# Estimate on Mobius function

Let $$\mu(n)$$ be the Mobius function, how to estimate $$\sum_{1\le i as $$x$$ goes to $$\infty$$? Are there some references on this?

If $$S$$ is your sum then

$$\left\lvert \sum_{1\leq n\leq x}\mu(n)\right\rvert^2 = 2S+ \sum_{1\leq n\leq x}\mu(n)^2.$$

The second sum on the right is $$(\frac{6}{\pi^2}+o(1))x$$, and hence estimating $$S$$ is equivalent to estimating $$\lvert \sum_{n\leq x}\mu(n)\rvert$$, a classical problem of analytic number theory.

In particular, assuming the Riemann Hypothesis, the left-hand side is $$O(x^{1+o(1)}$$), and hence (assuming RH)

$$S \ll x^{1+o(1)}.$$

Unconditionally we can show that $$S=o(x^2)$$, but cannot show $$S\ll x^{2-\epsilon}$$ for any $$\epsilon>0$$.

• "The best we can conclude unconditionally is $S = o(x^2)$". What do you mean by this? I'm no analytic number theorist, but I'd be surprised if $|\sum_{n \le x} \mu(n)| \ll \frac{x}{\log\log x}$ wasn't known unconditionally. – mathworker21 Jul 18 at 2:43
• Yes, sorry, I was being imprecise - I meant that in terms of improving the exponent nothing better is known. The standard proof of the prime number theorem can be adapted to show that $S\ll x^2 \exp(-O(\sqrt{\log x}))$, for example, and I believe better quantitative results are known. – Thomas Bloom Jul 18 at 9:43