Simple calculus series question; convergence of $\sum_{j = -\infty}^{\infty} \frac{1}{z - j}$ So I am having a brain fart and I cannot rigorous write (in my head) why the series in the question does not converge. I know it has to do with the harmonic series.
Is it because if $|z| \to \infty$, then the sum is $\sum \frac{-1}{j}$?
 A: Suppose we had $z=6.2$.  Let's look first at the part of the series where the indices are $\ge6.2$:
\begin{align}
& \frac 1 {7-6.2} + \frac 1 {8-6.2} + \frac 1 {9-6.2} + \frac 1 {10-6.2} + \cdots \\[8pt]
> {} & \frac 1 1 + \frac 1 2 + \frac 1 3 + \frac 1 4 + \cdots =\infty. 
\end{align}
Now the part where they're $<6.2$:
$$
\frac 1 {-0.8} + \frac 1 {-1.8} + \frac 1 {-2.8} + \frac 1 {-3.8} + \cdots
$$
The series of absolute values of this also diverges to $\infty$ by a similar comparison test.
What if we had $z=6.2+20i$?
$$
\left|\frac 1 {7-(6.2+20i)}\right| + \left|\frac 1 {8-(6.2+20i)}\right| + \left|\frac 1 {9-(6.2+20i)}\right| + \left|\frac 1 {10-(6.2+20i)}\right| + \cdots \text{ ?}
$$
$$
\left|\frac 1 {n-(6.2+20i)}\right| \ge \frac 1 {20+n-6.2}
$$
and the series whose terms are the latter diverges by comparison to a harmonic series.
A: This series does converge in the principal value sense; that is, as shown in this answer,
$$
\lim_{n\to\infty}\sum_{k=-n}^n\frac1{k+z}=\pi\cot(\pi z)\tag1
$$
Since the series does not converge absolutely, the order of the terms is important to the convergence. For example, we cannot break up the sum in $(1)$ as follows:
$$
\sum_{k=0}^\infty\frac1{k+z}+\sum_{k=1}^\infty\frac1{-k+z}\tag2
$$
because each of the sums in $(2)$ diverges. In fact, using the generalized Harmonic Numbers, $H_z$,
$$
\lim_{n\to\infty}\left(\sum_{k=0}^n\frac1{k+z}-\log(n)\right)=\gamma-H_{z-1}\tag3
$$
and
$$
\lim_{n\to\infty}\left(\sum_{k=1}^n\frac1{-k+z}+\log(n)\right)=H_{-z}-\gamma\tag4
$$
By the Reflection Formula for Harmonic Numbers, the sum of $(3)$ and $(4)$ is
$$
H_{-z}-H_{z-1}=\pi\cot(\pi z)\tag5
$$
which gives $(1)$.
