calculus - Possibility to simplify $sumlimits_{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.

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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.

Answer

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 :
$$\Gamma(1-x)\cdot\Gamma(x)=\frac{\pi}{\sin(\pi x)}$$