Let $a,b$ be two unequal integers. I have to find the sum below. $$ \sum_{k \in \mathbb{Z}}\frac1{(k+a)(k+b)} $$ I should use complex analysis, but I have no clue where to start. I only now that I can assume without loss of generality that $a = 0$ because $x \mapsto x -a$ is an invertible map on the set of integers. I would like to get I hint from you, just a hint.

  • 3
    $\begingroup$ I think you want the sum for $k\in \mathbb{Z}\setminus\{-a,-b\}$. $\endgroup$ – user37238 Jul 10 '14 at 12:21
  • $\begingroup$ Or $k, a, b \in \mathbb{N}$. $\endgroup$ – NovaDenizen Jul 10 '14 at 15:20
  • $\begingroup$ k should not equal -a or -b $\endgroup$ – Koenraad van Duin Jul 10 '14 at 19:53

Can you find $A$ and $B$ for which $$\frac{1}{(k+a)(k+b)}=\frac{A}{k+a}+\frac{B}{k+b}$$


If $b\gt a$, then $$ \begin{align} \sum_{\substack{k\in\mathbb{Z}\\k\not\in\{-a,-b\}}}\frac1{(k+a)(k+b)} &=\frac1{b-a}\lim_{n\to\infty}\sum_{\substack{k=-n\\k\not\in\{-a,-b\}}}^n\left(\frac1{k+a}-\frac1{k+b}\right)\\ &=\frac1{b-a}\lim_{n\to\infty}\left(\sum_{\substack{k=-n+a\\k\not\in\{0,a-b\}}}^{n+a}\frac1k-\sum_{\substack{k=-n+b\\k\not\in\{0,b-a\}}}^{n+b}\frac1k\right)\\ &=\frac1{b-a}\lim_{n\to\infty}\left(\sum_{\substack{k=-n+a\\k\ne0}}^{n+a}\frac1k-\sum_{\substack{k=-n+b\\k\ne0}}^{n+b}\frac1k+\frac2{b-a}\right)\\ &=\frac1{b-a}\lim_{n\to\infty}\left(\sum_{k=-n+a}^{-n+b-1}\frac1k-\sum_{k=n+a-1}^{n+b}\frac1k+\frac2{b-a}\right)\\[9pt] &=\frac2{(b-a)^2} \end{align} $$ Switching $a$ and $b$ gives the same result for $b\lt a$.

If $a=b$, then $$ \begin{align} \sum_{\substack{k\in\mathbb{Z}\\k\ne-a}}\frac1{(k+a)^2} &=\sum_{\substack{k\in\mathbb{Z}\\k\ne0}}\frac1{k^2}\\ &=2\sum_{k=1}^\infty\frac1{k^2}\\[6pt] &=\frac{\pi^2}{3} \end{align} $$


Consider this an addition to Lucian's post.

It can be shown that the quantity

$$\lim_{N\to\infty} \sum_{k = -N}^{k = N} f(k)$$

is equal to the negative of the sum of the residues of $\pi f(z) \cot(\pi z)$ at the poles of $f(z)$. This is the case when $f(z)$ is a rational function of the form $P(z)/Q(z)$ where the degree of $Q$ is greater than or equal to twice the degree of $P$. It also must be that $f$ has no poles on the integers.

In your case, $f(z) = \frac1{(z+a)(z+b)}$ so that the poles are just at $-a$ and $-b$.

This means that

$$\begin{align}\lim_{N\to\infty} \sum_{k = -N}^{k = N} \frac1{(k+a)(k+b)} &= -\left(\lim_{z\to -a} \frac{(z+a)\pi\cot(\pi z)}{(z+a)(z+b)} + \lim_{z\to -b} \frac{(z+b)\pi\cot(\pi z)}{(z+a)(z+b)}\right) \\&= -\lim_{z\to -a} \frac{\pi\cot(\pi z)}{(z+b)} + \lim_{z\to -b} \frac{\pi\cot(\pi z)}{(z+a)}\\&= -\frac{\pi\cot(a\pi )}{a-b} - \frac{\pi\cot(b\pi)}{a-b}\\&= -\pi\left(\frac{\cot(a\pi)-\cot(b\pi)}{a-b}\right)\end{align}$$

I will outline a proof of my first statement but will not show all of the steps.

First, show that $$\operatorname{Res}(\pi f(z)\cot(\pi z),k) = f(k), \;k \in \mathbb{Z}$$

Construct a square contour $\Gamma_N$ with verticies at $(N+1/2)(1+i)$, $(N+1/2)(-1+i)$, $(N+1/2)(-1-i)$ and $(N+1/2)(-1-i)$ in that order. Now show that $|\pi\cot(\pi z)|$ is bounded along this contour.

Using this, prove that

$$\lim_{n\to\infty} \int_{\Gamma_N}\! \pi f(z)\cot(\pi z)\, \mathrm{d}z = 0$$

The end result is a simple application of the residue theorem.

