Expansions of meromorphic functions I found in google books the following expansion 
$$\sum _{k=1}^{\infty } \left(\frac{1}{(x-\pi  k)^3}-\frac{1}{\pi  k+x}+\frac{1}{(\pi  k+x)^3}+\frac{1}{\pi  k-x}\right)+\frac{1}{x^3}-\frac{1}{x}=\cot ^3(x)$$
and
$$\sum _{k=1}^{\infty } \left(-\frac{2}{\pi ^4 k^4}+\frac{8}{3 \pi ^2 k^2}-\frac{4}{3 (x-\pi  k)^2}+\frac{1}{(x-\pi  k)^4}-\frac{4}{3 (\pi  k+x)^2}+\frac{1}{(\pi  k+x)^4}\right)+\frac{1}{x^4}-\frac{4}{3 x^2}+\frac{26}{45}=\cot ^4(x)$$
how it get it, it will be nice a method for to get this type of series.
 A: You need to find the Laurent expansion of $\cot^3(z)$ at $0$ : $\cot^3(z) = \frac{1}{z^3} + \frac{a}{z^2}+\frac{b}{z}+h(z)$ where $h(z)$ is analytic around $z=0$. With the $\pi$-periodicity of $\cot(z)$, you get that $$g(z) = \cot^3(z) - \sum_{k=-\infty}^\infty \frac{1}{(z-k\pi)^3} + \frac{a}{(z-k\pi)^2}+\frac{b}{z-k\pi}$$ is analytic around the real line, and since it is meromorphic on $\mathbb{C}$ and has no other poles, it is entire
(note that $\sum_{k=-\infty}^\infty \frac{1}{z-k\pi}$ doesn't converge, but grouping the terms $k,-k$ together, it does, and it is meromorphic).
Finally, since $\cot^3(z)$  is bounded as $\text{Im}(z) \to \pm \infty$, as $\displaystyle\sum_{k=-\infty}^\infty \frac{1}{(z-k\pi)^m}$, we have that $g(z)$ is a bounded entire function.
So that by the Liouville theorem : $g(z) = C$, and evaluating at $z=\pi/2$ gives
$C = g(\pi/2) = \cot^3(\pi /2) + \sum_{k=-\infty}^\infty \frac{a}{\pi^2 (2k+1)^{2}} = \frac{a}{\pi^2} 2(1-2^{-2})\zeta(2)$ and 
$$\cot^3(z) = C+\sum_{k=-\infty}^\infty \frac{1}{(z-k\pi)^3} + \frac{a}{(z-k\pi)^2}+\frac{b}{z-k\pi}$$
With $a = f'(0)=0, b = \frac{f''(0)}{2}= -1$ where $f(z) = z^3 \cot^3(z)$, you get the desired result, and everything works the same for $\cot^4(z)$ or 
$$\color{blue}{\text{any trigonometric function that is bounded away from its poles}}$$
