# How to find $\sum_{k \in \mathbb{Z}}\frac1{(k+a)(k+b)}$

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.

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

• If $a,b\in\mathbb{Z}$, then $\cot(a\pi)$ and $\cot(b\pi)$ are undefined. – 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: $$\sum_{k \in \mathbb{Z} \setminus \{-a \}} \frac{1}{k+a} =^{(*)} \sum_{k \in \mathbb{Z} \setminus \{ 0 \}} \frac{1}{k} =^{(**)} 0 \Rightarrow\sum_{k \in \mathbb{Z} \setminus \{ -a, -b \}} \frac{1}{k+a} = 0 - \frac{1}{-b+a} = \frac{1}{b-a}$$

Caution:

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 $N \to \infty$.

Solution:

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.

• 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. – Thomas Andrews Jul 10 '14 at 13:16
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 $$\dfrac2{(a-b)^2}$$