Possibility to simplify $\sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}} = \frac{\pi }{{\sin \pi a}}} $

Question

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 1

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

Answer 2

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 $2n$ 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 $2n+1$:

$$\operatorname*{Res}_{z = 2 n + 1} g(z) = \lim_{z\to 2 n + 1}(z-(2 n + 1))\pi \csc (\pi z)f(z) = \lim_{z\to 2 n + 1}\pi \left(\frac{z-2 n - 1}{\sin (\pi z)}\right)f(z) = -f(n)$$

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

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$$\sum_{k=-\infty}^\infty (-1)^k f(k)=-\sum \text{Residues of }\pi \csc (\pi z)f(z)$$

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

Answer 3

A related identity is proven in this answer using residue theory. Here is a real approach to that identity.

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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 $x$.

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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 $\frac2{k-x}+\frac2{k+x}\mapsto\frac{4\,\mathrm{d}t}t$
$\text{(4g):}$ $4\int_0^1\sum\limits_{k=0}^\infty\frac{2\,x^{2k}}{2k+1}\,\mathrm{d}x=4\sum\limits_{k=0}^\infty\frac2{(2k+1)^2}=\pi^2$

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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.

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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 $(7)$
$\text{(8d):}$ $\frac{1+\cos(\pi x)}{\sin(\pi x)}-\frac{\cos(\pi x)}{\sin(\pi x)}=\frac1{\sin(\pi x)}$

Answer 4

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}