For what values does $\sum_0^\infty\frac{z^n}{1+z^{2n}}$ converge?

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.

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What about using the root test? – azarel Jan 31 '12 at 1:57
If you write the series as $\sum_{n=0}^\infty 1/(z^n+z^{-n})$, you might be able to better discover a proof that it converges for $|z|>1$, and also a proof that it converges for $|z|<1$. – Greg Martin Jan 31 '12 at 2:03
Thanks Greg Martin, can you say a bit more about what you mean? So it converges every except possibly at $|z|=1$? When $|z|>1$ say, I get something like $$\left|\frac{1}{z^n+z^{-n}}\right|>\frac{1}{|z|^n+|z|^{-n}}<\frac{1}{|z|^n}<1$$ but I don't know how to use this. – Aishwini Jan 31 '12 at 2:14
If $|z|>1$ then $\left|\frac{1}{z^n+z^{-n}}\right|<\frac{1}{|z|^n-|z|^{-n}}<\frac{1}{|z|^n-1}< \frac{2}{|z|^n}$... The other case is similar... – N. S. Jan 31 '12 at 2:39
Incidentally, this is recognizably a Lambert series, and it is mentioned here: mathworld.wolfram.com/LambertSeries.html (with some slight futzing of the indices, you can get the sum in terms of theta functions) – deoxygerbe Jan 31 '12 at 5:19

2 Answers

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!

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Very interesting, thanks Leandro. :D – Aishwini Jan 31 '12 at 2:37

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.

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Cool, thanks for addressing the left over case. – Aishwini Jan 31 '12 at 2:38
I totally missed that. Simple and elegant argument. – student Jan 31 '12 at 2:39