition is that the eigenvalues of A are either equal to 1 or strictly less than 1 in absolute value. However, I don't know how to prove

Answer 1

If A is diagonizable, it means that there exists an invertable $P$ and a diagonal $D$ matrix s.t. $$A = PDP^{-1}$$ Then you can prove this by using $$A^j = (PDP^{-1})^j = PD^jP^{-1}$$