Radius Convergence

Question

$$\sum \frac{(-1)^nz^{2n+1}}{\log n} $$

I'm stuck on this problem. I tried the ratio test and the root test and i keep getting that the the series diverges. Am i doing something wrong? Any info will help.

Answer 1

$$ \left| \frac{\left( \dfrac{(-1)^{n+1} z^{2(n+1)+1} }{\log(n+1)} \right)}{ \left( \dfrac{(-1)^nz^{2n+1}}{\log n} \right) } \right| = \frac{\log n}{\log(n+1)} \cdot |z|^2 \to |z|^2 \quad \text{as } n\to\infty. $$ So this converges when $|z|<1,$ i.e. when $-1<z<1,$ and diverges when $|z|>1,$ i.e. when either $z>1$ or $z<-1.$

The case where $|z|=1$ is more involved, but the above gives you the radius.