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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 $\sum \frac{-1}{j}$?

Lemon
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  • It can't involve $|z|\to\infty$ because it should apply to each finite value of $z$ separately. – Michael Hardy Apr 28 '15 at 19:00
  • Well it can't be $j \to \infty$ either because the increasing index is $j$ – Lemon Apr 28 '15 at 19:01
  • I think the sum might converge is you look at symmetric partial sums. But if you insist that both sums $\sum_{j>0}, \sum_{j<0}$ converge, you'll get divergence by comparison with the harmonic series. – zhw. Apr 28 '15 at 19:06
  • Or the integral test right? – Lemon Apr 28 '15 at 19:08

2 Answers2

1

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.

Michael Hardy
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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 $(1)$.

robjohn
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