Taylor series for $\ln(\frac{1-z^2}{1+z^3})$

Question

Taylor series for $\ln(\frac{1-z^2}{1+z^3})$
I've tried to $$\ln(1-z^2)-\ln(1+z^3)=\sum (-1)^{3n-1}z^{2n}-\sum(-1)^{4n-1}z^{3n}$$
I didn't manage to make it one series any help is good

Answer 1

I think you mean $-\sum_{n\ge1}\frac1nz^{2n}+\sum_{n\ge1}\frac{(-1)^n}{n}z^{3n}$. The $z^{6m+j}$ coefficient can be calculated for each $j\in\{0,\,\cdots,\,5\}$:

• If $j=0$ both series contribute, so the coefficient is $-\frac{1}{3m}+\frac{1}{2m}=\frac{1}{6m}$
• If $j\in\{1,\,5\}$ neither series contributes, so the coefficient is $0$
• If $j\in\{2,\,4\}$ only the first series contributes, so the coefficient is $-\frac{1}{3m+j/2}$
• If $j=3$ only the second series contributes, so the coefficient is $-\frac{1}{2m+1}$

Answer 2

Starting from @J.G.'s answer, all symplify to $$\log\left(\frac{1-z^2}{1+z^3}\right)=\sum_{n=2}^\infty \frac{2 \cos \left(n\frac{\pi }{3}\right)-1}{n} z^n$$