# convergence of series to two distinct functions

I am trying to show something on the following series:

$\sum_{n=0}^\infty \frac{1}{z^n+z^{-n}}$ is converging to two different, holomorphic functions, one in $\{z:|z|<1\}$ and a different one in $\{z:|z|>1\}$.

But I have no idea of how to start. How do I do that?

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When $|z| <1$ $\frac{1}{z^n+z^{-n}}$ becomes like $z^n$ for large $n$ so the series converges approximately to $\frac{1}{1-z}$. For $|z| >1$ the series converges (approximately) to $\frac{z}{z-1}$. But I don't know a rigorous derivation. – user17090 Nov 20 '11 at 21:42
First: show it converges in $|z|>1$. After you do that, maybe you will have some more ideas! – GEdgar Nov 20 '11 at 21:42
@Ali - Thanks for pointing out my absurd response :) I'm not feeling well today. – alex.jordan Nov 20 '11 at 21:53
@alex: It is ok, don't mention. – user17090 Nov 20 '11 at 21:56

Note it's symmetric in $z$ and ${1 \over z}$, so if you know the sum for $|z| > 1$ then you have the sum for $|z| < 1$ via $f(z) = f(1/z)$.
Note that your series is ${\displaystyle \sum_{n = 0}^{\infty} {z^{-n} \over 1 + z^{-2n}}}$. Given any $z$ with $|z| > 1$, if $n$ is large enough, then $|z^{-2n}| = |z|^{-2n} < {1 \over 2}$, so that $|1 + z^{-2n}| \geq 1 - |z|^{-2n} > { 1 \over 2}$. So for such $n$ you have $$\bigg|{z^{-n} \over 1 + z^{-2n}}\bigg| < 2|z|^{-n}$$ So since the sum of $|z|^{-n}$ is convergent when $|z| > 1$, the original series is too.
Okay, so I have worked out the convergence, and also it diverges for any point $z,|z|=1$. But I cannot argue why the holomorphic functions must be distinct, or does that immediately follow from the fact that they are separated by the poles along the unit circle? But that is not waterproof yet. – Marie. P. Nov 24 '11 at 23:14