Convergence of the infinite product $\prod_{n = 1}^{\infty} \frac{z - \alpha_n}{z - \beta_n}$ I've been trying to solve this homework problem for a while but I can't seem to get any significant ideas about how to approach it, so I would really appreciate any hints that could help me solve it. 
The problem is exercise 8.14 from Steven Krantz and Robert Greene's book Function Theory of One Complex Variable. It goes as follows:

Suppose that 
  $$\sum |\alpha_n - \beta_n| < \infty$$
  Then determine the largest open set of $z$ for which
  $$\prod_{n = 1}^{\infty} \frac{z - \alpha_n}{z - \beta_n}$$
  converges normally.

What I've tried so far is writing the factors as 
$$\frac{z - \alpha_n}{z - \beta_n} = 1 + \frac{z - \alpha_n}{z - \beta_n} - 1 = 1 + \frac{\beta_n - \alpha_n}{z - \beta_n}$$
so as to put the infinite product in the form $\displaystyle{\prod (1 + f_n(z))}$ to try to apply the basic convergence criteria I have available  which says that this product would converge normally if the series
$$\sum |f_n(z)|$$ 
converges normally. Now I'm kind of stuck here because I think that maybe I would have to bound this sum with the sum $\sum |\alpha_n - \beta_n|$ but I'm not sure about how to proceed (assuming that this is the right way to follow).
So I would really appreciate some hints that would get me in the right track to solve this problem. Thank you very much.
 A: Hint:
Suppose that $z\in\mathbb{C}\setminus\overline{\{b_n\}}$, then there is an $\epsilon>0$ so that $|z-b_n|\ge\epsilon$ for all $n$. Then,
$$
\sum_n\left|\frac{b_n-a_n}{z-b_n}\right|\le\frac{1}{\epsilon}\sum_n|b_n-a_n|<\infty\tag{1}
$$
Inequality $(1)$ implies that
$$
\prod_n\frac{z-a_n}{z-b_n}=\prod_n\left(1+\frac{b_n-a_n}{z-b_n}\right)\tag{2}
$$
converges.
A: I think you are very close.
The convergence criteria you are using is correct - consider this:
We observe that 
$$|\beta_n - \alpha_n| = |\alpha_n - \beta_n| $$
so 
$$\left|\frac{\beta_n - \alpha_n}{z - \beta_n}\right| < |\alpha_n - \beta_n| $$
As you know that the absolute series converges as per the question's definition so $$\sum \left|\frac{\beta_n - \alpha_n}{z - \beta_n}\right| \leqslant \sum |\alpha_n - \beta_n| < \infty$$ so it converges by limit comparison theorem.
As per the largest open set, the condition needed for convergence is $\left|\frac{\beta_n - \alpha_n}{z - \beta_n} \right| \leqslant |\alpha_n - \beta_n| $ Solve for $z$.
