Possibility to simplify $\sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}} = \frac{\pi }{{\sin \pi a}}} $ Is there any way to show that
$$\sum\limits_{k =  - \infty }^\infty  {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}} = \frac{1}{a} + \sum\limits_{k = 1}^\infty  {{{\left( { - 1} \right)}^k}\left( {\frac{1}{{a - k}} + \frac{1}{{a + k}}} \right)}=\frac{\pi }{{\sin \pi a}}} $$
Where $0 < a = \dfrac{n+1}{m} < 1$
The infinite series is equal to
$$\int\limits_{ - \infty }^\infty  {\frac{{{e^{at}}}}{{{e^t} + 1}}dt} $$
To get to the result, I split the integral at $x=0$ and use the convergent series in $(0,\infty)$ and $(-\infty,0)$ respectively:
$$\frac{1}{{1 + {e^t}}} = \sum\limits_{k = 0}^\infty  {{{\left( { - 1} \right)}^k}{e^{ - \left( {k + 1} \right)t}}} $$
$$\frac{1}{{1 + {e^t}}} = \sum\limits_{k = 0}^\infty  {{{\left( { - 1} \right)}^k}{e^{kt}}} $$
Since $0 < a < 1$
$$\eqalign{
  & \mathop {\lim }\limits_{t \to 0} \frac{{{e^{\left( {k + a} \right)t}}}}{{k + a}} - \mathop {\lim }\limits_{t \to  - \infty } \frac{{{e^{\left( {k + a} \right)t}}}}{{k + a}} = \frac{1}{{k + a}}  \cr 
  & \mathop {\lim }\limits_{t \to \infty } \frac{{{e^{\left( {a - k - 1} \right)t}}}}{{k + a}} - \mathop {\lim }\limits_{t \to 0} \frac{{{e^{\left( {a - k - 1} \right)t}}}}{{k + a}} =  - \frac{1}{{a - \left( {k + 1} \right)}} \cr} $$
A change in the indices will give the desired series.
Although I don't mind direct solutions from tables and other sources, I prefer an elaborated answer. 

Here's the solution in terms of $\psi(x)$. By separating even and odd indices we can get
$$\eqalign{
  & \sum\limits_{k = 0}^\infty  {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}}  = \sum\limits_{k = 0}^\infty  {\frac{1}{{a + 2k}}}  - \sum\limits_{k = 0}^\infty  {\frac{1}{{a + 2k + 1}}}   \cr 
  & \sum\limits_{k = 0}^\infty  {\frac{{{{\left( { - 1} \right)}^k}}}{{a - k}}}  = \sum\limits_{k = 0}^\infty  {\frac{1}{{a - 2k}}}  - \sum\limits_{k = 0}^\infty  {\frac{1}{{a - 2k - 1}}}  \cr} $$
which gives
$$\sum\limits_{k = 0}^\infty  {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}}  = \frac{1}{2}\psi \left( {\frac{{a + 1}}{2}} \right) - \frac{1}{2}\psi \left( {\frac{a}{2}} \right)$$
$$\sum\limits_{k = 0}^\infty  {\frac{{{{\left( { - 1} \right)}^k}}}{{a - k}}}  = \frac{1}{2}\psi \left( {1 - \frac{a}{2}} \right) - \frac{1}{2}\psi \left( {1 - \frac{{a + 1}}{2}} \right) + \frac{1}{a}$$
Then
$$\eqalign{
  & \sum\limits_{k =  - \infty }^\infty  {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}}  = \sum\limits_{k = 0}^\infty  {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}}  + \sum\limits_{k = 0}^\infty  {\frac{{{{\left( { - 1} \right)}^k}}}{{a - k}}}  - \frac{1}{a} =   \cr 
  &  = \left\{ {\frac{1}{2}\psi \left( {1 - \frac{a}{2}} \right) - \frac{1}{2}\psi \left( {\frac{a}{2}} \right)} \right\} - \left\{ {\frac{1}{2}\psi \left( {1 - \frac{{a + 1}}{2}} \right) - \frac{1}{2}\psi \left( {\frac{{a + 1}}{2}} \right)} \right\} \cr} $$
But using the reflection formula one has
$$\eqalign{
  & \frac{1}{2}\psi \left( {1 - \frac{a}{2}} \right) - \frac{1}{2}\psi \left( {\frac{a}{2}} \right) = \frac{\pi }{2}\cot \frac{{\pi a}}{2}  \cr 
  & \frac{1}{2}\psi \left( {1 - \frac{{a + 1}}{2}} \right) - \frac{1}{2}\psi \left( {\frac{{a + 1}}{2}} \right) = \frac{\pi }{2}\cot \frac{{\pi \left( {a + 1} \right)}}{2} =  - \frac{\pi }{2}\tan \frac{{\pi a}}{2} \cr} $$
So the series become
$$\eqalign{
  & \sum\limits_{k =  - \infty }^\infty  {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}}  = \frac{\pi }{2}\left\{ {\cot \frac{{\pi a}}{2} + \tan \frac{{\pi a}}{2}} \right\}  \cr 
  & \sum\limits_{k =  - \infty }^\infty  {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}}  = \pi \csc \pi a \cr} $$
The last being an application of a trigonometric identity.
