= Solution
The even-power formula reduces every summand to a scalar multiple of $I$:
$$
\sum_{n=1}^\infty\frac{A^{4n}-2A^{2n}}n
=\left[\sum_{n=1}^\infty\frac{(1/16)^n}{n}
-2\sum_{n=1}^\infty\frac{(-1/4)^n}{n}\right]I.
$$
Both <series> have <absolute convergence>. Apply the <logarithm> expansion $\sum_{n\geq1}z^n/n=-\log(1-z)$ for $|z|<1$ to obtain
$$
\left[-\log\frac{15}{16}+2\log\frac54\right]I
=\boxed{\log\frac53\,I}.
$$
Thus the off-diagonal entries vanish and both diagonal entries have the required value.
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