Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

2 Answers 2

up vote 12 down vote accepted

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

share|improve this answer
    
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
    
Thank you very much. –  Pedro Tamaroff Feb 18 '12 at 1:21
    
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

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

share|improve this answer
1  
Pity I know nothing about complex analysis! –  Pedro Tamaroff Jul 20 '12 at 2:01
    
@Argon: a useful way! (+1) –  Chris's sis Aug 11 '12 at 12:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.