Sums Involving the Mobius Function

Question

Are there any good approximations for the following sums in terms of $n$? $$\sum_{k=1}^{n}\mu(k)$$ $$\sum_{k=1}^{n}\mu(k)\log^m(k)$$ $$\sum_{k=1}^{n}\frac{\mu(k)}{k}.$$ I realize that the third sum goes to $0$, I just don't know how fast. Thank you.

Answer 1

We can prove that the first and second sums grow slower than $n/(\log n)^A$ for any constant $A$, and the third sum is eventually smaller than $1/(\log n)^A$.

The Riemann hypothesis is equivalent to the first and second sums growing slower than $n^{1/2+\varepsilon}$ for any $\varepsilon>0$, and to the third sum being eventually smaller than $1/n^{1/2-\varepsilon}$. These exponents $1/2$ would be best possible.