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$\sum\limits_{n=1}^{\infty}\frac{1}{1+z^n}$, $|z|>1$.

There are two facts that my professor uses that I am confused about.

The first is: $|1+z^n| \geq ||z|^n-1|$, I believe this is true for any $|z|$.

The other is: $\frac{1}{|z|^n-1} \leq \frac{2}{|z|^n}$, I believe this is also true for any $|z|$.

Can anyone prove these statements for me?

Johnver
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  • in particular, that only holds for $2\leq |z|^n$. – MichaelChirico Apr 29 '15 at 03:33
  • Now I'm confused--are you asking to prove the two statements, or are you asking to prove convergence? – MichaelChirico Apr 29 '15 at 03:35
  • @TravisJ The left side and the right side also goes to zero since $|z|^n=|1+\epsilon|^n \to \infty$ for any $\epsilon >0$. – Mark Viola Apr 29 '15 at 04:17
  • @travis The term is $|z|^n-1=(1+\epsilon)^n -1 \to \infty$. – Mark Viola Apr 29 '15 at 12:19
  • @Dr.MV, yes, my mistake. But the inequality is still false when $|z|$ is close to $1$. For example, if $n=10$ and $|z|=1.05$ then LHS is about $1.5901$ and the RHS is about $1.2278$. – TravisJ Apr 29 '15 at 12:39
  • @TravisJ The inequality is not false. Fix $\epsilon$ first. Then, you can make $(1+\epsilon)^n$ arbitrarily large by choosing $n$ accordingly. – Mark Viola Apr 29 '15 at 12:43
  • @Dr.MV, I have since deleted the comment, but it was for fixed $n$ as $|z|$ is close to $1$. Close enough to 1 depends on what $n$ is. I am only looking at the inequality $\frac{1}{|z|^{n}-1}\leq \frac{2}{|z|^{n}}$. – TravisJ Apr 29 '15 at 12:51
  • That inequality holds only when $n>N$, where $N$ is adequately large. – Mark Viola Apr 29 '15 at 12:55

2 Answers2

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$|1+z^n| \geq |z^n| - 1 = |z|^n - 1$ is true due to the $\triangle$ inequality, and the second one is true if $|z|^n \leq 2|z|^n - 2 \iff 2 \leq |z|^n$, and this is true for some $n \geq N_0$ since $|z|^n \to +\infty$ as $n \to \infty$.

DeepSea
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3

The first statement is a version of the triangle inequality.

$$\begin{align} |x|&=|x-y+y|\le|x-y|+|y|\\ &\implies |x|-|y|\le|x-y|. \end{align}$$

Similarly $|y|-|x|\le|x-y|$. Hence, $$||x|-|y||\le|x-y|.$$

The other inequality is equivalent to $$|z|^n\le 2|z|^n-2$$ or $$|z|^n\ge 2.$$

Tim Raczkowski
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