3
$\begingroup$

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 ?

$\endgroup$
  • 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 MO Feb 9 at 23:11
  • 3
    $\begingroup$ 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 $L = \lim_{x\to 0, x > 0} \frac{\zeta(2+2x)}{\zeta(1+x)} = 0$ $\endgroup$ – reuns Feb 10 at 5:13
  • $\begingroup$ @reuns: I think this is what the OP was after. Please turn your comment into an answer! $\endgroup$ – GH from MO Feb 10 at 23:20
10
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.