 A: A related identity is proven in this answer using residue theory. Here is a real approach to that identity.

Convergence of the Principal Value
We will look at the principal value
$$
\begin{align}
f(x)
&=\lim_{n\to\infty}\sum_{k=-n}^n\frac1{k+x}\tag{1a}\\
&=\frac1x-\sum_{k=1}^\infty\frac{2x}{k^2-x^2}\tag{1b}
\end{align}
$$
$\text{(1b)}$ converges for all non-integer $x$.

Properties of $\boldsymbol{f(x)}$
$\bullet$ $f(x)$ has period $1$:
$$
\begin{align}
f(x)-f(x+1)
&=\lim_{n\to\infty}\left(\sum_{k=-n}^n\frac1{k+x}-\sum_{k=-n}^n\frac1{k+1+x}\right)\tag{2a}\\
&=\lim_{n\to\infty}\left(\sum_{k=-n}^n\frac1{k+x}-\sum_{k=-n+1}^{n+1}\frac1{k+x}\right)\tag{2b}\\
&=\lim_{n\to\infty}\left(\frac1{-n+x}-\frac1{n+1+x}\right)\tag{2c}\\[9pt]
&=0\tag{2d}
\end{align}
$$
Explanation:
$\text{(2a):}$ definition
$\text{(2b):}$ substitute $k\mapsto k-1$ in the right sum
$\text{(2c):}$  the sums telescope
$\text{(2d):}$  evaluate the limit
$\bullet$ $f(1/2)=0$:
$$
\begin{align}
f(1/2)
&=\lim_{n\to\infty}\sum_{k=-n}^n\frac1{k+1/2}\tag{3a}\\
&=\lim_{n\to\infty}\frac1{n+1/2}\tag{3b}\\[9pt]
&=0\tag{3c}
\end{align}
$$
Explanation:
$\text{(3a):}$ definition
$\text{(3b):}$ for $1\le j\le n$, the term with $k=j-1$ cancels the term with $k=-j$
$\phantom{\text{(3b):}}$ which leaves the term with $k=n$
$\text{(3c):}$ evaluate the limit
$\bullet$ $f(x)^2+\pi^2=-f'(x)$:
$$
\begin{align}
f(x)^2
&=\lim_{n\to\infty}\sum_{k=-n}^n\frac1{k+x}\sum_{j=-n}^n\frac1{j+x}\tag{4a}\\[3pt]
&=\sum_{k=-\infty}^\infty\frac1{(k+x)^2}+\lim_{n\to\infty}\sum_{\substack{|j|,|k|\le n\\j\ne k}}\left(\frac1{k+x}-\frac1{j+x}\right)\frac1{j-k}\tag{4b}\\
&=\sum_{k=-\infty}^\infty\frac1{(k+x)^2}+\lim_{n\to\infty}\sum_{\substack{|j|,|k|\le n\\j\ne k}}\frac2{k+x}\frac1{j-k}\tag{4c}\\
&=\sum_{k=-\infty}^\infty\frac1{(k+x)^2}+\lim_{n\to\infty}\sum_{k=1}^n\sum_{\substack{|j|\le n\\j\ne k}}\left(\frac2{k+x}+\frac2{k-x}\right)\frac1{j-k}\tag{4d}\\
&=\sum_{k=-\infty}^\infty\frac1{(k+x)^2}-\lim_{n\to\infty}\sum_{k=1}^n\left(\frac2{k+x}+\frac2{k-x}\right)(H_{n+k}-H_{n-k})\tag{4e}\\
&=-f'(x)-4\int_0^1\frac{\log\left(\frac{1+t}{1-t}\right)}{t}\,\mathrm{d}t\tag{4f}\\[12pt]
&=-f'(x)-\pi^2\tag{4g}
\end{align}
$$
Explanation:
$\text{(4a):}$ product of limits equals the limit of the products
