Can we prove that partial sum $|\sum_1^N \mu(n)/n| \le 1/2$ when $N > 1$?

Question

Here, $\mu(n)$ is Möbius function.

From this, we can see that partial sum $|\sum_1^N \frac{\mu(n)}{n}| \le 1$. when $N \in \mathbb{N}$.

We also know the series converges to $0$ and is not monotone.

Is it known if the partial sum $|\sum_1^N \frac{\mu(n)}{n}| \le \frac{1}{2}$ when $N > 1$ and $N \in \mathbb{N}$. I have confirmed that this bound holds for $1 < N \le 50,000,000$, and the absolute value of partial sum gets much much smaller that 0.5 when N gets to 50 million. But that doesn’t mean it will necessarily hold to all $N$s to infinity. I’m wondering if there is a way to prove it.

Answer 1

The convergence of $\sum_n \mu(n)/n $ is equivalent to the PNT and its proof gives an explicit decay. Yes checking to $10^7$ is enough but proving so is complicated.

(you can deduce the latter claim from https://en.wikipedia.org/wiki/Mertens_function#Known_upper_bounds )