Is it known if for any finite $N$, the partial sum $\sum_1^N \frac{\mu(n)}{n}$ is equal to $0$ or, alternatively, isn’t $0$ for any finite $N$? Where $\mu(n)$ is a Mobius function.see here.
We know that the sum converges to $0$ when $N \rightarrow \infty$see here. But I am wondering about the partial sum for finite $N$.
I have no ideas about how one could prove or disprove it.
Thanks in advance.
There is no $N$ for which the sum is zero.
First, recall that for any integer $N>1$, there is a prime number $p$ such that $N/2<p\le N$.
Let $S_n = \sum_{n=1}^N \mu(n)/n$, and let's consider the value of $p\cdot S_n$ modulo $p$: we can do that as, apart from the term $n=p$, the denominators are all non-zero modulo $p$. This gives us $$ p\cdot S_n = \sum_{n=1}^N \mu(n)\cdot\frac{p}{n} \equiv \mu(p) = -1 \pmod p $$ as the terms for $n\not=p$ all vanish modulo $p$ after multiplication by $p$.