Let $\lambda$ be the Lioville function of number theory.
I've heard several times that if $L=\sum_{n=1}^{\infty} \frac{\lambda(n)}{n} =O(1)$ then $L=0$ (the Prime Number Theorem). How can this be proved ? Or can someone kindly provide a reference ?
Let $\lambda$ be the Lioville function of number theory.
I've heard several times that if $L=\sum_{n=1}^{\infty} \frac{\lambda(n)}{n} =O(1)$ then $L=0$ (the Prime Number Theorem). How can this be proved ? Or can someone kindly provide a reference ?
$\sum_{n=1}^N\frac{\lambda(n)}{n}=O(1)$ can be proven in an elementary manner (with no analysis, real or complex). Indeed, observe we have $$\sum_{n=1}^N\lambda(n)\left\lfloor\frac{N}{n}\right\rfloor=\sum_{n=1}^N\lambda(n)\sum_{k\leq N,n\mid k}1=\sum_{k=1}^N\sum_{n\mid k}\lambda(n)=\sum_{k=1}^NQ(n),$$ where $Q(n)=1$ if $n$ is a perfect square, $Q(n)=0$ otherwise. Hence the right-hand side is clearly $O(N)$, while the left-hand side is $N\sum_{n=1}^N\frac{\lambda(n)}{n}+O(N)$. It follows immediately $\sum_{n=1}^N\frac{\lambda(n)}{n}=O(1)$.
On the other hand, as you note, $\sum_{n=1}^\infty\frac{\lambda(n)}{n}=0$ is equivalent to the prime number theorem. Hence there likely isn't an easy deduction of convergence from $O(1)$ bound, since that would give us an equally easy proof of PNT.