Doubt form the Book Theory of Riemann Zeta Function by Titchmarsh
I've problem on Theorem $5.8$ in the equation $5.8.2$. When $k=l$ the from Lemma $5.7$ we get, $$ \sum_{n=N+1}^bn^{-it}=O\left(N^{1-1/L}t^{1/\{(l+1)L\}}\log^{1/L}t\right).$$ But from here how the summation comes $\displaystyle \sum_{n\le N}n^{-it}$ in the equation $5.8.2$ ?
What's the meaning of "applying the result $O(\log t)$, times we obtain".
Can anyone clarify this please ?
Best Answer
The interval $[1,N]$ is split dyadically into $[\frac {N}{2}, N]$, $[\frac {N}{4},\frac {N}{2}]$ etc so in up to $[log_2(N)]$+$1$ intervals (it doesn't matter if the ends are not integral since we can apply Lemma 5.7 as the 2 integral ends of the respective interval are of the type $a \leq b \leq 2a$ and we stop when the lower end is at or below 1, so after at most $[log_2(N)]$+$1$ steps (it's actually that many unless $N$ is a power of 2 when we can use only $[log_2(N)]$ steps); obviously $[log_2(N)]$+$1$ is $O(logN)$ (which is $O(log(t))$ in our situation).
So as noted in the comment we apply Lemma 5.7 for each dyadic interval as above and we notice that we have an absolute $O\left(N^{1-1/L}t^{1/\{(l+1)L\}}\log^{1/L}t\right)$, and then a sum of the type $1$+$2$^$\frac {1-L}{L}$+...$2$^$\frac {k(1-L)}{L}$ from applying the estimates with $a=N$, $a=\frac{N}{2}$, etc and that sum is uniformly bounded by an absolute constant ($\sqrt2$/$(\sqrt2 - 1)$ for $L \geq 2$ as it is a piece of a geometric series etc, so we absorb the $O(logN)$ sum into a constant, hence we do not need to increase the power of the $log(t)$ in the big $O$ at the end; same for the other sum