Show that the series
$\sum_{n=1}^{\infty} ({1\over 1 - z^{n+1}} - {1\over 1-z^{n}})$
represents two analytic functions in the regions $|z| \lt 1$ and $|z| \gt 1$, and these functions are not analytic continuations of each other.
Okay, so because the function is written as a series it is pretty obvious it is analytic. This is what I think I need to do: Split the power series into two individual power series that can be represented as separate analytic continuations.
The problem I am having is that neither of the fractions in the series can be represented in a way that would let me work with a familiar power series-I've tried separating to get $\sum_{n=1}^{\infty} {1\over 1 - z^{n+1}}$ - $\sum_{n=1}^{\infty} {1\over 1-z^{n}}$ and then individually integrating each series which gives me log of something which is not useful and differentiating which just gives me something more complicated. If I combine the two terms then seperate, I get $\sum_{n=1}^{\infty} ({z^{n+1}\over (1 - z^{n+1})(1-z^{n})} - {z^{n}\over (1-z^{n})(1-z^{n+1})})$ which is slightly similar to $\sum_{n=1}^{\infty} {z^{n}}$ which I know can be analytically continued over $\frac{1}{1-z}$
I would appreciate any suggestions, thank you.
The original sum is a telescoping series that can be written as
$lim_{n \to \infty}({1 \over1-z^{n+1}}-{1 \over1-z})$
This leads into two functions
a) $1- {1\over1-z}$ for $|z| < 1$
b) $- {1\over1-z}$ for $|z|>1$
If we expand these two in $\Bbb C-{\{1\}}$ then restricting one or the other in $D(0,1)$ or $\Bbb C-\hat D(0,1)$ then we can see that they are not the same.