Proving that $\sum_{k\le x}\frac{k}{\varphi(k)}=Ax+O(\log x)$

Question

I'm hoping to show that $\sum_{k\le x}\frac{k}{\varphi(k)}=Ax+O(\log x)$, for some positive constant $A$ but I keep getting stuck.

I know that $\sum_{k\le x}\frac{1}{\varphi(k)} = O(\log x)$ and $\frac{k}{\varphi(k)}=\sum_{d|k}\frac{\mu^2(d)}{\varphi(d)}$ and have used these in my work so far and I think I've correctly gotten the sum to look like $x\sum_{d\le x}\frac{\mu^2(d)}{d\varphi(d)}+O(\log x)$, but I'm not sure how to proceed (or if this is even correct).

Any help would be greatly appreciated

Answer 1

If $a_n=\sum_{d|n} b_d$ with $|b_d|\le 1/d$ then $$\sum_{n\ge x}a_n= \sum_{d\le x} b_d \lfloor x/d \rfloor = \sum_{d\le x} b_d (\frac{x}d+O(1)) = x (O(1/x)+\sum_{d\ge 1} \frac{b_d}d) + O(\sum_{d\le x} \frac1d)$$