Rudin's Real and Complex Analysis Chapter 15 Q.11

Question

I'm having a go at Rudin's Real and Complex Analysis Q15.11 (this is just for fun).

Under what conditions on a sequence of real numbers $y_n$ does there exist a bounded holomorphic function in the open right half plane which is not identically zero but which has a zero at each point $1 + i y_n$? In particular, can this happen if

1. $y_n = \log(n)$;
2. $y_n = \sqrt{n}$;
3. $y_n = n$;
4. $y_n = n^2$.

Using the following conformal mapping $z \mapsto \frac{z-1}{z+1}$, we can reduce to solving the same question on the open unit disk, for which the Blaschke condition gives a criteria.

For example, here's my solution for part 4. (for which the answer appears to be yes) :

$$ \begin{array}{rcl} \displaystyle \sum\limits_{n=1}^\infty \left(1 - \left|\frac{1 + in^2 - 1}{1 + in^2 + 1} \right| \right) & = & \displaystyle \sum\limits_{n=1}^\infty \left(1 - \left|\frac{ in^2}{2 + in^2} \right| \right) \\ & = & \displaystyle \sum\limits_{n=1}^\infty \frac{\big|2 + in^2\big| -n^2}{\big|2 + in^2\big|} \\ & \leq & \displaystyle \sum\limits_{n=1}^\infty \frac{2 + n^2 -n^2}{\big|2 + in^2\big|} \\ & = & \displaystyle 2 \sum\limits_{n=1}^\infty \frac{1}{\big|2 + in^2\big|} \\ & \leq & \displaystyle 2 \sum\limits_{n=1}^\infty \frac{1}{n^2} \\ & < & \infty \end{array} $$ (By the way, if someone could explain to me in a comment how to align displayed equations here on MSE I'd be grateful)

I did not try to solve 1. and 2. yet if the answer to 3. is negative than I can use this knowledge to deduce that 1. and 2. have negative answer by the comparison test.

Here's what I got so far for 3. : $$\begin{array}{rcl} \displaystyle \sum\limits_{n=1}^\infty \left(1 - \left|\frac{1 + in - 1}{1 + in + 1} \right| \right) & = & \displaystyle \sum\limits_{n=1}^\infty \left(1 - \left|\frac{ in}{2 + in} \right| \right) \\ & = & \displaystyle \sum\limits_{n=1}^\infty \left(1- \frac{n}{\sqrt{4 + n^2}}\right). \end{array}$$

I've trying using the ratio test but things get quite nasty. Is there a shortcut I'm missing ?

Answer 1

$$ 1- \frac{n}{\sqrt {4+n^2}} = \frac{\sqrt {4+n^2} - n}{\sqrt {4+n^2}}.$$

Multiply top and bottom by $\sqrt {4+n^2} + n$ (ye olde conjugate trick) to get

$$\frac{4}{\sqrt {4+n^2}(\sqrt {4+n^2} + n)} \le \frac{4}{n^2}.$$

Hence the series in question converges, and thus there exists such a bounded holomorphic function in the right half plane.