Im trying to compute the following
$\sum_{\substack{q \leq x \\ q \ square free}} q^{-1/2}$
I know that $\sum_{\substack{q \leq x \\ q \ square free}} 1 = \frac{6}{\pi^2}x + \mathcal{O}(\sqrt{x})$
and
$\sum_{n \leq x } n^{-1/2} = 2\sqrt{x}+c+\mathcal{O}\Big(\frac{1}{\sqrt{x}}\Big)$ where $c$ is some constant. Is there there any way to combine the two to get a result for the first summation?
On
We present by way of enrichment an evaluation using analytic number theory which provides a closed form as a target of more elementary methods. Observe that the Dirichlet series of the indicator of squarefree numbers is given by
$$L(s) = \prod_p \left(1+\frac{1}{p^s}\right) = \prod_p \frac{1-1/p^{2s}}{1-1/p^s} = \frac{\zeta(s)}{\zeta(2s)}.$$
Observe furthermore that
$$\frac{1}{n^{s+1/2}} = \frac{n^{-1/2}}{n^s}$$
and hence the Dirichlet generating function for this problem is
$$M(s) = \frac{\zeta(s+1/2)}{\zeta(2s+1)}.$$
The dominant simple pole here is at $s=1/2$ and we may apply the Wiener-Ikehara theorem, getting for the asymptotics
$$\frac{x^{1/2}}{1/2} \mathrm{Res}_{s=1/2} \frac{\zeta(s+1/2)}{\zeta(2s+1)} = 2\sqrt{x} \frac{6}{\pi^2} = \frac{12}{\pi^2} \sqrt{x}.$$
The next terms from the underlying Mellin transform integral
$$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} M(s) \frac{x^s}{s} \; ds$$
originate with the non-trivial zeros of the Riemann Zeta function.
If we define $$c(x)=\sum_{\substack{q \leq x \\ q \ square free}} 1,$$ the sum is $$\sum_{n\le x}\frac1{\sqrt{n}}(c(n)-c(n-1))=\sum_{n\le x}\left(\frac1{\sqrt{n}}-\frac1{\sqrt{n+1}}\right)c(n)+\frac{c(\lfloor x\rfloor)}{\sqrt{\lfloor x\rfloor+1}},$$ that's called partial summation. I think it should be easy to move on from here.