8
$\begingroup$

Was playing around the series $\sum_0^\infty\frac{z^n}{1+z^{2n}}$, where $z$ is complex, trying to figure out where it converges.

Assuming $|z|>1$ $ \frac{|z^n|}{|1+z^{2n}|}>\frac{|z^n|}{1+|z^{2n}|}>\frac{|z^n|}{2|z|^{2n}}=\frac{1}{2|z|^{n}} $ but that didn't get me far, since the series whose terms are the last number above is a convergent series.

What is a better way to approach this series to determine where it converges? Thanks.

  • 0
    Incidentally, this is recognizably a Lambert series, and it is mentioned here: http://mathworld.wolfram.com/LambertSeries.html (with some slight futzing of the indices, you can get the sum in terms of theta functions)2012-01-31

2 Answers 2

8

Given $z \in \mathbb{C}$ with $|z| > 1$, choose any $r > 1$. We have

$|1 + z^{2n}| \geq \left|1 - |z|^{2n}\right| > \frac{|z|}{r}^{2n}$ for large enough $n$. The first inequality comes from $|a + b| \geq ||a| - |b||$, which is always true, and the second one comes from the limit $\lim_{n \rightarrow \infty} \frac{\left|1 - |z|^{2n}\right|}{|z|^{2n}} = 1,$ and the choice of $r$, which makes $1/r$ less than $1$. Now, applying the root test as azarel suggested, we get $\limsup_n \sqrt[n]{\left|\frac{z^n}{1 + z^{2n}}\right|} < \limsup_n \frac{|z|}{\frac{|z|^2}{\sqrt[n]{r}}} = \frac{1}{|z|} < 1.$ Therefore the series converges for $|z| > 1$.

As Greg Martin noted, this series may be written as $\sum_{n = 0}^{\infty} \frac{1}{z^n + z^{-n}}$ which shows that its behavior must be the same for $|z| > 1$ or $|z| < 1$, so that we get convergence on $\mathbb{C}\backslash\mathbb{S}^1$. The behavior for $|z| = 1$ seems more complicated, though, so I'll leave to someone more experienced to answer. Hope this helps!

  • 0
    Very interesting, thanks Leandro. :D2012-01-31
7

The cases $|z|\ne1$ having been disposed of, note that when $|z|=1$, the denominator is bounded above (in absolute value) by 2, while the numerator is (in absolute value) 1, so the quotient is bounded below (in absolute value) by $1/2$. In particular, the terms are not going to zero, so the series diverges.

  • 0
    I totally missed that. Simple and elegant argument.2012-01-31