Through chapter V of Titchmarsh's book "The theory of the Riemann Zeta function" it is used a "counting technique" that I am not understanding. In particular, Theorem 5.12, p 106, uses something like: if $b\leq 2a$, with $0<a<t$ and $\sum_{a<n\leq b}n^{-s-it}=O(f(t))$ then $\sum_{n\leq t}n^{-s-it}=O(f(t)\log(t))$, as $t\to\infty$. My question is: how this $\log(t)$ appears? That is how is it related to $b\leq 2a$ and $0<a<t$?
2026-04-02 14:18:12.1775139492
Chapter V: Titchmarsh's book "The theory of the Riemann Zeta function"
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You need to apply the result, as also stated in the book, multiple times: Take $b_n=t 2^{-n}$ and $a_n = t 2^{-(n+1)}$ (e.g. $b_0 =t$, $a_0 =t/2$, $b_1 =t/2$, $a_1 =t/4$, and so on) and apply this until $a_n \le 1$. This happens exactly when $n \le \log(t)/\log(2)$ and $\log(t)/\log(2) \le (n+1)$. Thus, we only need to apply the result $ \log(t)$-times.