$\text{(4b):}$ the left sum contains the terms with $j=k$
$\phantom{\text{(11):}}$ apply partial fractions to the right sum
$\text{(4c):}$ take advantage of the symmetry in $j$ and $k$
$\text{(4d):}$ for $k=0$, the sum in $j$ is $0$
$\phantom{\text{(4d):}}$ for $k\lt0$, if we substitute $(j,k)\mapsto(-j,-k)$,
$\phantom{\text{(4d):}}$ we get the sum with $k\gt0$ and $k+x\mapsto k-x$
$\text{(4e):}$ the sum in $j$ telescopes to $H_{n-k}-H_{n+k}$
$\text{(4f):}$ the left sum is $f'(x)$
$\phantom{\text{(4f):}}$ the right sum is a Riemann Sum with $\frac kn\mapsto t$ and $\frac1n\mapsto\mathrm{d}t$
$\phantom{\text{(4f):}}$ $H_{n+k}-H_{n-k}\mapsto\log\left(\frac{1+t}{1-t}\right)$ and $\frac2{k-x}+\frac2{k+x}\mapsto\frac{4\,\mathrm{d}t}t$
$\text{(4g):}$ $4\int_0^1\sum\limits_{k=0}^\infty\frac{2\,x^{2k}}{2k+1}\,\mathrm{d}x=4\sum\limits_{k=0}^\infty\frac2{(2k+1)^2}=\pi^2$

Conclude that $\boldsymbol{f(x)=\pi\cot(\pi x)}$
We can separate $(4)$ and integrate:
$$
\begin{align}
\int\frac{\pi\,\mathrm{d}f}{f^2+\pi^2}&=-\int\pi\,\mathrm{d}x\tag{5a}\\
\tan^{-1}(f/\pi)&=C-\pi x\tag{5b}\\[9pt]
f&=\pi\tan(C-\pi x)\tag{5c}
\end{align}
$$
$(3)$ allows us to compute $C=\pi/2$, giving
$$
f(x)=\pi\cot(\pi x)\tag6
$$
for $x\in(0,1)$. $(2)$ removes this restriction on $x$, validating $(6)$ for all non-integer $x$. That is,
$$
\sum_{k=-\infty}^\infty\frac1{k+x}=\pi\cot(\pi x)\tag7
$$
when taken in the principal value sense.

Answer to the Question
$$
\begin{align}
\sum_{k=-\infty}^\infty\frac{(-1)^k}{k+x}
&=\sum_{k=-\infty}^\infty\frac2{2k+x}-\sum_{k=-\infty}^\infty\frac1{k+x}\tag{8a}\\
&=\sum_{k=-\infty}^\infty\frac1{k+x/2}-\sum_{k=-\infty}^\infty\frac1{k+x}\tag{8b}\\[6pt]
&=\pi\cot(\pi x/2)-\pi\cot(\pi x)\tag{8c}\\[15pt]
&=\pi\csc(\pi x)\tag{8d}
\end{align}
$$
Explanation:
$\text{(8a):}$ the alternating sum is twice the sum of the even terms
$\phantom{\text{(8a):}}$ minus the sum of all the terms
$\text{(8b):}$ adjust the terms of the left sum to apply $(7)$
$\text{(8c):}$ apply $(7)$
$\text{(8d):}$ $\frac{1+\cos(\pi x)}{\sin(\pi x)}-\frac{\cos(\pi x)}{\sin(\pi x)}=\frac1{\sin(\pi x)}$
A: EDIT: The classical demonstration of this is obtained by expanding in Fourier series the function $\cos(zx)$ with $x\in(-\pi,\pi)$.
