Infinite product $\prod_k (1-z/a_k)(1-z/b_k)^{-1}$ converges uniformly on compacts of$\{\mathbb Im z>0\}$ and is Herglotz Hi everyone I find the following exercise of infinite product functions.I'd stuck in this exercise.

Let $\{a_k\}$ and $\{b_k\}$ sequences of positive real numbers such that $b_k<a_k<b_{k+1}$ for all $k$ in $\mathbb N$. Prove that the infinite product $\prod_k (1-z/a_k)(1-z/b_k)^{-1}$ converges uniformly on compacts of $\mathbb C_+:=\{z:\mathbb{Im}z>0\}$ and the function $f(z)=\prod_k (1-z/a_k)(1-z/b_k)^{-1}$ is Herglotz.

Let $K$ a compact set in $\mathbb C_+$ in order to show that the infinite product converges uniformly on $K$ it will suffices to show that $\sum_k  [(1-z/a_k)(1-z/b_k)^{-1}-1]$ converges absolutely and uniformly on $K$. 
\begin{align*}[(1-z/a_k)(1-z/b_k)^{-1}-1]=\frac{z}{b_k-z}\left(1-\frac{b_k}{a_k}\right) \tag{1}\end{align*} 
Since $d=d(K,\mathbb C_+^c)>0$ this is a lower bound for $|b_k-z|$ and so for the denominator of (1) in norm. The $z$ in the numerator is bounded by some $M>0$ on $K$, since $K$ is compact and the identity function is continuous. Then 
\begin{align*}|[(1-z/a_k)(1-z/b_k)^{-1}-1]|\le \frac{M}{d}\left(1-\frac{b_k}{a_k}\right) \tag{2}\end{align*}
But from here I'm stuck. I'd appreciate if someone can help me I didn't know how to use the form in which the numbers $\{a_k\}$ and $\{b_k\}$ are defined. 
 A: You want to show that $\sum_k c_k(z)$ converges absolutely and uniformly on
any compact set $K$ in the upper half-plane, where
$$
 c_k(z) = \frac{1-z/a_k}{1-z/b_k} - 1 = \frac{z}{z - b_k}\frac{a_k-b_k}{a_k} \, .
$$
Consider first the case that the sequences $(a_k)$, $(b_k)$ are bounded,
and therefore have a common limit $A$. Since
$$
 \left| \frac{z}{z - b_k} \right | \le \frac{|z|}{\text{Im}(z)}
$$
is bounded on $K$ by some constant $M$, we have
$$
 |c(z)| \le M \frac{a_k-b_k}{a_k} \le M \frac{b_{k+1}-b_k}{b_1}
$$
and therefore
$$
 \sum_{k=1}^N |c(z)| \le \frac{M}{b_1} \sum_{k=1}^N (b_{k+1}-b_k)
 = \frac{M}{b_1} (b_{N+1} - b_1) \le \frac{M}{b_1} (A - b_1)
$$
which implies the absolute and uniform convergence 
of $\sum_k c_k(z)$ on $K$.
Now consider the case that the sequences $(a_k)$, $(b_k)$ are
unbounded and therefore converge to $+\infty$.
Let $M$ be an upper bound for $|z|$ on $K$. Then for all sufficiently
large $k \ge k_0$, $b_k > 2M \ge 2|z|$ and therefore
$$
\left| \frac{z}{z - b_k} \right | \le \frac{|z|}{b_k - |z|}
 \le \frac{M}{b_k/2} \, .
$$
Then
$$
\sum_{k=k_0}^N |c(z)| \le 2M \sum_{k=k_0}^N  \frac{a_k-b_k}{a_kb_n}
 = 2M \sum_{k=k_0}^N \bigl( \frac{1}{b_k} - \frac{1}{a_k} \bigr)
 \le 2M \sum_{k=k_0}^N \bigl( \frac{1}{b_k} - \frac{1}{b_{k+1}} \bigr) \\
 = 2M \bigl( \frac{1}{b_{k_0}} - \frac{1}{b_{N+1}} \bigr)
 \le 2M \frac{1}{b_{k_0}}
$$
which again implies the absolute and uniform convergence 
of $\sum_k c_k(z)$ on $K$.

To prove that the product maps the upper half-plane into itself (i.e.
is a "Herglotz function"), observe that the argument of
$$
\frac{1-z/a_k}{1-z/b_k} = \frac{b_k}{a_k} \frac{a_k-z}{b_k-z}
$$
for $z = x + iy$ is the angle under which the segment 
$[b_k, a_k]$ is seen from $z$:
$$
 \arctan \frac{a_k-x}{y} -  \arctan \frac{b_k-x}{y} < 
\arctan \frac{b_{k+1}-x}{y} -  \arctan \frac{b_k-x}{y} \, .
$$
The sum of the arguments of all factors in the product can therefore
be estimated by a "telescoping sum" which is less than
$\frac {\pi}{2} -\frac {-\pi}{2} = \pi$.
