This is a question for a proof in Theorem 4.2.
Let $(a_n)$ and $(b_n)$ be some numbers from a relation satisfying that $$ \sum_{n=1}^{\infty}\frac{1+|a_n|}{\sqrt{b_n b_{n-1}}}<\infty $$ Put $\epsilon_{n}:=\frac{|z|+|a_n|}{\sqrt{b_n b_{n-1}}}$. In the proof, it says that ".. and by assumption $\epsilon_n<1$ for $n$ sufficiently large depending on $z$". Why is this the case?
We know that the terms in the summation tends to $0$, because it converges, so we can make it as small as we want. So I think it's something about finding the upper bound of $$ \epsilon_n=\frac{1+|a_n|}{\sqrt{b_n b_{n-1}}}+\frac{|z|-1}{\sqrt{b_n b_{n-1}}} $$ for large $n$. If this is true, I do not know what to do for the last terms on the right side
Since the fraction $$ \frac{1+|a_n|}{\sqrt{b_n b_{n-1}}} $$ approaches $0$ as $n$ approaches infinity, and noting that the numerator is always at least $1$, it follows that the denominator approaches infinity.
Hence for a fixed value of $z$, the fraction $$ \frac{|z|-1}{\sqrt{b_n b_{n-1}}} $$ approaches $0$ as $n$ approaches infinity.