The answer to the question at Proving that pole of infinite series "goes to" $+\infty$ or $-\infty$ wonders me how to do it if I cant pull out the numerator from the summation because it depends on $k$.
Lets define the infinite series:
$$f(x) = \sum\limits_{k=1}^{\infty} \frac{x^{k+s}}{(1-x^k)^2}, \quad s \in \mathbb{N}_0$$
So, it has poles at $x = \pm 1$. From the plot, I see that:
$$\lim\limits_{x \to -1}f(x) = \begin{cases}+\infty & \text{if } s \text{ is even}\\-\infty & \text{if } s \text{ is odd}\end{cases}$$
If I try to apply the answer of previous question, then:
$$\sum\limits_{k=1}^{\infty} \frac{x^{k+s}}{(1-x^k)^2} = x^s \sum\limits_{k=1}^{\infty} \frac{x^{k}}{(1-x^k)^2}$$
But now, the numerator is not always positive anymore as it depends on $k$.
So the question is, how to prove that:
$$\lim\limits_{x \to -1}\sum\limits_{k=1}^{\infty} \frac{x^{k}}{(1-x^k)^2} = +\infty$$
So the signs of the single terms in the summation are alternating.
The alternating signs still don't matter: notice that when $x\rightarrow-1$, $x^k$ is alternating $\pm1$.
This means, when $k$ is odd, the $(1-x^k)\approx2$ and the term is just $-1/4$.
When $k$ is even, it's more fun and you get $$\lim_{x\rightarrow-1}\frac{x^k}{(1-x^k)^2}=+\infty$$
This is one term going to infinity. The other terms go to constant $-1/4$s, so clearly the sum goes to $+\infty$.