# How do I show that $f(z) \equiv \sum_{n \in \mathbb{Z}} \frac{1}{(z-n)^2}$ is a meromorphic function?

Let $$f(z)=\sum_{n=-\infty}^\infty \frac{1}{(z-n)^2}.$$
Show $f$ is meromorphic on $\mathbb{C}$ with double poles at each integer.

I think I got it to be meromorphic. I fixed an integer $m$ and considered the series $$\sum_{n=m+1}^\infty \frac{1}{(z-n)^2}\qquad\text{and}\qquad\sum_{n=-\infty}^{-m+1} \frac{1}{(z-n)^2}.$$
I showed these were analytic which implies the whole series is meromorphic. How do I show $f$ has poles at each integer?

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Those two series are not analytic (in $\mathbb C$), really. They are only meromorphic. You can clearly see their poles! – Mariano Suárez-Alvarez Aug 3 '12 at 21:11

To show your series defines a meromorphic function you can do the following:

• Show that the series converges uniformly in every compact subset of $\mathbb C\setminus\mathbb Z$. This implies that the series defines on that set a holomorphic function.

• Now let us show that it is meromorphic with poles at the integers. To do this, it is enough to show that for each $n\in\mathbb Z$ we can write $f(z)=g(z)+\frac{1}{(z-n)^2}$ with $g(z)$ a function which is holomorphic in a neighborhood of $n$. Can you do this?

Let us do the first one.

Let $N\in\mathbb N$ and let $\Omega_N=\{z\in\mathbb C\setminus\mathbb Z:|z|<N\}$. If $z\in\Omega_N$ and $n\in\mathbb Z$ we have $|z-n|\geq |n|-|z|\geq |n|-N$, so that the series $\sum\limits_{\substack{n\in\mathbb Z\\|n|>N}}\frac{1}{(z-n)^2}$ is majorated, term by term, by the series $\sum\limits_{\substack{n\in\mathbb Z\\|n|>N}}\frac{1}{(n-N)^2}$, which converges. Weierstrass' criterion tells us then that the series $\sum\limits_{\substack{n\in\mathbb Z\\|n|>N}}\frac{1}{(z-n)^2}$ in fact converges absolutely and uniformly on $\Omega_N$. It follows that the last series defines an holomorphic function on $\Omega_N$, and then so does $$\sum\limits_{\substack{n\in\mathbb Z\\|n|>N}}\frac{1}{(z-n)^2}+\sum\limits_{\substack{n\in\mathbb Z\\|n|\leq N}}\frac{1}{(z-n)^2}=\sum\limits_{n\in\mathbb Z}\frac{1}{(z-n)^2}.$$ Doing this for all $N\in\mathbb Z$ shows that the series, in fact, defines a function which is holomorphic in $\mathbb C\setminus\mathbb Z$.

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One way to show this is meromorphic is via Morera's theorem, which is as follows.

Suppose that for every simple closed curve $\gamma$ in some domain, such that $\gamma$ does not wind around any point in $\mathbb{C}$ that is not in the domain, $\displaystyle\int_\gamma f(z) \, dz=0$. Then $f$ is holomorphic in that domain.

Now look at $$\int_\gamma f(z)\,dz = \int_\gamma \sum_{n\in\mathbb{Z}} \frac{dz}{(z-n)^2}.$$

Either Fubini's theorem or Tonelli's theorem implies that the last expression above is equal to $$\sum_{n\in\mathbb{Z}} \int_\gamma \frac{dz}{(z-n)^2}.$$ (Fubini's theorem implies the order of integration can be reversed in iterated integrals in which the integral of the absolute value is finite. The sum is an instance of a Lebesgue integral with respect to counting measure. Tonelli's theorem gets the same conclusion in case the function being integrated is everywhere non-negative, regardless of whether its value is finite or not.) The last integral above is $0$ since $\gamma$ doesn't wind around $n$. Hence the conclusion of Morera's theorem holds.

This shows that $f$ is holomorphic in $\mathbb{C}\setminus\mathbb{Z}$, and hence meromorphic if it has a pole at each point in $\mathbb{Z}$.

For any $n_0\in\mathbb{Z}$, we have $$f(z) = \frac{1}{(z-n_0)^2} + \sum_{\begin{smallmatrix}n\in\mathbb{Z}\\ n\ne n_0\end{smallmatrix}} \frac{1}{(z-n)^2}.$$

The second term can be shown to be holomorphic in $\mathbb{Z}\setminus\{n_0\}$ by the method used above. The first term has a pole of order $2$ at $n_0$.

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To compute that integral you need to bound the sum somehow in order to interchange the sum and the integral: you need to do it in absolute value. That amounts to the same bound I did in my answer. – Mariano Suárez-Alvarez Aug 3 '12 at 23:20
Do you mean in order to prove that the series converges? – Michael Hardy Aug 3 '12 at 23:22
In what way would your argument differ for the sum $\sum\frac{1}{(z-n)}$? – Mariano Suárez-Alvarez Aug 3 '12 at 23:24