If $A$ and $B$ are two matrices, $$A=\begin{bmatrix}-1 & -\frac{1}{2} \\ \frac{1}{2} & 1\end{bmatrix} \\ B=\begin{bmatrix}1 & 3 \\ -\frac{1}{9} & -\frac{1}{3}\end{bmatrix}$$ calculate $L=\lim_{n\to\infty}\sum_{k=0}^{n}S_k,$ where $S_k$ is the sum of elements of $A^{2k+1}+B^{2k+1}$
Naturally, what I tried first is to handcheck if there is a pattern for raising $A$ and $B$ to a power. What I got is: $$A^{2k} = \left(\frac{3}{4}\right)^kI_2$$ and $$B^k=\left(\frac{2}{3}\right)^{k-1}B$$
Then:
$$S_k= \text{Sum-of-elemets}(A^{2k}\cdot A + B^{2k+1}) = \text{Sum-of-elemets}(\left(\frac{3}{4}\right)^k \cdot I_2 \cdot A + \left(\frac{2}{3}\right)^{2k} \cdot B)$$
The sum of $A$'s elements is $0$, thus the left term is $0$. The sum of $B$'s elements is $\frac{32}{9}$, so:
$$S_k = \left(\frac{2}{3}\right)^k \cdot \frac{32}{9}$$
Finally: $$L = \lim_{n\to\infty}\sum_{k=0}^n S_k = \lim_{n\to\infty}\sum_{k=0}^n \left(\frac{2}{3}\right)^k \cdot \frac{32}{9} = \frac{32}{9} \cdot \lim_{n\to\infty} \frac{1-\left(\frac{2}{3}\right)^{n+1}}{1-\frac{2}{3}} = \frac{32}{9} \cdot 3 $$
However, I don't think this is correct. My workbook says the answer is $\frac{32}{5}$, so it seems I'm close but not quite there. I am also not sure at all how there's a $5$ in this whole deal. That being said, any ideas where I am wrong? I wasn't not able to find that out.