Infinite Series $\sum\limits_{n=-\infty}^{\infty}\frac{1}{(x+n\pi)^m}$ How can we find a closed form for the following infinite series for any $m\in\mathbb N$?
$$\sum_{n=-\infty}^{\infty}\frac{1}{(x+n\pi)^m}$$

To obtain convergence in the specific case $m=1$ let's rewrite your series $$f_m(x):=\sum_{n=-\infty}^{\infty}\frac{1}{(x+n\,\pi)^m}$$ 
as :
\begin{align}
f(x)&:=\frac 1x+\sum_{n=1}^\infty\frac{1}{x+n\,\pi}+\frac{1}{x-n\,\pi}\\
&=\frac 1x+\sum_{n=1}^\infty\frac{2x}{x^2-(n\,\pi)^2}\\
\\
&=\cot(x)\\
\end{align}
This was deduced by Euler (proof using Fourier series for $\,x:=\pi\,z$).
It is straightforward to obtain : $\;\displaystyle f_{m+1}(x)=\frac {(-1)^{m}}{m!}\left(\frac d{dx}\right)^mf(x)\;$ so that
\begin{align}
f_{m+1}(x)&=\frac {(-1)^m\;\cot^{(m)}(x)}{m!} \\
\end{align}
The first results for $\;m!\cdot f_{m+1}(x)\,$ are in the following table (supposing $\,t:=\tan(x)$) :
\begin{array} {l|c} 
m!\cdot f_{m+1}&(-1)^m\,\cot^{(m)}(x)\\
\hline
0!\cdot f_1&\frac 1{t}\\
1!\cdot f_2&\frac {1+1t^2}{t^2}\\
2!\cdot f_3&\frac {2+2t^2}{t^3}\\
3!\cdot f_4&\frac {6+8t^2+2t^4}{t^4}\\
4!\cdot f_5&\frac {24+40t^2+16t^4}{t^5}\\
\end{array}
Since $\;\left(\tan(x)^n\right)'=+n\,\tan(x)^{n-1}+n\,\tan(x)^{n+1}\;$
while $\;\left(\cot(x)^n\right)'=-n\,\cot(x)^{n-1}-n\,\cot(x)^{n+1}\;$
the triangle of coefficients obtained at the numerator will be the same than the triangle for the derivatives of the $\tan$ function (OEIS A101343 with more references from Knuth like the paper 'Computation of tangent, Euler, and Bernoulli numbers', Foata and others).
This allows to propose the general formula :
$$f_{m+1}(x)=\frac {\sum_{k=0}^{\lfloor (m+1)/2\rfloor}\;c_{m,k}\;\tan(x)^{2k}}{m!\;\tan(x)^{m+1}}=\frac 1{m!}\sum_{k=0}^{\lfloor (m+1)/2\rfloor}\;c_{m,k}\;\cot(x)^{m+1-2k}$$
with $\;\;\displaystyle c_{m,k}:=(m-2(k-1))\;c_{m-1,k-1}+(m-2k)\;c_{m-1,k}\;$ and $\;c_0,k=\delta_k^0$
An 'explicit formula' was proposed at OEIS : $\;c_{m,k}=\operatorname{tr}_{n,k}+\operatorname{tr}_{n,k-1}\;$ with $\operatorname{tr}$ defined by :
$$\operatorname{tr}_{n,i}=\sum_{j=j_0}^{2i} \binom{j+n-2\,i-1}{n-2\,i-1}\,(j+n-2\,i)!\;2^{2\,i-j}(-1)^{j-i}\;\operatorname{Stirling}_2(n,j+n-2\,i)$$
(with a Stirling number of the second kind at the right and $j_0=1$ if $2\,i>n$ and $j_0=0$ else)