  • $\begingroup$ If $a,b\in\mathbb{Z}$, then $\cot(a\pi)$ and $\cot(b\pi)$ are undefined. $\endgroup$ – robjohn Jul 10 '14 at 21:04

I dropped the spoiler markup, as too many open answers have been posted already.

Hint 2:

Like fellow user Michael sugggested, a decomposition helps.

Using partial fraction decomposition: $$ 1 = (k+b) A + (k+a) B = (A + B) k + b A + a B \iff \\A = -B \wedge (b-a) A = 1 \iff \\A = \frac{1}{b - a} \wedge B = \frac{1}{a - b} $$

for $a \ne b$.

Hint 3:

Note the task asks for $a, b \in \mathbb{N}$ and $a \ne b$.

Now looking at the basic sums: $$ \sum_{k \in \mathbb{Z} \setminus \{-a \}} \frac{1}{k+a} =^{(*)} \sum_{k \in \mathbb{Z} \setminus \{ 0 \}} \frac{1}{k} =^{(**)} 0 \Rightarrow\sum_{k \in \mathbb{Z} \setminus \{ -a, -b \}} \frac{1}{k+a} = 0 - \frac{1}{-b+a} = \frac{1}{b-a} $$


Fellow user Thomas Andrews raised concerns, that the index transformation in equation $(*)$ might not be valid. And I assumed in equation $(**)$ that terms $\frac{1}{k}$ and $-\frac{1}{k}$ cancel out pairwise, but that might be too naive as well.

Proof attempt: I assumed $$ \sum_{k\in \mathbb{Z}\setminus \{-a\}}\frac{1}{k+a} = \lim_{N\to\infty} S_N $$

with $$ S_N = \sum_{k=-N}^{-a-1}\frac{1}{k+a}+\sum_{-a+1}^N\frac{1}{k+a} $$

We have $$ S_N = \sum_{k=-N+a}^{-1}\frac{1}{k}+\sum_{k=1}^{N+a}\frac{1}{k}=\sum_{k=1}^{N-a}-\frac{1}{k}+\sum_{k=1}^{N+a}\frac{1}{k}=\sum_{k=N-a+1}^{N+a}\frac{1}{k}=\sum_{k=1}^{2a}\frac{1}{N-a+k} $$

and for $N>a$ (and $a \in \mathbb{N}$) we get $$ 0 < S_N < \frac{2a}{N-a}, $$

so $S_N$ will vanish for $N \to \infty$.


The above leads to $$ \sum_{k \in \mathbb{Z} \setminus \{ -a, -b \}} \frac{1}{(k+a)(k+b)} =\frac{1}{(b-a)^2} + \frac{1}{(a-b)^2}=\frac{2}{(a-b)^2} $$

for $a \ne b$.

BTW, the case $a = b$ boils down to $$ \sum_{k \in \mathbb{Z}\setminus \{a\}}\frac{1}{(k+a)^2} = \sum_{k \in \mathbb{Z}\setminus \{0\}}\frac{1}{k^2} = 2 \sum_{k = 1}^\infty\frac{1}{k^2} = 2 \, \zeta(2) = \frac{\pi^2}{3} $$ using $\zeta(2)$, see A013661.

  • $\begingroup$ You can't say $\sum_{k\in \mathbb Z\setminus-a} f(k+a)=\sum_{k\in\mathbb Z\setminus 0} f(k)$ unless the sum is absolutely convergent, in general, I believe. $\endgroup$ – Thomas Andrews Jul 10 '14 at 13:16
  • $\begingroup$ I cant read your answer $\endgroup$ – Rene Schipperus Jul 10 '14 at 21:49

Too long for a comment : In general, $~\displaystyle\sum_{k \in \mathbb{Z}}\frac1{(k+a)(k+b)}~=~-\pi\cdot\frac{\cot(a\pi)-\cot(b\pi)}{a-b}$

  • For $a=b\not\in\mathbb Z$, this becomes $\bigg[\dfrac\pi{\sin(\pi a)}\bigg]^2$

  • For $a=b\in\mathbb Z$, upon eliminating the problematic term $k=-a$, this becomes $2~\zeta(2)=\dfrac{\pi^2}3$

  • For $a\in\mathbb Z$ and $b\not\in\mathbb Z$, upon eliminating the problematic term $k=-a$, this becomes $\dfrac1{(a-b)^2}+\dfrac{\pi\cdot\cot(b\pi)}{a-b}$

  • For $a\neq b$ integers, upon eliminating the problematic terms $k=-a$ and $k=-b$, this becomes $\dfrac2{(a-b)^2}$

Unfortunately, the only ways in which I personally am able to prove any of this is by manipulating Euler's infinite product formulas for the sine and cosine functions, as well as his integral expression for generalized harmonic numbers, along with some good, old fashioned l'Hopital.


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