I have a few questions about Taylor and Laurent series expansions.
Is it possible to compute a laurent series of a real function with poles?
If $f(x)=\frac{1}{(x^{2}-1)^{2}}$ is a real function with a singularity at $x=\pm 1$.
How do I compute its Taylor series expansion and also determine whether the series converges?
A Taylor series for $f$ is
$$f(x) = \sum_{k=0}^\infty a_k(x-x_0)^k,$$ so we need to specify $x_0$. Here is one approach to compute the Taylor series centered at $x_0=0:$
Consider the following identity involving a geometric series: $$ \sum_{k=0}^n r^k = \frac{1-r^{n+1}}{1-r}, \quad (r \ne 0).$$ When $|r|<1,$ the limit of this series as $n\to\infty$ converges.
$$\frac{1}{1-x} = 1+x+x^2 +x^3 + \cdots, \quad (|x|<1).$$
$$\frac{1}{(1-x^2)^2} = 1+x^2+x^4 +x^6 + \cdots, \quad (|x|<1).$$
$$\begin{aligned} \frac{1}{(1-x^2)^2} &= [1+x^2+x^4 +x^6 + \cdots]\cdot[1+x^2+x^4 +x^6 + \cdots], \quad (|x|<1)\\ &= 1 + 2x^2 + 3x^4 + 4x^6 + 5x^8+ \cdots , \quad (|x|<1) \end{aligned}$$
This particular series converges because it is the product of convergent series. Each of these converge by comparison with the geometric series. There are other tests for convergence of a series including the ratio test and the root test. See here (and the associated links) for more info:
https://mathworld.wolfram.com/RadiusofConvergence.html
A Laurent series has also terms with negative powers of $x-x_0$. There are a number of $\textit{obvious}$ choices for $x_0$, for example $x_0\in\{0,1,-1\}.$
Let's write down the Laurent series for $x_0=0$ and $|x|>1.$
$$\begin{aligned} \frac{1}{(1-x^2)^2}&=\frac{1}{x^4} \cdot \frac{1}{(1-\frac{1}{x^2})^2}\\ &=\frac{1}{x^4}\left[1+\frac{2}{x^2}+\frac{3}{x^4}+\frac{4}{x^6}+\frac{5}{x^8}+\cdots\right]\\ &=\frac{1}{x^4}+\frac{2}{x^6}+\frac{3}{x^8}+\frac{4}{x^{10}}+\frac{5}{x^{12}}+\cdots, \quad (|x|>1). \end{aligned} $$