<s>[INST] Finding all $z\in \mathbb{C}$ such that the series $\sum\limits_{n=1}^{\infty} \frac{1}{1+z^n}$ converges I am trying to find out all $z\in \mathbb{C}$ such that the series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{1+z^n}$ converges. 
I notice that for $\left|z\right|\leq 1$, we have $\left|1+z^n\right|\leq 1+\left|z\right|^n\leq 1 + 1=2$ and hence $\limsup_{n\to \infty}\frac{1}{1+z^n}\ge 1/2$ which means that $\frac{1}{1+z^n}$ does not go to zero and so the series does not converge for $|z|\leq 1$. 
$\left|\frac{1}{1+z^n}\right|=\sqrt{1/\left(1+2\Re(z^n)+\left|z\right|^{2n}\right)}$ and I suspect $1/\left(1+2\Re(z^n)+\left|z\right|^{2n}\right)^{1/2n}$  goes to $1/\left|z\right|$ but I am unable to prove that. 

Edit: 
Suppose $|z|>1$. Suppose $z=r(\cos\theta + i \sin \theta)$ with $r>1$. Then $\displaystyle |1+z^n|=\sqrt{1+2r^n\cos n\theta+r^{2n}}>(r^n-1)$, so $\displaystyle \frac{1}{|1+z^n|}<\frac{1}{r^n-1}$. We will try to prove that $\displaystyle \sum \frac{1}{r^n-1}$ is convergent which will give us our desired result by the comparison test. Note that $\displaystyle \frac{r^n-1}{r^{n+1}-1}=\frac{r^{n-1}+\dots+1}{r^n+\dots+1}=1- (r-1)\frac{r^n}{r^{n+1}-1}$ which goes to $1/r<1$ and hence $\displaystyle \sum \frac{1}{r^n-1}$ is convergent by the ratio test.
 [/INST] Suppose $|z|>1$. Suppose $z=r(\cos\theta + i \sin \theta)$ with $r>1$. Then $\displaystyle |1+z^n|=\sqrt{1+2r^n\cos n\theta+r^{2n}}>(r^n-1)$, so $\displaystyle \frac{1}{|1+z^n|}<\frac{1}{r^n-1}$. We will try to prove that $\displaystyle \sum \frac{1}{r^n-1}$ is convergent which will give us our desired result by the comparison test. Note that $\displaystyle \frac{r^n-1}{r^{n+1}-1}=\frac{r^{n-1}+\dots+1}{r^n+\dots+1}=1- (r-1)\frac{r^n}{r^{n+1}-1}$ which goes to $1/r<1$ and hence $\displaystyle \sum \frac{1}{r^n-1}$ is convergent by the ratio test.
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