You can bound $\|A_k\| \leq C(n)\max_{i,j} |(A_k)_{ij}|$ for some function of the dimension only $C(n)$, because all norms are equivalent in finite dimension.

This maximum is attained on a (positive) diagonal entry, because of positive definiteness.

Then you have $$\sum_k \|A_k\| \leq C(n) \sum_k \max_i (A_k)_{ii} \leq C(n)\sum_k \sum_i (A_k)_{ii} = C(n) \sum_i (\sum_k (A_k)_{ii}),$$ which is finite because $\sum_k (A_k)_{ii}$ is finite for each $i$.

You can bound $\|A_k\| \leq C(n)\max_{i,j} |(A_k)_{ij}|$ for some function of the dimension only $C(n)$, because all norms are equivalent in finite dimension. If I am not mistaken $C(n)=\sqrt{n}$, but it doesn't really matter here.

This maximum is attained on a (positive) diagonal entry, because of positive definiteness.

Then you have $$\sum_k \|A_k\| \leq C(n) \sum_k \max_i (A_k)_{ii} \leq C(n)\sum_k \sum_i (A_k)_{ii} = C(n) \sum_i (\sum_k (A_k)_{ii}),$$ which is finite because $\sum_k (A_k)_{ii}$ is finite for each $i$.