Analysis Problem of the Day 43

Today’s problem appeared as Problem 12 on the UCLA Spring 2022 Analysis Qual:

Problem 2: a) Show that $\sum_{n \in \mathbb{Z}} \frac{1}{(z-n)^2} = \frac{\pi^2}{\sin^2(\pi z)}.$

b) Deduce Basel’s identity $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}.$

Solution: Since we are attempting to describe a meromorphic function as an infinite series, we may appeal to the powerful Mittag-Leffler theorem, which states that any meromorphic function admits a normally convergent expansion of the form $f(z) = h(z)+\sum_a (R_a(z) + p_a(z)),$ where $h$ is holomorphic, $a$ ranges over the poles of $f, R_a$ is the principal part of $f$ at $z=a,$ and $p_a(z)$ is a polynomial that makes the series normally convergent. In particular, under the condition that $f$ is defined at $0,$ only has simple poles, and is uniformly bounded on a set of circles $\{|z|=R_i\}$ with $R_i \to \infty,$ then one may take $h(z)=f(0),$ $R_a(z) = \frac{a_{-1}}{z-a}$ and $p_a(z) = \frac{a_{-1}}{a},$ where $a_{-1}$ is the residue of $f$ at $z=a.$

We now note that the right hand side is the derivative of $f(z)=-\pi \cot(\pi z).$ Considering $g(z)=f(z)+\frac{1}{z},$ which is well-defined at 0 (since $f$ has a simple pole at $z=0$ ) with

$g(0)=\lim_{z \to 0} -\pi \cot(\pi z)+\frac{1}{z}=\lim_{z \to 0} \frac{-\pi z \cos(\pi z)}{\cos(\pi z)}=-1$

by L’Hospital’s, we now note that for $R_k = k+\frac12,$ $g$ is uniformly bounded on $|z|=R_k,$ since it suffices by periodicity to only consider the pole at $0$ and $\cot(\pi z)$ is bounded for $|\text{Im }z| \to \infty.$ Thus, $g$ satisfies the conditions of the strong version of Mittag-Leffler, and since at $z=n,$ the residue of $g$ is

$\lim_{z \to n} \frac{\pi(n-z) \cos(\pi z)}{\sin(\pi z)} = \lim_{z \to n} \frac{-\pi \cos(\pi z)}{\pi \cos(\pi z)} = -1$

by L’Hospital’s rule, it follows that the following normally convergent series expansion holds:

$-\pi \cot \pi z +\frac{1}{z}= -1+\sum_{n \in \mathbb{Z} \setminus \{0\}} -\frac{1}{z-n}-\frac{1}{n}.$

Taking the derivatives on both sides, we obtain the desired expression, and in fact get normal convergence of the sum on $\mathbb{C} \setminus \mathbb{Z}$ (since the derivative of a normally convergent sequence of holomorphic functions also converges normally).

b) Subtracting $\frac{1}{z^2}$ from both sides in (a) and plugging in $z=0$ yields we get

$2 \sum_{n=1}^\infty \frac{1}{n^2} = \lim_{z \to 0} \frac{\pi^2}{\sin^2(\pi z)} -\frac{1}{z^2}.$

Using the Taylor expansion of $\sin \pi z,$ we get

$\sin^2 (\pi z) = \left(\pi z - \frac{\pi^3 z^3}{6}+o(z^3)\right)^2 = \pi^2 z^2 - \frac{\pi^4 z^4}{3} + o(z^4),$

$\frac{\pi^2 \sin^2(\pi z)}{\sin^2(\pi z)} = \frac{1}{z^2-\frac{\pi^2}{3} z^4 +o(z^4)} = \frac{1}{z^2} \cdot \frac{1}{1-\frac{\pi^2}{3} z^2 +o(z^2)} = \frac{1}{z^2} \left(1+\frac{\pi^2}{3} z^2 + o(z^2)\right).$

Thus,

$\lim_{z \to 0} \frac{\pi^2}{\sin^2(\pi z)} -\frac{1}{z^2}= \frac{\pi^2}{3},$

yielding Basel’s identity $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}.$

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