# On the convergence of $\sum_{n=1}^{\infty} \frac{\lambda(n)}{n}$ and the Prime Number Theorem

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 ?

• 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). – GH from MO Feb 9 '19 at 23:11
• 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 $L = \lim_{x\to 0, x > 0} \frac{\zeta(2+2x)}{\zeta(1+x)} = 0$ – reuns Feb 10 '19 at 5:13
• @reuns: I think this is what the OP was after. Please turn your comment into an answer! – GH from MO Feb 10 '19 at 23:20

$$\sum_{n=1}^N\frac{\lambda(n)}{n}=O(1)$$ 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.