Help bounding the partial sum of reciprocals of odd square-free integers between $\frac{n}{2}$ and $n$. After a sieving process, I am left with the partial sum of reciprocals of odd square-free integers between $\frac{n}{2}$ and $n$. I want to know if there are sharp bounds for this type of sum in the literature, and how they could be derived.
I have tried the following, which is wrong. As the partial sum of reciprocals of square-free integers up to $n$ is $\frac{6}{\pi^2}\log(n)+O(1)$, the partial sum of reciprocals of square-free integers between $\frac{n}{2}$ and $n$ is the difference between the partial sum of reciprocals of square-free integers up to $n$, and the partial sum of reciprocals of square-free integers up to $\frac{n}{2}$; and so, we have that $$\frac{6}{\pi^2}\left(\log(n) -\log\left(\frac{n}{2}\right)\right)+O(1)=$$ $$=\frac{6}{\pi^2}\left(\log(n) -(\log(n)-\log(2))\right)+O(1)=$$ $$=\frac{6}{\pi^2}\left(\log(2)\right)+O(1)$$ However, this is wrong, I get a constant and the sum of reciprocals of square-free integers between $\frac{n}{2}$ and $n$ is a divergent series. And thus I have not even attempted to bound the partial sum of the odd ones, which is the one I need.
Thanks in advance!
Sum of reciprocals of square-free integers from $n/2$ to $n$ should be less than or equal to $$\sum_{n/2}^n\left(\frac1n\right)$$ which is less than or equal to $\frac2n\times(n/2+1)=1+2/n$. So I don't understand how it could be divergent.