We have a symmetric positive definite matrix $A \in \Bbb R^{m \times m}$. How to calculate the following?
$$ \lim_{n \to \infty} \mbox{tr} \big( (\underbrace{A\cdot A\cdots A}_n)^{-1} - (\underbrace{A\cdot A\cdots A}_n + \underbrace{A\cdot A\cdots A}_{2n})^{-1} \big)$$
The main issue for me is how to transform the right-hand part of the expression containing inverse of the sum of matrices. The rest is more or less clear.
Note that as $A$ is symmetric positive definite we can use the spectral theorem to write $$ A = \sum_i \lambda_i P_i $$ where $\lambda_i > 0$ are the eigenvalues (up to multiplicity) and $\{P_i\}_i$ form a set of orthogonal projectors onto the corresponding eigenspaces. Note that $\mathrm{Tr}[P_i] = d_i$ where $d_i$ is the dimension of the eigenspace.
From this form you can see that for example $$ A^{k} = \sum_i \lambda_i^k P_i\, $$ for any $k \in \mathbb{Z}$. And furthermore $$ (A^n + A^{2n})^{-1} = \sum_i \frac{1}{\lambda_i^n + \lambda_i^{2n}} P_i $$ Putting everything together we arrive at $$ \mathrm{Tr}[A^{-n} - (A^n + A^{2n})^{-1}] = \sum_i d_i \left(\frac{1}{\lambda_i^n} - \frac{1}{\lambda_i^n + \lambda_i^{2n}}\right) = \sum_i d_i \frac{\lambda_i^{2n}}{\lambda_i^{2n} + \lambda_i^{3n}} $$
Looking at each term in the sum individually they split into two cases. If $\lambda_i > 1$ then the term tends to $0$ otherwise if $0 < \lambda_i \leq 1$ then the term tends to $d_i$. Overall we find that $$ \lim_{n \to \infty} \mathrm{Tr}[A^{-n} - (A^n + A^{2n})^{-1}] = \sum_{i : \lambda_i \leq 1} d_i $$ i.e., the total dimension of the eigenspaces corresponding to eigenvalues of $A$ that are no larger than $1$.