# conclusion of simple pole of $-\frac{\zeta'(s)}{\zeta(s)}$

This is from Davenport's MNT(Page 85). I am unsure how they concluded this.

Since $$-\frac{\zeta'(s)}{\zeta(s)}$$ has a simple pole with residue $$1$$, we have for $$1 < \sigma \leq 2$$

$$-\frac{\zeta'(\sigma)}{\zeta(\sigma)} < \frac{1}{\sigma-1}+A$$ where A denotes a positive absolute constant(not necessarily the same at each occurrence)

Questions

1-I am unsure how they got the inequality.

2-I am not sure what they meant by the comment of the constant $$A$$.

My thought process is that, since $$-\frac{\zeta'(\sigma)}{\zeta(\sigma)}$$ has a simple pole with residue $$1$$, then $$-\frac{\zeta'(s)}{\zeta(s)}-\frac{1}{s-1}$$ is an analytic function but I am unsure what to do next. I would appreciate if someone can guide further.

As for your second question, Davenport is saying that he will use $$A$$ to denote some constant (whose exact value is unimportant to the arguments) multiple times in the book, but he doesn't want you to think that means that $$A$$ takes the same value each time. Some might replace these occurrences of $$A$$ by $$O(1)$$.