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I feel like I can use the result $$\sum_{n=x}^N \frac{a_n}{n^s} = A(N)N^{-s} + s \int_x^N A(t)t^{-s-1}dt$$ where $s=1/2$ to verify the $\ll$ approximation.

If I pick $A(n)=\sum_{x\leq t\leq n}\chi(t)$, then I know $|A(n)|\ll 1$. I think that will make the first term become $O(N^{-1/2})$, but am not sure how I should go about handling the integral term. Thanks!

Thanks for the help in advance!

Emma Glaslow
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    $A(t)=\sum_{x\leq m\leq t}\chi(m) = O(1)$ because $\chi$ is $q$-periodic with zero-mean. Thus $$A(N)N^{-s} + s \int_x^N A(t)t^{-s-1}dt = O(N^{-s})+O(s x^{-\Re(s)}/\Re(s))$$ – reuns Dec 07 '20 at 01:25
  • $O(N^{-s})+O(s x^{-\Re(s)}/\Re(s))=O(x^{-s})$ which is what you asked in the title – reuns Dec 07 '20 at 01:58
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    Why have you edited the question to remove the Dirichlet character? The sum $\sum_{n>N} n^{-1/2}$ doesn't even converge. – Wojowu Dec 08 '20 at 11:12

2 Answers2

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$$\sum_{n=x}^N \frac{a_n}{n^s} = A(N)N^{-s} + s \int_x^N A(t)t^{-s-1}dt$$ gives for $\Re(s) >0$ $$|\sum_{n=x}^\infty \frac{\chi(n)}{n^s}| =| s \int_x^\infty (\sum_{x\leq n\leq t}\chi(n))t^{-s-1}dt|\le |s| \int_x^\infty q t^{-\Re(s)-1}dt= \frac{q|s| x^{-\Re(s)}}{\Re(s)}$$

reuns
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  • Can I connect this result in any way to help prove https://math.stackexchange.com/q/3936812/849262? I'm still not quite following the answer provided. – Emma Glaslow Dec 07 '20 at 02:41
  • Since $\chi(n)$ is zero whenever $\gcd(n,q)>1$, that $q$ bound can become $\varphi(q)$ – TravorLZH Dec 07 '20 at 02:58
  • Can you give a reference or help clear up some questions and elaborate regarding your answer for the other question? – Emma Glaslow Dec 07 '20 at 03:12
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You have to edit the title as I think you actually meant $\sum_{n>N} \chi(n) n^{-1/2} \ll N^{-1/2} $. While this is valid for a Dirichlet character that is non-principal, it fails when $\chi$ is principal. Letting $$ S(y):= \sum_{1\leq n \leq y } \chi(n) $$ we get $$ \sum_{1\leq n \leq x}\chi(n) n^{-1/2} =\frac{S(x)}{x^{1/2}}+ \frac{1}{2} \int_1^x \frac{S( u ) }{u^{3/2 } } \mathrm d u.$$ Using this for $x=N$ and $x=M$ and subtracting we get $$ \sum_{N< n \leq M}\chi(n) n^{-1/2} =\frac{S(M)}{M^{1/2}}-\frac{S(N)}{N^{1/2}}+ \frac{1}{2} \int_N^M \frac{S( u ) }{u^{3/2 } } \mathrm d u.$$ By the estimate $S(u ) =O(1)$ this becomes $$\sum_{N< n \leq M}\chi(n) n^{-1/2} = O\left( \frac{1 }{M^{1/2}} \right) -\frac{S(N)}{N^{1/2}} + \frac{1}{2} \int_N^M \frac{S( u ) }{u^{3/2 } } \mathrm d u .$$ As $M\to \infty$ this converges because the integral converges absolutely since it is $$ O\left( \int_N^M \frac{ 1 }{u^{3/2 } } \mathrm d u \right)= O\left( \frac{2}{N^{1/2}} - \frac{2}{M^{1/2}} \right)= O(N^{-1/2}) .$$ Hence, the limit satisfies $$\sum_{n=N+1 }^\infty \chi(n) n^{-1/2} = O\left( \frac{1 }{M^{1/2}} \right) -\frac{S(N)}{N^{1/2}} O(N^{-1/2}) ,$$ which, by $M >N$ and the estimate $|S(u)| =O(1)$ is $ O(N^{-1/2}) $.

Mr. No
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