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I've encountered an inequality pertaining to the following expression:

$\frac{1}{z} + \sum_{n=1}^{\infty}\frac{2z}{z^2 - n^2}$, where $z$ is a complex number.

After writing $z$ as $x + iy$ we have the inequality when $y \gt 1$ and $|x| \le \frac{1}{2} $:

$|\frac{1}{z} + \sum_{n=1}^{\infty}\frac{2z}{z^2 - n^2}| \le C + C\sum_{n=1}^{\infty}\frac{y}{y^2+n^2}$

The "proof" of the inequality is given as follows:
$\frac{1}{z} + \sum_{n=1}^{\infty}\frac{2z}{z^2 - n^2} = \frac{1}{x+iy} +\sum_{n=1}^{\infty}\frac{2(x+iy)}{x^2 - y^2 - n^2 + 2ixy}$

But I fail to see how the inequality follows.

Mark
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2 Answers2

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As $$1\leq y\leq|x+iy|\leq{y\over2}+y={3\over2} y$$ we have $${1\over |x+iy|}\leq 1$$ and $$|n^2+y^2-x^2-2ixy|\geq |n^2+y^2-x^2|\geq\Bigl(1-{1\over8}\Bigr)(n^2+y^2)\qquad(n\geq1)\ .$$ It follows that $$\left|{1\over z}+\sum_{n=1}^\infty{2z\over z^2-n^2}\right|\leq 1+\sum_{n=1}^\infty {3y \over{7\over8}(n^2+y^2)}=1+{24\over7}\sum_{n=1}^\infty {y \over n^2+y^2}\ .$$ Therefore the stated inequality is true with $C=4$, say.

Christian Blatter
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The following is not the simplest way to solve this question but it may provide additional insight.

The key observation is that both sums have a closed-form representation. To see this, we first need to show that for $w = \sigma + it$ $$ |\pi \cot(\pi w)| \le \pi \coth(\pi t).$$ This is because $$ |\pi \cot(\pi w)| = \pi \left| \frac{e^{i\pi\sigma-\pi t}+e^{-i\pi\sigma+\pi t}} {e^{i\pi\sigma-\pi t}-e^{-i\pi\sigma+\pi t}} \right|\le \pi \frac{e^{\pi t}+e^{-\pi t}}{e^{\pi t}-e^{-\pi t}} = \pi \coth(\pi t)$$ for $t>0.$

Similarly, when $t<0$, $$ |\pi \cot(\pi w)| = \pi \left| \frac{e^{i\pi\sigma-\pi t}+e^{-i\pi\sigma+\pi t}} {e^{i\pi\sigma-\pi t}-e^{-i\pi\sigma+\pi t}} \right|\le \pi \frac{e^{\pi t}+e^{-\pi t}}{e^{-\pi t}-e^{\pi t}} = -\pi \coth(\pi t)$$

Now we use a classic technique to evaluate the two sums, introducing the functions $$ f_1(w) = \pi \cot(\pi w) \frac{2z}{z^2-w^2} \quad \text{and} \quad f_2(w) = \pi \cot(\pi w) \frac{z}{z^2+w^2},$$ with the conditions that $z$ not be an integer for $f_1(z)$ and not $i$ times an integer for $f_2(z).$ We choose $\pi \cot(\pi w)$ because it has poles at the integers with residue $1$.

The key operation of this technique is to compute the integrals of $f_1(z)$ and $f_2(z)$ along a circle of radius $R$ in the complex plane, where $R$ goes to infinity. Now by the first inequality we certainly have $|\pi\cot(\pi w)| < 2\pi$ for $R$ large enough. The two terms $\frac{2z}{z^2-w^2}$ and $\frac{z}{z^2+w^2}$ are both $\theta(1/R^2)$ so that the integrals are $\theta(1/R)$ and vanish in the limit. (Here we have used the two bounds on $|\pi \cot(\pi z)|$ that we saw earlier.) This means that the sum of the residues at the poles add up to zero.

Now let $$ S_1 = \sum_{n=1}^\infty \frac{2z}{z^2-n^2} \quad \text{and} \quad S_2 = \sum_{n=1}^\infty \frac{y}{y^2+n^2}.$$ By the Cauchy Residue Theorem, $$ \frac{2}{z} - 2\pi\cot(\pi z) + 2S_1 = 0 \quad \text{and} \quad \frac{1}{y} - \pi\coth(\pi y) + 2S_2 = 0.$$ Solving these, we obtain $$ S_1 = -\frac{1}{z} + \pi\cot(\pi z) \quad \text{and} \quad S_2 = - \frac{1}{2} \frac{1}{y} + \frac{1}{2} \pi\coth(\pi y).$$

Starting with the left side of the original inequality we finally have $$\left| \frac{1}{z} + S_1\right| < \pi \coth(\pi y) = 2 S_2 + \frac{1}{y} .$$ It follows that $C=2$ is an admissible choice.

Marko Riedel
  • 61,317