"Compute $$\sum_1^{\infty} \frac{1}{{n^2}{(n+1)^2}}$$ using contour integration"

I have used the function $F(z) = \frac {\pi cot\pi z}{z^2(z+1)^2}$

Which has double poles at $z=0$ and $z=-1$

For the pole at $z=0$, if I calculate the residue by taking the limit of $\frac{dF}{dz}$ as $z \to 0 $, I end up with a $cosec(0)$ term, which is $\infty$

Instead I can try to calculate the residue using the Laurent series, and about $z=0$ again I find;

$\frac {\pi cot(\pi z)}{z^2{(z+1)^2}} = \frac{\pi}{z^2} [(\pi z)^{-1} - \frac {1}{3}(\pi z) - \frac {1}{45}(\pi z)^3 ...][1 - 2z +3z^2 -...]$

And I find the residue to be $3-\frac{1}{3} \pi^2$

To compute the residue at $z=-1$, however, I can't compute the expansion of $\pi cot(\pi z)$ about $z=-1$ using normal Taylor expansion methods, because if:

$g(z) = \pi cot(\pi z)$, then $g(-1) = \frac{1}{0}$

and that's where I'm stuck - computing the residue at $z=-1$.

Once I've found the residue, computing the series is simple:

$$\sum_1^{\infty} \frac{1}{{n^2}{(n+1)^2}} = \frac{1}{2} \sum_{Res}$$

Any help would be greatly appreciated!

  • 2
    $\begingroup$ The poles at $0$ and $-1$ are triple poles, since $\cot (\pi z)$ has a simple pole at every integer. $\endgroup$ Jan 22, 2016 at 9:32
  • $\begingroup$ @DanielFischer I was under the impression that because of the way the series is manipulated using a contour integral, the poles created by the $\pi n$ at$ cot(\pi z)$ were already accounted for? $\endgroup$
    – emc3636
    Jan 22, 2016 at 11:15
  • 1
    $\begingroup$ The series gets you half the sum of the residues at the points in $\mathbb{Z}\setminus \{0,-1\}$. The singularities at $0$ and $-1$ must be treated separately. $\endgroup$ Jan 22, 2016 at 11:17

2 Answers 2



$$f(a) = \sum_{n=-\infty}^{\infty} \frac1{(n^2+a^2)[(n+1)^2+a^2]} $$

The sum we want will be $\frac12 \lim_{a \to 0} [f(a) - \frac{2}{a^2 (1+a^2)}]$. (These represent the $n=0$ and $n=-1$ terms in the sum that become singular as $a \to 0$.) We may evaluate this sum using the residue theorem. Recall that, from considering the integral

$$\lim_{N \to \infty} \oint_{C_N} dz \, \cot{\pi z} \, h(z) = 0$$

where $C_N$ is the square with vertices $(\pm (1 \pm i) N/2$. The integral is also equal to $i 2 \pi$ times the sum of the residues, which all must sum to zero. Thus

$$\sum_{n=-\infty}^{\infty} h(n) = -\pi \sum_k \operatorname*{Res}_{z=z_k}[ \cot{\pi z} h(z) ] $$

where the $z_k$ are the non-integer poles of $h$. In this case, we see that

$$f(a) = -\pi \sum_k \operatorname*{Res}_{z=z_k} \frac{\cot{\pi z}}{(z^2+a^2)[(z+1)^2+a^2]}$$

In this case, we have these poles at $\pm i a$ and $-1\pm i a$. Thus,

$$\begin{align}f(a) &= -\pi \left [\frac{-i \coth{\pi a}}{i 2 a (1+i 2 a)} + \frac{i \coth{\pi a}}{-i 2 a (1-i 2 a)} + \frac{\cot{\pi (-1+i a)}}{i 2 a (1-i 2 a)} + \frac{\cot{\pi (-1-i a)}}{-i 2 a (1+i 2 a)} \right ] \\ &= \frac{\pi}{a} \left (\frac{\coth{\pi a}}{1-i 2 a} + \frac{\coth{\pi a}}{1+i 2 a} \right )\\ &= \frac{2 \pi \coth{\pi a}}{a (1+4 a^2)} \end{align}$$

The sum is

$$\begin{align}\sum_{n=1}^{\infty} \frac1{n^2 (n+1)^2} &= \frac12 \lim_{a \to 0} \left [ \frac{2 \pi \coth{\pi a}}{a (1+4 a^2)}- \frac{2}{a^2 (1+a^2)} \right ] \\ &= \frac12 \lim_{a \to 0} \left [\frac{2}{a^2} \left (1+\left (\frac{\pi^2}{3} - 4 \right ) a^2+\cdots \right ) - \frac{2}{a^2} (1-a^2+\cdots) \right ] \end{align} $$

The singular pieces cancel, and we finally have

$$\sum_{n=1}^{\infty} \frac1{n^2 (n+1)^2} = \frac{\pi^2}{3} - 3$$


First of all, $z=0$ and $z=-1$ are triple poles. Secondly, you can either use that $$ \frac{1}{1+z} = \frac{1}{1-(-z)} = \sum_{k=0}^\infty (-z)^k = \sum_{k=0}^\infty (-1)^k\,z^k $$ and $$ \frac{1}{(1+z)^2} = - \frac{d}{dz} \left( \frac{1}{1+z} \right) = - \sum_{k=1}^\infty k(-1)\,z^{k-1} $$ to get enough terms in your Laurent series. Alternatively: the residue at $z=0$ is given by $$ \frac{1}{2!} \lim_{z\to 0} \frac{d^2}{dz^2} \big( z^3 F(z) \big) $$ and similarly for $z=-1$.

The residue at $z=0$ turns out to be $-\frac13\pi^2+3$ and the same at $z=-1$.


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