Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Did you missed the $\frac{1}{z}$ in the last line? – Pragabhava Oct 12 '12 at 0:28
Yeah sorry I'll add it! – Mark Oct 12 '12 at 0:43
Hint: Use the triangle inequality, and the fact that $|y| \le |z|$ to obtain the first $C$. Also, is the term $\frac{y}{y^2 + n^2}$ correct? Shouldn't be $\frac{y}{y^2 - n^2}$? – Pragabhava Oct 12 '12 at 0:47
The term is correct, it is not a minus sign in the denominator. I'll think about the hint though. – Mark Oct 12 '12 at 13:23
Are the two C's the same constant? – zyx Oct 18 '12 at 3:27
up vote 2 down vote accepted

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.