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 $$\sum_{n=1}^{\infty} \frac{{\bf A}^{4n} - 2{\bf A}^{2n}}{n} = \ln\left(\frac{5}{3}\right)I= \begin{bmatrix} \ln\frac{5}{3} & 0\\ 0 & \ln\frac{5}{3} \end{bmatrix}$$