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Exercise 2.4.4 from M. Ram Murty's Problems in Analytic Number Theory asks us to show that for $\chi$ a nontrivial Dirichlet character $(\operatorname{mod} q)$, we have the estimate $$\sum_{n\leq x}\frac{f(n)}{\sqrt n}=2L(1,\chi)\sqrt{x}+O(1)$$ where $f(n):=\sum_{d\vert n}\chi(n)$. I'm having trouble getting the correct bound on the error term.

Using Abel's summation formula and Dirichlet's hyperbola method, I found that $$\sum_{n\leq x}\frac{f(n)}{\sqrt n}=\frac{A(x)}{\sqrt x}+\frac{1}{2}\int_1^x\frac{A(t)}{t^{\frac{3}{2}}}dt$$ where $$A(x)=xL(1,\chi)-x\sum_{n> y}\frac{\chi(n)}{n}+O(y)+O\left(\frac{x}{y}\right)\text{ for any }y>0$$ Per my reading course professor, the optimal choice of $y$ for minimizing the terms $O(y),O(x/y)$ would be $y=\sqrt x$, but this yields $$A(x)=xL(1,\chi)-x\sum_{n> \sqrt x}\frac{\chi(n)}{n}+O(x^{1/2})$$ and the $O(x^{1/2})$ term, when integrated in $\frac{1}{2}\int_1^x\frac{A(t)}{t^{3/2}}dt$, yields a $O(\log x)$ term and thus poses a problem for $\sum_{n\leq x}f(n)/\sqrt{n}$ since all the terms that are not $2L(1,\chi)\sqrt{x}$ should be bounded. What am I missing?

Alann Rosas
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  • Have you tried applying the hyperbola method directly to the sum over $f(n)/\sqrt n$? – TravorLZH Oct 17 '23 at 18:25
  • maybe an approach involving Perron's formula works, all we need to do is to bound $L$ and $\zeta$. – LurchiDerLurch Oct 17 '23 at 18:42
  • @TravorLZH I have not. How would that go about, though? The hyperbola method applies to Dirichlet convolutions, and I don’t see how $f(n)/\sqrt{n}$ is expressible as such a convolution. – Alann Rosas Oct 17 '23 at 19:00

1 Answers1

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From the definition, we have

$$ S=\sum_{n\le x}{f(n)\over\sqrt n}=\sum_{ab\le x}{\chi(a)\over\sqrt{ab}}. $$

Applying the hyperbola method, there is

\begin{aligned} S &=\sum_{a\le\sqrt x}{\chi(a)\over\sqrt a}\sum_{b\le x/a}{1\over\sqrt b}+\sum_{b\le\sqrt x}{1\over\sqrt b}\sum_{a\le\sqrt x}{\chi(a)\over\sqrt a}-\sum_{a\le\sqrt x}{\chi(a)\over a}\sum_{b\le\sqrt x}{1\over\sqrt. b} \\ &=S_1+S_2-S_3. \end{aligned}

By summation techniques, it can be proven that when $\chi$ is a fixed non-principal character, there are constants $C_1$ and $C_2$ such that

$$ \sum_{n\le y}{1\over\sqrt n}=2y^{\frac12}+C_1+O(y^{-\frac12}),\quad\sum_{n\le y}{\chi(a)\over\sqrt a}=C_2+O(y^{-\frac12}) $$

and

$$ \sum_{n\le y}{\chi(n)\over n}=L(1,\chi)+O(y^{-1}). $$

Therefore, we have

\begin{aligned} S_1 &=\sum_{a\le\sqrt x}{\chi(a)\over\sqrt a}\left[2\sqrt{\frac xa}+C_1+O\left(\sqrt{\frac ax}\right)\right] \\ &=2\sqrt x\sum_{a\le x}{\chi(a)\over a}+C_2\sum_{a\le\sqrt x}{\chi(a)\over\sqrt a}+O\left({1\over\sqrt x}\sum_{a\le\sqrt x}|\chi(a)|\right) \\ &=2\sqrt x L(1,\chi)+O(1), \end{aligned}

$$ S_2=\sum_{b\le\sqrt x}{1\over\sqrt b}\left[C_2+O\left(\sqrt{\frac bx}\right)\right]=2C_2\sqrt x+O(1), $$

and

$$ S_3=\left[2x^{\frac12}+C_1+O(x^{\frac12})\right]\left[C_2+O(x^{-\frac12})\right]=2C_2\sqrt x+O(1). $$

Combining everything, we obtain

$$ S=2\sqrt xL(1,\chi)+O(1), $$

which is the desired result.

TravorLZH
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  • Thank you for answering. Think you could explain the first equality $\sum_{n\leq x}f(n)/\sqrt{n}=\sum_{ab\leq x}\chi(a)/\sqrt{ab}$? I'm not sure how to get this. – Alann Rosas Oct 18 '23 at 05:36
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    $$\sum_{n \leq x} \frac{\sum_{a \mid n} \chi(a)}{\sqrt{n}} = \sum_{n \leq x} \sum_{ab = n} \frac{\chi(a)}{\sqrt{ab}} = \sum_{ab \leq x} \frac{\chi(a)}{\sqrt{ab}}.$$ – Peter Humphries Oct 18 '23 at 15:09