$\left| \frac{1}{z}+ \sum_{n=1}^\infty \frac{2z}{z^2-n^2}\right| \le C + C \sum_{n=1}^\infty \frac{y}{y^2+n^2}$ for $y>1$ and $|x|\le 1/2$.

Question

Suppose $|x| \le \frac{1}{2}$, and $y>1$ for $z=x+iy$.

Then we have $$\frac{1}{z} + \sum_{n=1}^\infty \frac{2z}{z^2-n^2} = \frac{1}{x+iy} + \sum_{n=1}^\infty \frac{2(x+iy)}{x^2-y^2-n^2+2ixy};$$ From this I would like to get the bound

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

It is easy to deal with the first term since $|x+iy| \ge |y| > 1$.

However, I am having difficulty with the denominator of the second term. We have $|x^2-y^2-n^2+2ixy| \ge |x^2-y^2-n^2|$. How could I get rid of $x^2$ to get a lower bound of a multiple of $y^2+n^2$?

Answer 1

The idea is that under these restrictions, $z^2 \approx -y^2$ lies near the negative real axis, so that $|z^2 -n^2| \approx y^2 + n^2$.

Now the precise estimate: You have $x^2 \le \frac 14 \le \frac{y^2}{4}$, so that $$ |z^2-n^2| \ge |\operatorname{Re}(z^2-n^2)| = y^2+n^2-x^2 \ge \frac 34 (y^2+n^2) $$ and $$ |2z| = 2 \sqrt{x^2+y^2} \le 2y \sqrt{\frac 5 4} $$ It follows that $$ \frac{2|z|}{|z^2-n^2|} \le \frac{4\sqrt{5}}{3} \frac{y}{y^2-n^2} $$ and therefore $$ \Bigg| \frac{1}{z}+ \sum_{n=1}^\infty \frac{2z}{z^2-n^2}\Bigg| \le C + C \sum_{n=1}^\infty \frac{y}{y^2+n^2} $$ with $C = \frac{4\sqrt{5}}{3}$.