Chapter V: Titchmarsh's book “The theory of the Riemann Zeta function”












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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 $bleq 2a$, with $0<a<t$ and $sum_{a<nleq b}n^{-s-it}=O(f(t))$ then $sum_{nleq t}n^{-s-it}=O(f(t)log(t))$, as $ttoinfty$. My question is: how this $log(t)$ appears? That is how is it related to $bleq 2a$ and $0<a<t$?






real-analysis sequences-and-series complex-analysis summation







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  • $begingroup$
    Here's a link to the book if anyone wants it.
    $endgroup$
    – Mason
    Dec 4 '18 at 19:53
















0












$begingroup$


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 $bleq 2a$, with $0<a<t$ and $sum_{a<nleq b}n^{-s-it}=O(f(t))$ then $sum_{nleq t}n^{-s-it}=O(f(t)log(t))$, as $ttoinfty$. My question is: how this $log(t)$ appears? That is how is it related to $bleq 2a$ and $0<a<t$?






real-analysis sequences-and-series complex-analysis summation







share|cite|improve this question









$endgroup$












  • $begingroup$
    Here's a link to the book if anyone wants it.
    $endgroup$
    – Mason
    Dec 4 '18 at 19:53














0












0








0





$begingroup$


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 $bleq 2a$, with $0<a<t$ and $sum_{a<nleq b}n^{-s-it}=O(f(t))$ then $sum_{nleq t}n^{-s-it}=O(f(t)log(t))$, as $ttoinfty$. My question is: how this $log(t)$ appears? That is how is it related to $bleq 2a$ and $0<a<t$?






real-analysis sequences-and-series complex-analysis summation







share|cite|improve this question









$endgroup$




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 $bleq 2a$, with $0<a<t$ and $sum_{a<nleq b}n^{-s-it}=O(f(t))$ then $sum_{nleq t}n^{-s-it}=O(f(t)log(t))$, as $ttoinfty$. My question is: how this $log(t)$ appears? That is how is it related to $bleq 2a$ and $0<a<t$?






real-analysis sequences-and-series complex-analysis summation




real-analysis sequences-and-series complex-analysis summation






share|cite|improve this question













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asked Dec 4 '18 at 19:39









GoldSoundzGoldSoundz

1098




1098












  • $begingroup$
    Here's a link to the book if anyone wants it.
    $endgroup$
    – Mason
    Dec 4 '18 at 19:53


















  • $begingroup$
    Here's a link to the book if anyone wants it.
    $endgroup$
    – Mason
    Dec 4 '18 at 19:53
















$begingroup$
Here's a link to the book if anyone wants it.
$endgroup$
– Mason
Dec 4 '18 at 19:53




$begingroup$
Here's a link to the book if anyone wants it.
$endgroup$
– Mason
Dec 4 '18 at 19:53










1 Answer
1






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer









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  • $begingroup$
    Thanks! I think I got it.
    $endgroup$
    – GoldSoundz
    Dec 4 '18 at 20:59











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! I think I got it.
    $endgroup$
    – GoldSoundz
    Dec 4 '18 at 20:59
















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! I think I got it.
    $endgroup$
    – GoldSoundz
    Dec 4 '18 at 20:59














1












1








1





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 '18 at 20:46









p4schp4sch

4,995217




4,995217












  • $begingroup$
    Thanks! I think I got it.
    $endgroup$
    – GoldSoundz
    Dec 4 '18 at 20:59


















  • $begingroup$
    Thanks! I think I got it.
    $endgroup$
    – GoldSoundz
    Dec 4 '18 at 20:59
















$begingroup$
Thanks! I think I got it.
$endgroup$
– GoldSoundz
Dec 4 '18 at 20:59




$begingroup$
Thanks! I think I got it.
$endgroup$
– GoldSoundz
Dec 4 '18 at 20:59


















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