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 (see my work below), 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 $$A(x)=xL(1,\chi)-x\sum_{n> y}\frac{\chi(n)}{n}+O(y)+O\left(\frac{x}{y}\right)\text{ for any }y>0$$ The optimal choice of $y$ for minimizing the terms $O(y),O(x/y)$ would seem to 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. Am I missing something? Is there a better choice for $y$?
My work:
From Abel's summation formula, we can write $$\sum_{n\leq x}\frac{f(n)}{\sqrt n}=\frac{A(x)}{\sqrt x}+\frac{1}{2}\int_1^x\frac{A(t)}{t^{\frac{3}{2}}}dt$$ where $A(x):=\sum_{n\leq x}f(n)$. Noting that $f=\chi\ast 1$, and taking $g(n)=\chi(n)$ and $h(n)=1$, we can then use Dirichlet's hyperbola method (written as presented in Murty's book) $$\sum_{n\leq x}f(n)=\sum_{n\leq y}g(n)H\left(\frac{x}{n}\right)+\sum_{n\leq \frac{x}{y}}h(n)G\left(\frac{x}{n}\right)-G(y)H\left(\frac{x}{y}\right)\text{ }(x,y>0)$$ with $G(x):=\sum_{n\leq x}\chi(n)$ and $H(x):=\sum_{n\leq x} 1=\lfloor x/y\rfloor$ to write \begin{align} A(x) &= \sum_{n\leq y}\chi(n)\left\lfloor\frac{x}{n}\right\rfloor+\sum_{n\leq \frac{x}{y}}\left(\sum_{k\leq\frac{x}{n}}\chi(k)\right)-\left(\sum_{n\leq y}\chi(n)\right)\left\lfloor\frac{x}{y}\right\rfloor\\\\ &= \sum_{n\leq y}\chi(n)\left\lfloor\frac{x}{n}\right\rfloor+\sum_{n\leq \frac{x}{y}}\left(\sum_{k\leq\frac{x}{n}}\chi(k)\right)+O(1)\left\lfloor\frac{x}{y}\right\rfloor\text{ (since $\sum_{n\leq y}\chi(n)$ is bounded)}\\\\ &= \sum_{n\leq y}\chi(n)\left\lfloor\frac{x}{n}\right\rfloor+\sum_{n\leq \frac{x}{y}}\left(\sum_{k\leq\frac{x}{n}}\chi(k)\right)+O\left(\frac{x}{y}\right)\text{ (since $\lfloor x/y\rfloor=x/y + O(1)$)}\\\\ &= \sum_{n\leq y}\chi(n)\left\lfloor\frac{x}{n}\right\rfloor+O\left(\frac{x}{y}\right)+O\left(\frac{x}{y}\right)\text{ (follows from $\sum_{k\leq \frac{x}{n}}\chi(k)$ bounded independently of $x$ and $n$)}\\\\ &= \sum_{n\leq y}\chi(n)\left\lfloor\frac{x}{n}\right\rfloor+O\left(\frac{x}{y}\right) \end{align} We can then work on the first term by writing $$\sum_{n\leq y}\chi(n)\left\lfloor\frac{x}{n}\right\rfloor=\sum_{n\leq y}\left(\chi(n)\frac{x}{n}+O(1)\chi(n)\right)=x\sum_{n\leq y}\frac{\chi(n)}{n}+O(y)=xL(1,\chi)-x\sum_{n> y}\frac{\chi(n)}{n}+O(y)$$ so $$A(x)=xL(1,\chi)-x\sum_{n> y}\frac{\chi(n)}{n}+O(y)+O\left(\frac{x}{y}\right)$$
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.