Proof that $\sum_{i=1}^{\infty} \frac{1}{1+z^n}$ converges if $|z|\gt 1$ with $z$ complex.

Question

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 $n \ge N$.

Answer 1

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.

Answer 2

Use $|u+v| \ge ||u|-|v||$ with $u=z^n, v=1$. Note that if $z=re^{it}$ with $r>1$ then $z^n=r^ne^{nit}$ so $|z^n|=|z|^n$.

Answer 3

I think I got the answer now.

Since $$\frac{1}{|z|^n-1}=\frac{1}{(|z|^n-1)|z|^n}+\frac{1}{|z|^n}$$ it follows that $$\sum_{i=1}^{\infty}\frac{1}{|z|^n-1}=\sum_{i=1}^{\infty}\frac{1}{(|z|^n-1)|z|^n}+\sum_{i=1}^{\infty}\frac{1}{|z|^n}$$.

The first summation on the right side converges by Dirichlet's test and the second converges because $|z| \ge 1$.

It follows (by the comparison test) that $$\sum_{i=1}^{\infty}\frac{1}{1+z^n}$$ also converges.