The Probabilistic argument

zetafunction

the question is if we suppose that a function f(n) can take only two values +1 and -1 both with equal probability and define the summatory

[tex] \sum_{n=0}^{x}f(n) =A(x) [/tex]

how can one prove that [tex]A(x)= O(x^{1/2+e}) [/tex]
?? if we set A(n)=M(n) the Mertens function and since

[tex] \sum_{n=0}^{\infty}(M(n)-M(n-1))n^{-s} =1/ \zeta (s) [/tex]

then is RH true by this argument? ,

matt grime

A(x) is a random variable, so you oughtn't you to be asking for something like E(A(x)), or E(A(x)^2)?

CRGreathouse

Are you really arguing that the value of the Mertens function is independent of its argument? That seems like a non-starter to me.

At best you have a heuristic suggesting that RH 'should' be true, but there are plenty of those.

gel

That is true, with probability one, if f(n) are independent. In fact, the law of the iterated logarithm says that with probability one it is of order [itex]O(\sqrt{x\log\log x})[/itex].

as CRGreathouse mentions, this does not follow because the Mertens function is deterministic and not random. It's expected to share many properties of a random sequence, but that hasn't been proven.