On Multiplicative number theory : Classical theory (Montgomery, Vaughan) pp.185, there is an exercise concerning sums below.
$$ \sum_{n=1}^{\infty}\frac{\mu(n)}{\phi(n)} = 0 $$ $$ \sum_{n=1}^{\infty}\frac{\mu(n)\log(n)}{\phi(n)} = 0$$
They say these formulas can be proved by same method as they proved the PNT.
I think I can get some asymptotic bound for $$ \sum_{n \le x}\frac{\mu(n)}{\phi(n)} $$ $$ \sum_{n \le x}\frac{\mu(n)\log(n)}{\phi(n)} .$$
if I follow the solution of this exercise.
I'm curious about above asymptotic bounds.
Actually authors quoted Hardy, G. H. (1921). Note on Ramanujan’s trigonometrical function c_q(n), and certain series of arithmetical functions, Proc. Cambridge Philos. Soc. 20, 263–271, but it's unable to find on internet.
I am a beginner in this branch of mathematics so please feel free to correct any mistakes.
We have $$\begin{align}\sum_{n\le x}\frac{\mu(n)}{\phi(n)}&=\sum_{n\le x}\left(\mu(n)\sum_{d\mid n}\frac1{\mu(d)}\frac dn\right)=\sum_{n\le x}\sum_{d\mid n}\frac{d\mu(n)}{n\mu(d)}\\&=\sum_{d\le x}\sum_{q\le x/d}\frac{\mu(q)}q=\mathcal{O}\left(\sum_{d\le x}\sum_{q\le x/d}\frac1q\right)\\&=\mathcal{O}\left(\sum_{d\le x}\log\frac xd+C+\mathcal{O}\left(\frac dx\right)\right)\\&=\mathcal{O}\left(\sum_{d\le x}\log\frac xd\right)\\&=\mathcal{O}\left(\log x\right)\end{align}$$