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I'm trying solve a tricky problem where I need to find a function $f(z)$ such that the residue at every positive integer $n$ is equal to $\frac{1}{n(n+1)}$

For example, the function would have a pole at $z=3$ with the residue being equal to $\frac{1}{12}$ and so on.

My guess is that the function would look like $\frac{1}{\sin(\pi z)}$ times some other function $g(z)$ to generate poles at integer values of $z$.

To summarize: I need help finding a function $f(z)$, undefined at positive integers, such that $\lim_{z\rightarrow n}f(z)(z-n)=\frac{1}{n(n+1)}$

(I don't even know if a function like this is even possible but I hope so)

Noa Arvidsson
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2 Answers2

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The function with the prescribed residues can be constructed as $$ \begin{align} \sum_{k=1}^\infty\frac1{(z-k)k(k+1)} &=\sum_{k=1}^\infty\left(\frac{\frac1{z(z+1)}}{z-k}+\frac{\frac1z}k-\frac{\frac1{z+1}}{k+1}\right)\tag{1a}\\ &=\sum_{k=1}^\infty\frac1{z(z+1)}\left(\frac1k-\frac1{k-z}+z\left(\frac1k-\frac1{k+1}\right)\right)\tag{1b}\\ &=\frac{H_{-z}}{z(z+1)}+\frac1{z+1}\tag{1c}\\[3pt] &=\frac{z+H_{-z}}{z(z+1)}\tag{1d} \end{align} $$ Explanation:
$\text{(1a):}$ Partial Fractions
$\text{(1b):}$ reorganize
$\text{(1c):}$ $H_x=\sum\limits_{k=1}^\infty\left(\frac1k-\frac1{k+x}\right)$ and $\sum\limits_{k=1}^\infty\left(\frac1k-\frac1{k+1}\right)=1$
$\text{(1d):}$ simplify

where $H_x$ are the Extended Harmonic Numbers. The relationship between the Extended Harmonic Numbers and the digamma function can be found in this answer.

robjohn
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Update: I've found the function $\frac{\pi e^{i\pi z}}{z(z+1)\sin(\pi z)}$ works!

Noa Arvidsson
  • 339
  • Of course that function has poles also at the negative integers. Perhaps you should clarify in your question that that this is allowed. – Martin R Oct 18 '22 at 11:17