Taylor series expansion of $\frac{(1- \cos x)}{x^2} $ According to my calculations, the Taylor series expansion for $1- \cos x$ around $x=\pi$ is 
$$2 - \frac{(x-\pi)^2}{2} + \frac{(x-\pi)^4}{24} - \frac{(x-\pi)^6}{720} + \frac{(x-\pi)^8}{40320} -\dotsb
$$
In solving the Taylor series expansion of $\frac{(1- \cos x)}{x^2}$ around $x= \pi$, would it be the same if I'll just multiply $\frac{1}{x^2}$ to the series written above? If that's not possible, how can I solve the series expansion of the quotient of two functions without doing ridiculously long sessions of product rule differentiation? 

As $x \to \pi$, you may express $\dfrac1{x^2}$ as 
$$
\begin{align}
\dfrac1{x^2}&=\dfrac1{(\pi+\color{red}{x-\pi})^2}
\\\\&=\dfrac1{\pi^2} \cdot \dfrac1{\left(1+\frac{\color{red}{x-\pi}}\pi\right)^2}
\\\\&=\dfrac1{\pi^2} \cdot \sum_{n=0}^\infty\frac{(-1)^n(n+1)}{\pi^n}\left({\color{red}{x-\pi}}\right)^n
\end{align}
$$ then multiplying by
$$
1-\cos x=2+\sum_{n=1}^\infty\frac{(-1)^n}{(2n)!}\left({\color{red}{x-\pi}}\right)^{2n}
$$ one gets, as $x \to \pi$,

$$
\dfrac{1-\cos x}{x^2}=\frac{2}{\pi^2}-\frac{4 (x-\pi)}{\pi^3}+\left(\frac{6}{\pi^4}-\frac{1}{2\pi^2}\right) (x-\pi)^2-\left(\frac{8}{\pi^5}-\frac{1}{\pi^3}\right) (x-\pi)^3+\cdots.
$$

