11.2 Evaluating Series

Home | Assessment | Notes | Index | Worksheets | Blackboard

We can also use Cauchy's residue theorem to sum series. The link between the two is the function $g(z)=\mathsf{cot}(\pi z)$ which we define on $\mathbb{C}\setminus \mathbb{Z}$ by $g(z)=\mathsf{cos}(\pi z)/\mathsf{sin}(\pi z)$. Since $\mathsf{sin}(\pi z)$ has a simple zeroe at every integer and $\mathsf{cos}(\pi z)$ is non-zero at every integer our function $g$ has a simple pole at every integer. It also has no other poles. We can calculate that $$\mathsf{Res}(g,n)={\displaystyle \frac{\mathsf{cos}(\pi n)}{\pi \mathsf{cos}(n\pi )}}={\displaystyle \frac{1}{\pi }}$$ using our lemma for residues of simple poles of ratios.

Suppose now that we are given a series $$\sum _{n=1}^{\mathrm{\infty }}{a}_n$$ that we wish to sum. Suppose that the terms $a_n$ are all non-zero and that we can find a meromorphic function $f:\mathbb{C}\to \mathbb{C}$ with $f(n)=a_n$ for all $n\in \mathbb{N}$. We can calculate that $$\mathsf{Res}(fg,n)={\displaystyle \frac{f(n)\mathsf{cos}(\pi n)}{\pi \mathsf{cos}(\pi n)}}={\displaystyle \frac{f(n)}{\pi }}={\displaystyle \frac{a_n}{\pi }}$$ so that $$2i\sum _{n=1}^N{a}_n=2\pi i\sum _{n=1}^N\mathsf{Res}(fg,n)$$ will contribute to the conclusion of Cauchy's residue theorem whenever $\mathrm{\Gamma }$ is a contour with a winding number of $1$ around each of $1,\dots ,N$. Depending on what other poles $f$ has - and we will see some examples later - there may be other contributions as well.

The contour ${\mathrm{\Gamma }}_N$ we will use will be a square centered at the origin with sides parallel to the axes and of length $2N+1$. We want the contour integral to vanish as $N\to \mathrm{\infty }$ leaving only the residues of $g$ that give the series, and the additional residues of $f$ that will tell us the value of the series.

Figure 1: The square contour ${\mathrm{\Gamma }}_N$ contains the poles of $g$ at $-N,\dots ,N$ and possibly poles of $f$ as well.

For the contour integral to vanish as $N\to \mathrm{\infty }$ we will need to estimate the integral $$\underset{{\mathrm{\Gamma }}_N}{\int }f(z)\mathsf{cot}(\pi z)\mathsf{d}z$$ as $N\to \mathrm{\infty }$. As we will see in the example, the following lemma tells us that strong enough decay in $|f(z)|$ as $|z|\to \mathrm{\infty }$ will suffice.

Lemma

There is $M>0$ such that $|\mathsf{cot}(\pi z)|\le M$ whenever $N\in \mathbb{N}$ and $z$ lies on ${\mathrm{\Gamma }}_N$.

Example

Calculate $$\sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{1}{n^2}}$$ using the lemma and the strategy above.

Solution:

Take $f(z)=1/z^2$ on $\mathbb{C}\setminus \{0\}$ so that $f(\pm n)=1/n^2$ for all $n\in \mathbb{N}$. For every non-zero $n\in \mathbb{Z}$ we have $$\mathsf{Res}(fg,n)={\displaystyle \frac{1}{n^2\pi }}$$ as above. At $-n$ we have the same residue. Thus $$\underset{{\mathrm{\Gamma }}_N}{\int }{\displaystyle \frac{1}{z^2}}{\displaystyle \frac{\mathsf{cos}(\pi z)}{\mathsf{sin}(\pi z)}}\mathsf{d}z=4\pi i\sum _{n=1}^N{\displaystyle \frac{1}{n^2\pi }}+2\pi i\mathsf{Res}(fg,0)$$ and it remains to both dispense with the contour integral and calculate the reimaining residue.

For the contour integral, note that $$|\underset{{\mathrm{\Gamma }}_N}{\int }{\displaystyle \frac{1}{z^2}}\mathsf{cot}(\pi z)\mathsf{d}z|\le \ell ({\mathrm{\Gamma }}_N){\displaystyle \frac{M}{N^2}}=M{\displaystyle \frac{4(2N+1)}{N^2}}$$ from our lemma on the cotangent function and the fact that $|z|\ge N$ for all $z$ on ${\mathrm{\Gamma }}_N$. This quantity goes to zero as $N\to \mathrm{\infty }$ so $$0=4\pi i\sum _{n\in \mathbb{N}}{\displaystyle \frac{1}{n^2\pi }}+2\pi i\mathsf{Res}(fg,0)$$ and it remains to calculate the residue at the origin.

The pole at the origin has order three. The residue is therefore $$\underset{z\to 0}{lim}{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{z^3\mathsf{cos}(\pi z)}{z^2\mathsf{sin}(\pi z)}}\right)}^{(2)}$$ and $${\displaystyle \frac{z}{\mathsf{sin}z}}=1+{\displaystyle \frac{z^2}{6}}+\cdots$$ so $${\displaystyle \frac{z}{\mathsf{sin}(\pi z)}}={\displaystyle \frac{1}{\pi }}{\displaystyle \frac{\pi z}{\mathsf{sin}(\pi z)}}={\displaystyle \frac{1}{\pi }}(1+{\displaystyle \frac{{\pi }^2{z}^2}{6}}+\cdots )$$ and $$\begin{array}{rl}& {\displaystyle \frac{z}{\mathsf{sin}(\pi z)}}\mathsf{cos}(\pi z)\\ =& {\displaystyle \frac{1}{\pi }}(1+{\displaystyle \frac{{\pi }^2{z}^2}{6}}+\cdots )(1-{\displaystyle \frac{{\pi }^2{z}^2}{2!}}+\cdots )\\ =& ={\displaystyle \frac{1}{\pi }}-{\displaystyle \frac{\pi }{3}}{z}^2+\cdots \end{array}$$ so that $\mathsf{Res}(fg,0)=-\pi /3$. Plugging into what we had earlier gives $$\sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{1}{n^2}}={\displaystyle \frac{{\pi }^2}{6}}$$ concluding the example.