I am interested in necessary conditions under which the series
$$\sum_{n=0}^\infty A^n \Omega (A^n)^{\top}$$
converges wrt. a submultiplicative matrix norm, where A is diagonalizable and $\Omega$ is positive definite. I believe that the eigenvalues of A must be strictly less than 1, but I am failing to find a reason why.
If the eigenvalues of $A$ are strictly less than $1$, then we have $$ \left\|\sum_{n=0}^\infty A^n \Omega A^{nT}\right\| \leq \sum_{n=0}^\infty \|A^n\| \|\Omega\| \|A^{Tn}\| $$ which converges, since $\lim_{n \to \infty} (\|\Omega\| \|A^{n}\|\|A^{nT}\|)^{1/n} = \rho(A) \rho(A^T) = \rho(A)^2$, where $\rho$ denotes the spectral radius.
On the other hand, if $x$ is a (possibly complex) eigenvector of $A^T = A^*$ with an eigenvalue $\lambda$, with $|\lambda| \geq 1$, then the series $$ x^*\left(\sum_{n=0}^\infty A^n \Omega A^{nT}\right)x = \sum_{n=0}^\infty \bar \lambda^n x^*\Omega \lambda^n x = (x^*Ax) \sum_{n=0}^\infty |\lambda|^{2n} $$ fails to converge. Here, $A^*$ denotes the conjugate-transpose of $A$.
Both of these apply whether or not $A$ is diagonalizable.