Let's detail Smirnov's proof (in "Course of Higher Mathematics 2 VI.1 Fourier series") :
$\cos(zx)$ is an even function of $x$ so that the $\sin(kx)$ terms disappear and the Fourier expansion is given by :
$$\cos(zx)=\frac{a_0}2+\sum_{k=1}^{\infty} a_k\cdot \cos(kx),\ \text{with}\ \ a_k=\frac2{\pi} \int_0^{\pi} \cos(zx)\cos(kx) dx$$
Integration is easy and $a_0=\frac2{\pi}\int_0^{\pi} \cos(zx) dx= \frac{2\sin(\pi z)}{\pi z}$ while
$a_k= \frac2{\pi}\int_0^{\pi} \cos(zx) \cos(kx) dx=\frac1{\pi}\left[\frac{\sin((z+k)x)}{z+k}+\frac{\sin((z-k)x)}{z-k}\right]_0^{\pi}=(-1)^k\frac{2z\sin(\pi z)}{\pi(z^2-k^2)}$
so that for $-\pi \le x \le \pi$ :
$$
\cos(zx)=\frac{2z\sin(\pi z)}{\pi}\left[\frac1{2z^2}+\frac{\cos(1x)}{1^2-z^2}-\frac{\cos(2x)}{2^2-z^2}+\frac{\cos(3x)}{3^2-z^2}-\cdots\right]
$$
Setting $x=0$ returns your equality :
$$
\frac1{\sin(\pi z)}=\frac{2z}{\pi}\left[\frac1{2z^2}-\sum_{k=1}^{\infty}\frac{(-1)^k}{k^2-z^2}\right]
$$
while $x=\pi$ returns the $\mathrm{cotg}$ formula :
$$
\cot(\pi z)=\frac1{\pi}\left[\frac1{z}-\sum_{k=1}^{\infty}\frac{2z}{k^2-z^2}\right]
$$
(Euler used this one to find closed forms of $\zeta(2n)$) 
The $\cot\ $ formula is linked to $\Psi$ via the Reflection formula :
$$\Psi(1-x)-\Psi(x)=\pi\cot(\pi x)$$
The $\sin$ formula is linked to $\Gamma$ via Euler's reflection formula :
$$\Gamma(1-x)\cdot\Gamma(x)=\frac{\pi}{\sin(\pi x)}$$ 
A: Set $x=0$ in the Fourier series of $\cos(ax)$:
$$\cos(ax)=\frac{2a\sin(\pi a)}{\pi}\left[\frac1{2a^2}+\sum_{k=1}^\infty\frac{(-1)^k\cos(kx)}{a^2-k^2}\right],\quad a\notin\mathbb{Z}$$
we get
\begin{align}
\frac{\pi}{\sin(\pi a)}&=\frac1a+\sum_{k=1}^\infty\frac{2a(-1)^k}{a^2-k^2}\\
&=\frac1a+\sum_{k=1}^\infty\frac{(-1)^k}{a-k}+\sum_{k=1}^\infty\frac{(-1)^k}{a+k}\\
&=\frac1a+\sum_{k=-\infty}^{-1}\frac{(-1)^k}{a+k}+\sum_{k=1}^\infty\frac{(-1)^k}{a+k}\\
&=\sum_{k=-\infty}^{\infty}\frac{(-1)^k}{a+k}
\end{align}
A: This is a very elegant and quick way to evaluate this sum with complex analysis.  Consider
$$g(z) = \pi \csc (\pi z)f(z)$$
$\csc$ has poles at $2 \pi n$ and $2 \pi n + \pi$ for $n \in \mathbb Z$.  Assuming $f(z)$ has no poles at any  integer, the residue of $g(z)$ at $2\pi n$ is 
$$\operatorname*{Res}_{z = 2 n} g(z) = \lim_{z\to 2 n}(z-2 n)\pi \csc (\pi z)f(z) = \lim_{z\to 2 n}\pi \left(\frac{z-2 n}{\sin (\pi z)}\right)f(z) = f(n)$$
and at $2 \pi n + \pi$:
$$\operatorname*{Res}_{z = 2  n + 1} g(z) = \lim_{z\to 2  n + 1}(z-(2 n + 1))\pi \csc (\pi z)f(z) = \lim_{z\to 2 n + 1}\pi \left(\frac{z-2 n - 1}{\sin (\pi z)}\right)f(z) = -f(n)$$
Let $C_N$ be the square contour with the verticies $\left(N+\frac{1}{2}\right)(1+i)$, $\left(N+\frac{1}{2}\right)(-1+i)$, $\left(N+\frac{1}{2}\right)(-1-i)$ and $\left(N+\frac{1}{2}\right)(1-i)$.
By residue theorem, we have
$$\int_{C_N}g(z)\,dz = \sum_{n=-N}^N (-1)^n f(n) + S$$
where $S$ is the sum of the residues of the poles of $f$. Now, seeing that  the left side vanishes as $N \to \infty$ (see here), we have

$$\sum_{k=-\infty}^\infty (-1)^k f(k)=-\sum \text{Residues of }\pi \csc (\pi z)f(z)$$
Clearly the only singularity of $f(z)=\frac{1}{a+z}$ is at $z_0=-a$.  Thus
$$\operatorname*{Res}_{z=z_0} \,(\pi \csc (\pi z)f(z))=\lim_{z \to z_0} (z-z_0)\pi \csc (\pi z)f(z)=\lim_{z \to -a} \pi \csc (\pi z)\frac{z+a}{z+a}=-\pi \csc (\pi a)$$
Thus
$$\sum_{k=-\infty}^\infty (-1)^k f(k)=-\operatorname*{Res}_{z=z_0}\,(\pi \csc (\pi z)f(z))=-(-\pi \csc (\pi a))=\frac{\pi}{\sin (\pi a)}$$
QED 
