Proof that $\sum_{i=1}^{\infty} \frac{1}{1+z^n}$ converges if $|z|\gt 1$ with $z$ complex. The hint given is to compare it to a geometric series with $r=\frac{1}{|z|}$.
I'm trying to show that for some value of $N$ $$\frac{1}{|1+z^n|} \le \frac{1}{|z|^n}$$ whenever $n \ge N$.

Since $|z^n+1|=\bigl|z^n-(-1)\bigr|\geqslant|z^n|-1=|z|^n-1$,$$\left|\frac1{z^n+1}\right|\leqslant\frac1{|z|^n-1}.$$And since$$\lim_{n\to\infty}\frac{\frac1{|z|^n-1}}{\frac1{|z|^n}}=1,$$the comparison test tells us that that the series $\sum_{n=1}^\infty\frac1{z^n+1}$ converges absolutely.
