# 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.

• It seems that using contour integration with the residue theorem should work here. Have you tried that? – savick01 Feb 17 '12 at 22:22
• @savick01 I know nothing about Complex Analisys. I'd prefer the use of the Digamma Function, which I'm familiar with. – Pedro Tamaroff Feb 17 '12 at 22:23
• I suppose you are interpreting that infinite sum as $\frac{1}{a} + \sum_{k=1}^{\infty} (-1)^k(\frac{1}{a+k} + \frac{1}{a-k})$. – Aryabhata Feb 17 '12 at 22:31
• @Aryabhata What do you mean? – Pedro Tamaroff Feb 17 '12 at 22:32
• @Peter: The order in which you combine them would matter I would think. – Aryabhata Feb 17 '12 at 22:33

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

• Very nice. Guess I'll have to study Fourier series... What is the convergence domain of the given series? Do you have a link to Smirnov's book? – Pedro Tamaroff Feb 17 '12 at 23:23
• check my solution with $\psi(x)$, that's what I wanted! – Pedro Tamaroff Feb 18 '12 at 5:29
• @Peter: Nice! So we just had to observe (:-)) that $\tan(\frac a2)+\cot(\frac a2)= \frac 2{\sin(a)}\$ and use the Reflection formula twice! – Raymond Manzoni Feb 18 '12 at 9:55
• Can you tell me what is the domain of convergence of the series you found by Fourier analysis? – Pedro Tamaroff Feb 20 '12 at 21:34
• @Peter: I think that the formula is valid for any $z\$ not in $\mathbb{Z}\$ but please don't ask me for a rigorous proof! To get a reasonable approximation of the left term taking the sum for $k\$ from $1$ to a little more than $|z|\$ should be enough. For $z\in \mathbb{Z}\$ only one term will remain! – Raymond Manzoni Feb 20 '12 at 22:53

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

• Pity I know nothing about complex analysis! – Pedro Tamaroff Jul 20 '12 at 2:01
• @Argon: a useful way! (+1) – user 1591719 Aug 11 '12 at 12:09

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}\tag1\\ &=\frac1x-\sum_{k=1}^\infty\frac{2x}{k^2-x^2}\tag2 \end{align} $$(2)$$ 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)\tag3\\ &=\lim_{n\to\infty}\left(\sum_{k=-n}^n\frac1{k+x}-\sum_{k=-n+1}^{n+1}\frac1{k+x}\right)\tag4\\ &=\lim_{n\to\infty}\left(\frac1{-n+x}-\frac1{n+1+x}\right)\tag5\\[9pt] &=0\tag6 \end{align} Explanation:
$$(3)$$: definition
$$(4)$$: substitute $$k\mapsto k-1$$ in the right sum
$$(5)$$: the sums telescope
$$(6)$$: 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}\tag7\\ &=\lim_{n\to\infty}\frac1{n+1/2}\tag8\\[9pt] &=0\tag9 \end{align} Explanation:
$$(7)$$: definition
$$(8)$$: for $$1\le j\le n$$, the term with $$k=j-1$$ cancels the term with $$k=-j$$
$$\phantom{\text{(8):}}$$ which leaves the term with $$k=n$$
$$(9)$$: 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{10}\\[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{11}\\ &=\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{12}\\ &=\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{13}\\ &=\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{14}\\ &=-f'(x)-4\int_0^1\frac{\log\left(\frac{1+x}{1-x}\right)}{x}\,\mathrm{d}x\tag{15}\\[12pt] &=-f'(x)-\pi^2\tag{16} \end{align} Explanation:
$$(10)$$: product of limits equals the limit of the products
$$(11)$$: the left sum contains the terms with $$j=k$$
$$\phantom{\text{(11):}}$$ apply partial fractions to the right sum
$$(12)$$: take advantage of the symmetry in $$j$$ and $$k$$
$$(13)$$: for $$k=0$$, the sum in $$j$$ is $$0$$
$$\phantom{\text{(14):}}$$ for $$k\lt0$$, if we substitute $$(j,k)\mapsto(-j,-k)$$,
$$\phantom{\text{(14):}}$$ we get the sum with $$k\gt0$$ and $$k+x\mapsto k-x$$
$$(14)$$: the sum in $$j$$ telescopes to $$H_{n-k}-H_{n+k}$$
$$(15)$$: the left sum is $$f'(x)$$
$$\phantom{\text{(15):}}$$ the right sum is a Riemann Sum
$$(16)$$: $$4\int_0^1\sum\limits_{k=0}^\infty\frac{2\,x^{2k}}{2k+1}\,\mathrm{d}x=4\sum\limits_{k=0}^\infty\frac2{(2k+1)^2}=\pi^2$$

Conclude that $$\boldsymbol{f(x)=\pi\cot(\pi x)}$$

We can separate $$(16)$$ and integrate: \begin{align} \int\frac{\pi\,\mathrm{d}f}{f^2+\pi^2}&=-\int\pi\,\mathrm{d}x\tag{17}\\ \tan^{-1}(f/\pi)&=C-\pi x\tag{18}\\[9pt] f&=\pi\tan(C-\pi x)\tag{19} \end{align} $$(9)$$ allows us to compute $$C=\pi/2$$, giving $$f(x)=\pi\cot(\pi x)\tag{20}$$ for $$x\in(0,1)$$. $$(6)$$ removes this restriction on $$x$$, validating $$(20)$$ for all non-integer $$x$$. That is, $$\sum_{k=-\infty}^\infty\frac1{k+x}=\pi\cot(\pi x)\tag{21}$$ 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{22}\\ &=\sum_{k=-\infty}^\infty\frac1{k+x/2}-\sum_{k=-\infty}^\infty\frac1{k+x}\tag{23}\\[6pt] &=\pi\cot(\pi x/2)-\pi\cot(\pi x)\tag{24}\\[15pt] &=\pi\csc(\pi x)\tag{25} \end{align} Explanation:
$$(22)$$: the alternating sum is twice the sum of the even terms
$$\phantom{\text{(22):}}$$ minus the sum of all the terms
$$(23)$$: adjust the terms of the left sum to apply $$(21)$$
$$(24)$$: apply $$(21)$$
$$(25)$$: $$\frac{1+\cos(\pi x)}{\sin(\pi x)}-\frac{\cos(\pi x)}{\sin(\pi x)}=\frac1{\sin(\pi x)}$$