given $x \in \mathbb{R}$, $k \in \mathbb{N}$\ Let $A$ be the set of positive integers at most $x$ but coprime to $k$ Let $\mu$ be the mobius function, show the following
$$\displaystyle\sum_{n \in A}\dfrac{1}{n} = \Big(\sum_{d|k}\dfrac{\mu(d)}{d}\Big)\log x + O(1)$$
Can someone give me some hint on this. I tried to rewrite $LHS$ with $$\displaystyle\sum_{n \in A}\dfrac{1}{n} = \sum_{n=1}^{\lfloor{x}\rfloor} \dfrac{1}{n}\Big(\sum_{d|gcd(n,k)}\mu(d)\Big)$$
Then use Abel summation, but I feel like this is not the right direction.
We use the fact that \[1_{(n,k) = 1} = \sum_{d \mid (n,k)} \mu(d).\] It follows that \[\sum_{\substack{n \leq x \\ (n,k) = 1}} \frac{1}{n} = \sum_{n \leq x} \sum_{d \mid (n,k)} \frac{\mu(d)}{n}.\] The condition $d \mid (n,k)$ is the same as the two conditions $d \mid k$ and $n \equiv 0 \pmod{d}$. So we may interchange the order of summation, yielding \[\sum_{\substack{n \leq x \\ (n,k) = 1}} \frac{1}{n} = \sum_{d \mid k} \mu(d) \sum_{\substack{n \leq x \\ n \equiv 0 \pmod{d}}} \frac{1}{n}.\] We make the change of variables $m = \frac{n}{d}$, so that the inner sum is \[\frac{1}{d} \sum_{m \leq \frac{x}{d}} \frac{1}{m}.\] Now we use the fact that this sum is $\log \frac{x}{d} + \gamma_0 + O\left(\frac{d}{x}\right)$. So the original sum is \[\sum_{d \mid k} \frac{\mu(d)}{d} \log x - \sum_{d \mid k} \frac{\mu(d) \log d}{d} + \gamma_0 \sum_{d \mid k} \frac{\mu(d)}{d} + O\left(\frac{1}{x} \sum_{d \mid k} |\mu(d)|\right).\]