Let $A$ be a nonnegative square matrix (every entry is non-negative, $A\ge 0$). Look at these two conditions:
(1) $A^n\to0$ as $n\to\infty$
(2) $(I-A)^{-1}\ge 0$
I can show that (1) implies (2). In this case, the inverse is $I+A+A^2+\dots$.
My question is
"Is the converse true? i.e. does (2) imply (1) as well?"
Yes. Let $x$ be the Perron vector of $A$. Then $0\le(I-A)^{-1}x=(1-\rho(A))^{-1}x$. Hence $(1-\rho(A))^{-1}>0$ and $\rho(A)<1$. In turn, $A^n\to0$ as $n\to\infty$.