Prob. 6(d), Chap. 3, in Baby Rudin: Given $z \in \mathbb{C}$, does the series $\sum \frac{1}{1+z^n}$ converge or diverge?

Question

Here's Prob. 6(d), Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Investigate the behavior (convergence or divergence) of $\sum a_n$ if $$a_n = \frac{1}{1+z^n}$$ for complex values of $z$.

My effort:

If $\left\vert z \right\vert > 1$, then we note that $$ \left\vert a_n \right\vert = \frac{1}{\left\vert 1 + z^n \right\vert} \leq \frac{1}{\left\vert z \right\vert^n - 1} < \frac{1}{\left\vert z\right\vert^n},$$ and the series $$\sum_{n=0}^\infty \frac{1}{\left\vert z\right\vert^n} = \frac{1}{1- \frac{1}{\left\vert z\right\vert} } = \frac{ \left\vert z \right\vert }{ \left\vert z \right\vert - 1 } < +\infty,$$ which implies that our series converges if $\left\vert z \right\vert > 1$. Am I right?

If $\vert z \vert < 1$, then $\lim_{n \to \infty} \vert z \vert^n = 0$ and so $$\left\vert a_n \right\vert = \frac{1}{\left\vert 1 + z^n \right\vert} \geq \frac{1}{1 + \left\vert z \right\vert^n} \to 1 \ \mbox{ as } \ n \to \infty,$$ so $$\lim_{n\to\infty} a_n \neq 0,$$ which implies that our series diverges. Am I right?

If $\vert z \vert = 1$, then we note that $$\left\vert a_n \right\vert = \frac{\vert z \vert^{-n}}{ \left\vert z^{-n} + 1 \right\vert} = \frac{1}{ \left\vert z^{-n} + 1 \right\vert } \geq \frac{1}{\vert z \vert^{-n} + 1 } = \frac{1}{2},$$ which implies that $$\lim_{n\to\infty} a_n \neq 0,$$ showing that our series diverges. Am I right?

@DonAntonio which one? Do you mean the following? $$\lim_{n\to\infty}a_n \neq 0.$$ — Saaqib Mahmood, Dec 25 '16 at 12:38
The last inequality in the your first line of equations is wrong. You can get $\frac{1}{|z|^n-1}\le\frac2{|z^n|}$ if $n$ is large enough. — Lutz Lehmann, Dec 25 '16 at 12:54
@SaaqibMahmuud After ...you wrote "effort". Lutz already made this clear. — DonAntonio, Dec 25 '16 at 13:12
Linked duplicate https://math.stackexchange.com/questions/1257017/proving-the-convergence-of-sum-limits-n-1-infty-frac11zn-for-z only deals with $|z|>1$ — Henry, Oct 18 '23 at 08:19

score 3 · Accepted Answer · edited Dec 25 '16 at 14:15

3

Case I: $|z|>1$

Clearly, $|z^n+1|>|z|^n-1\ge\frac{1}{2}|z|^n$, whenever $|z|^n\ge 2$; which eventually happens, say, for $n\ge n_0(z)$, and hence $$ \sum_{n=1}^\infty\frac{1}{|1+z^n|}=\sum_{n=1}^{n_0(z)}\frac{1}{|1+z^n|}+ \sum_{n>n_0(z)}\frac{1}{|1+z^n|}\le \sum_{n=1}^{n_0(z)}\frac{1}{|1+z^n|}+ \sum_{n>n_0(z)}\frac{2}{|z|^n}<\infty. $$

Case II: $|z|<1$

Clearly $\frac{1}{1+z^n}\to 1\ne 0$. Hence divergence.

Case II: $|z|=1$

Clearly $\frac{1}{1+z^n}\not\to 0$. Hence divergence.

edited Dec 25 '16 at 14:15

Saaqib Mahmood

26,762

answered Dec 25 '16 at 13:55

Yiorgos S. Smyrlis

83,933

I've edited your answer. Do you agree with my edit? – Saaqib Mahmood Dec 25 '16 at 14:16
It fine. Thanks. – Yiorgos S. Smyrlis Dec 25 '16 at 14:44

score 2 · Answer 2 · answered Dec 25 '16 at 13:43

An idea: suppose $\;|z|>1\;$ , then apply the limit comparison test:

$$\frac{\left|\frac1{1+z^n}\right|}{\left|\frac1{z^n}\right|}=\frac{|z|^n}{|z^n+1|}\xrightarrow[n\to\infty]{}1\implies\sum_{n=1}^\infty\frac1{1+z^n} \;\;\text{converges as}\;\;\sum_{n=1}^\infty\frac1{z^n}\;\;\text{does}$$

Can you see how to use exactly the same as above to get divergence if $\;|z|\le1\;$ ?

Prob. 6(d), Chap. 3, in Baby Rudin: Given $z \in \mathbb{C}$, does the series $\sum \frac{1}{1+z^n}$ converge or diverge?

2 Answers2

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