<s>[INST] 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.
 [/INST] 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 
</s>