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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.

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It'll diverge... unless you intended $n$ to be the index instead of $m$. – J. M. Nov 2 '11 at 15:05
Thanks, J.M. and mixedmath. You are quite right in spotting the typo. Thanks! Edited. Assuming that $|R|<1$, what would the series converge to? Thanks. – saxa Nov 2 '11 at 15:11
It can't converge for $|R|<1$ since your exponent goes from $-\infty$. It might converge for $|R|=1$ when $z\neq 1,-1$, in which case the value will be $\log{|\frac{1+z}{1-z}|}$ – Thomas Andrews Nov 2 '11 at 15:22
@saxa: If $|R|<1$ the negative index terms will diverge. So your only hope is $|R|=1$. Even so, the $n=0$ terms are troubling. Generally for a series like this, you can use $\sum\frac{z^n}{n}=\sum \frac{1}{z}\int z^n\;dz$ if it converges – Ross Millikan Nov 2 '11 at 15:24
Actually, this sort of sum shows up in the theory of vertex algebras all the time. Usually it would be taken to be $\log | \frac{1+z}{1-z} |$, with the understanding that one is not actually performing the sum, one is simply setting the commutator(?) of two current operators to be this function, which makes the rest of the mathematics consistent. – Craig Nov 2 '11 at 15:35
up vote 2 down vote accepted

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.

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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:

$$f(z)=\frac{\pi i}{2}\ \text{for} \ |z|=1,\ \text{Im}(z)>0\ \text{and} \ f(z)=-\frac{\pi i}{2}\ \text{for} \ |z|=1,\ \text{Im}(z)<0.$$

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If $z=1$ everything does not cancel, because $\frac{1}{n}$ is not the same as $\frac{1}{2n}$ – Thomas Andrews Nov 2 '11 at 15:41
@ThomasAndrews: Well depending on how you look at it, yes it does. The series $\sum_{n=-M,\ n\neq =0}^{n=M}\frac{z^n}{n}$ literally equals zero for $z=1$.... It has a singularity, but sure it is not clear what value should be assigned there. – Eric Naslund Nov 2 '11 at 15:53
So, you are defining the difference of two diverging series to be $0$? That seems odd. – Thomas Andrews Nov 2 '11 at 16:40
Also notice that if $|z|=1$, then $z^{-1}=\bar{z}$. So the only meaningful values for these series are real numbers when $|z|=1$, but your answer is imaginary. You've made a mistake somewhere. – Thomas Andrews Nov 2 '11 at 16:42
Also, if $z=-1$, the first series (of $z^n/n$) converges, but $z^{2n}/2n$ diverges. – Thomas Andrews Nov 2 '11 at 16:47

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