Prove the series $\sum_{n\geq1} \frac{1}{1+z^n}$ is convengent if $|z|>1$.

Question

Consider $$\sum_{n\geq1} \frac{1}{1+z^n}$$ with $z\in\mathbb{C}$; I have prove that series is NOT convergent when $|z|\leq1$ to it I applied Weierstrass' method. But I do not know how I can prove that series is convergent when $|z|>1$, someone could help?, thanks

Answer 1

For $|z|\geq a>1$, you may observe that $$ \left|\sum_{n\geq1} \frac{1}{1+z^n}\right|\leq\sum_{n\geq1}\frac{1}{\left| 1+z^n\right|}\leq \sum_{n\geq1} \frac{1}{|z|^n} \left(1+\frac1{(a+1)n}\right)\leq 2\sum_{n\geq1} \frac{1}{|z|^n} $$ and the latter series is a convergent geometric series where we have used $a^n\geq 1+(a+1)n$.

Answer 2

Hint when denominator becomes more than 1 the total mod will become less than 1 so we can prove that the given number is perfect fraction and we are done with the proof.