Lets define the infinite series:
$$f(x) = \sum\limits_{k=1}^{\infty} \frac{x^s}{(1-x^k)^2}, \quad s \in \mathbb{N}_0$$
For me, it is clearly to see that it has two poles: at $x = -1$ and at $x = 1$. At the pole, the denominator has a root, so the pole is in general undefined because it would be a division by zero.
If I look at the plot, I can see how the plot behaves if I near from the right or left side to the pole. But is there a relative easy way (like finding the poles) to show that
$$\lim\limits_{x \to -1}f(x) = \begin{cases}+\infty & \text{if } s \text{ is even}\\-\infty & \text{if } s \text{ is odd}\end{cases}$$
Best Answer
Just pull out the common term $x^s$ which does not depend on $k$:
$$f(x)=x^s\sum_{k=1}^{\infty}\frac{1}{(1-x^k)^2}$$
and as long as you know one value of $s$ makes $f(x)$ go to $\infty$ as $x\rightarrow-1$, then clearly increasing $s$ just means multiplying by something close to $-1$, i.e. flipping the sign.