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)


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.


1 Answer 1


To continue your good idea: analytic functions are continuous and hence bounded on any compact set.

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 $O(1)$.

  • $\begingroup$ I got it, Thank you so much $\endgroup$ May 22, 2023 at 17:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .