PNT implies Mobius cancellation.

Let us assume we have

$\displaystyle \sum_{n \le x} \Lambda(n) \sim x$

How do we prove the following?

$\displaystyle M(x) =\sum_{n \le x} \mu(n) = o(x )$

Here is how a proof goes:

Using the identity

$\displaystyle \left(\frac{1}{\zeta(s)}\right)^{\prime} = \frac{-\zeta^{\prime}(s)}{\zeta(s)}\cdot\frac{1}{\zeta(s)},$

and comparing the Dirichlet coefficients, we get

$\displaystyle -\mu(n) \log n=\sum_{d \mid n} \mu(d) \Lambda\left(\frac{n}{d}\right)$

Now summing over ${n}$ ,

$\displaystyle \sum_{n \le x}-\mu(n) \log n= \sum_{d \le x }\mu(d) \sum_{m \le \frac{x}{d}} \Lambda\left(m\right)$

Using PNT $\displaystyle \sum_{n \le x} \Lambda(n) = (1+ \epsilon)x$ for sufficiently large ${x> x_0(\epsilon)}$ and the Chebyshev bound $\displaystyle \sum_{n \le x} \Lambda(n) \ll x$ for smaller ${x}$ , we get

$\displaystyle \sum_{n \le x}-\mu(n) \log n = o(x \log x)$

Replacing ${\log n }$ by ${\log x}$ introduced an error of ${O(x)}$ because $\sum_{n \le x} \log \left(x/n \right) \ll x$

therefore

$\displaystyle \sum_{n \le x}-\mu(n) \log x = o(x \log x),$

and cancelling the ${\log x}$ we get

$\displaystyle M(x) = \sum_{n \le x} \mu(n) = o(x)$

Question: Why do we need to consider ${\mu(n) \log n }$ instead of starting from ${\mu(n)}$ ? One reason is the ${\mu(n) \log n}$ is the natural device (by that identity) that connects ${\mu(n)}$ to primes ${\Lambda(n)}$ But what is the arithmetic meaning of this ${\log }$ factor?