How to compute the residue of $f(z)=\frac{n/z}{z^{n}-1}$ to be $-n$ at $z=0$? [duplicate]

Question

How to get residue of $f(z)=\frac{n/z}{z^{n}-1}$ with correct residue of $-n$ at zero?

It works it out just fine at 1 with:

Residue[(n/z/(z^n - 1)), {z, 1}]

Answer 1

Regarding the comment: the residue at z=1 is clearly wrong

If a function, $f(z)$, has a pole of order $k$ at $z=z_0$ then

$$ \begin{equation} \text{Res}(f,z_0)=\frac{1}{(k-1)!}\frac{d^{k-1}}{dz^{k-1}} \left[(z-z_0)^k f(z) \right]\Bigg|_{z=z_0} \end{equation} $$

Ok, so let's consider the function -which is the example in the OP with $n=1$

$$ \begin{equation} f = - \frac{1}{z(1-z)} \end{equation} $$

The above has a pole of order $1$ at $z=0$ and a pole of order $1$ at $z=1$. Hence, we compute explicitly

$$ \text{Res}(f,0) = \frac{1}{(1-1)!} \left((z-0)^1(-)\frac{1}{z}\frac{1}{1-z} \right)\Bigg|_{z=0} = - 1 $$

and also

$$ \text{Res}(f,1) = \frac{1}{(1-1)!} \left((z-1)^1(-)\frac{1}{z}\frac{1}{1-z} \right)\Bigg|_{z=1} = 1 $$

And likewise for higher values of $n$.

So, when Mathematica computes

Residue[n/(z (z^n - 1)) /. n -> 1, {z, 0}]
Residue[n/(z (z^n - 1)) /. n -> 1, {z, 1}]

as

-1

1

it is indeed correct.

Then, all the methods agree.

@Artes suggested

FindSequenceFunction[Table[Residue[n/(z (z^n - 1)), {z, 0}], {n, 10}],
  n]

which returns

-n

and is the same as

Assuming[n ∈ Integers && n >= 1, 
 Simplify@SeriesCoefficient[n/(z (z^n - 1)), {z, 0, -1}]]

from the linked question. Credits are to @Carl Woll.

Hopefully the above offers some clarity and will stop the war of the comments.