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 ?
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2$\begingroup$ I think you meant to ask: if $L$ exists (i.e. the series converges), then it equals to zero. Compare this with Theorem 426 in Hardy-Wright's book: if the ratio of $\pi(x)$ and $x/\log x$ tends to a limit, then the limit equals $1$ (which is the Prime Number Theorem). $\endgroup$– GH from MOCommented Feb 9, 2019 at 23:11
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4$\begingroup$ If $\sum_{n=1}^\infty a(n) n^{-1}=L$ converges then summation by parts shows $\lim_{x\to 0, x > 0} \sum_{n=1}^\infty a_n n^{-1-x} = L$, that is for $a(n) =\lambda(n) $ if $L$ exists then $L = \lim_{x\to 0, x > 0} \frac{\zeta(2+2x)}{\zeta(1+x)} = 0$ $\endgroup$– reunsCommented Feb 10, 2019 at 5:13
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$\begingroup$ @reuns: I think this is what the OP was after. Please turn your comment into an answer! $\endgroup$– GH from MOCommented Feb 10, 2019 at 23:20
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$\sum_{n=1}^N\frac{\lambda(n)}{n}=O(1)$ 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.
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1$\begingroup$ See also mathoverflow.net/questions/198404/… (in particular Lucia's answer) $\endgroup$– GagarCommented Feb 10, 2019 at 15:23