How to sum this complex series? How might I find $\sum\limits_{n=-\infty}^\infty {z^n\over n}-\sum\limits_{n=-\infty}^\infty {z^{2n}\over 2n}$ where $z=Re^{i\theta}$? Thanks.
 A: First, a few things: The index of summation appears to be wrong, and I hope we are not summing when $n=0$.  I assume we are looking at $$f(z)=\sum_{n\in\mathbb{Z}\backslash \{0\}}\frac{z^n}{n}-\sum_{n\in\mathbb{Z}\backslash \{0\}}\frac{z^{2n}}{2n}.$$  
We have to be very careful.  First, if $|z|\neq 1$, this will diverge by the divergence test.  Both $z=1$ and $z=-1$ are  singularities, and it is not clear what kind.  From now on, lets assume that $|z|=1$ and $z\neq \pm 1$.  
The Answer: The result is tricky and we will have $f(z)=\frac{\pi i}{2}$ when $|z|=1,\ \text{Im}(z)>0$ and $f(z)=-\frac{\pi i}{2}$ when $|z|=1,\ \text{Im}(z)>0$.  This can be extended to the unit disk, but it is very discontinuous.  The reason for this is subtleties with branch cuts, and the problems that they cause.
Proof: All the series conditionally converge, and we can justify rewriting $f(z)$ as
$$f(z)=\sum_{n=1}^\infty \frac{z^n}{n}-\sum_{n=1}^\infty\frac{z^{-n}}{n}-\sum_{n=1}^\infty \frac{z^{2n}}{2n}+\sum_{n=1}^\infty \frac{z^{-2n}}{2n}.$$  
Using the fact that $$\sum_{n=1}^\infty \frac{z^n}{n}=-\log(1-z)$$ we see that the above is $$-\log(1-z)+\log\left(1-z^{-1}\right)+\frac{1}{2}\log\left(1-z^{2}\right)-\frac{1}{2}\log\left(1-z^{-2}\right).$$ 
Caution with branches: Notice that $f(z)=-f(z^{-1})$, and since we are working on the unit circle and $z^{-1}=\overline{z}$ we see that the function will take opposite values on the top and bottom halves.
Combining some terms this is
$$\frac{1}{2}\log\left(\frac{\left(1-z^{-1}\right)^2\left(1-z^{2}\right)}{\left(1-z\right)^2\left(1-z^{-2}\right)}\right)=\frac{1}{2}\log\left(\frac{\left(1-z^{-1}\right)\left(1+z\right)}{\left(1-z\right)\left(1+z^{-1}\right)}\right)=\frac{1}{2}\log(-1).$$  Hence the answer is in some sense $\frac{1}{2}\log(-1)$, but not quite.  We have to be very careful with Branch Cuts.  When we combined the above terms, the argument secretely changed by a whole factor of $2\pi$ because we were not keeping track.  If we very carefully work things out, we arrive at:
$$f(z)=\frac{\pi i}{2}\ \text{for} \ |z|=1,\ \text{Im}(z)>0\ \text{and} \ f(z)=-\frac{\pi i}{2}\ \text{for} \ |z|=1,\ \text{Im}(z)<0.$$
A: To get rid of the problem with the index $n=0$, we can rewrite the series, as you may have intended,
$$
\sum_{k=-\infty}^\infty\frac{z^{2k+1}}{2k+1}\tag{1}
$$
If $|z|\not=1$, the double-ended series diverges, so I will assume that $R=1$. Using the series
$$
\sum_{k=0}^\infty\frac{z^{2k+1}}{2k+1}=\frac{1}{2}\log\left(\frac{1+z}{1-z}\right)\tag{2}
$$
the series $(1)$ becomes
$$
\begin{align}
\sum_{k=-\infty}^\infty\frac{e^{i(2k+1)\theta}}{2k+1}
&=\sum_{k=0}^\infty\frac{e^{i(2k+1)\theta}-e^{-i(2k+1)\theta}}{2k+1}\\
&=\frac{1}{2}\log\left(\frac{1+e^{i\theta}}{1-e^{i\theta}}\right)-\frac{1}{2}\log\left(\frac{1+e^{-i\theta}}{1-e^{-i\theta}}\right)\\
&=\frac{1}{2}\log\left(\frac{e^{-i\theta/2}+e^{i\theta/2}}{e^{-i\theta/2}-e^{i\theta/2}}\right)-\frac{1}{2}\log\left(\frac{e^{i\theta/2}+e^{-i\theta/2}}{e^{i\theta/2}-e^{-i\theta/2}}\right)\\
&=\frac{1}{2}\log(i\cot(\theta/2))-\frac{1}{2}\log(-i\cot(\theta/2))\\
&=\left\{\begin{array}{}+\frac{\pi}{2}i&\text{ when }0<\theta<\pi\\-\frac{\pi}{2}i&\text{ when }-\pi<\theta<0\end{array}\right.\tag{3}
\end{align}
$$
This agrees with Eric Naslund's answer.
