Infinite sum of antisymmetric matrix? I have an antisymmetric matrix ${\bf A} =
\begin{bmatrix} 
0 & \frac{1}{2} \\
- \frac{1}{2} & 0
\end{bmatrix}
$, and I’d like to prove that
$$\sum_{n=1}^{\infty} \frac{{\bf A}^{4n} - 2{\bf A}^{2n}}{n} = 
\begin{bmatrix}
ln\frac{5}{3} & 0\\
0 & ln\frac{5}{3}
\end{bmatrix}
$$
I have solved for ${\bf A}^{2n}$ easily, and I recognise the Taylor series of the natural log in the summation. I am however unable to link the two together, and I can never get a perfect $\frac {5}{3}$ in the final matrix. Thank you!

Note that 
$${\bf A^2} =
\begin{bmatrix} 
-1/4 & 0 \\
0 & -1/4
\end{bmatrix} = (-1/4) I \quad \text{and} \quad 
{\bf A^4} =
\begin{bmatrix} 
1/16 & 0 \\
0 & 1/16
\end{bmatrix} = (1/16) I$$
So now we can factor out the identity matrix and reduce the problem to solving
$$\sum_{n=1}^{\infty}\frac{(1/16)^n - 2(-1/4)^n}{n}$$
We can, as you realized, use the taylor series of the natural log summation, namely,
$$-\ln(1-x)=\ln(\frac{1}{1-x})=\sum_{n=1}^{\infty}\frac{x^n}{n}$$
And from this we can directly solve the problem.
$$\left(\sum_{n=1}^{\infty}\frac{(1/16)^n}{n}\right) -2\left(\sum_{n=1}^{\infty}\frac{(-1/4)^n}{n}\right)$$
$$=\ln\left(\frac{1}{1-1/16}\right)-2\ln\left(\frac{1}{1+1/4}\right)$$
$$=\ln\left(\frac{16}{15}\right) -\ln\left(\frac{16}{25}\right)=\ln\left(\frac{5}{3}\right)$$
Thus 
$$\sum_{n=1}^{\infty} \frac{{\bf A}^{4n} - 2{\bf A}^{2n}}{n} = \ln\left(\frac{5}{3}\right)I=
\begin{bmatrix}
\ln\frac{5}{3} & 0\\
0 & \ln\frac{5}{3}
\end{bmatrix}$$
