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## Main Question or Discussion Point

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? ,

[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? ,

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