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

$$\sum_{n=0}^{x}f(n) =A(x)$$

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

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

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]

then is RH true by this argument?

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]

how can one prove that $$A(x)= O(x^{1/2+e})$$

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 $O(\sqrt{x\log\log x})$.

then is RH true by this argument? ,

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.