Let $G(x)=\sum_{k\leq x}\frac{\mu(k)}{k}$, where $\mu$ is the Mobius function. From this question and its answer, its mention the Riemann hypothesis is equivalent to $G(x)=O(x^{-\frac{1}{2}+\epsilon})$.
I have never heard of this, the closet thing I have heard is the Riemann hypothesis is equivalent to $M(x)=O(x^{\frac{1}{2}+\epsilon})$ for all $\epsilon>0$, where $M(x)=\sum_{k\leq x}\mu(k)$, the Mertens function. I have tried to prove the above fact but did not succeed. So I would like to ask for reference for the above equivalent form of the Riemann hypothesis(the $G(x)$ one), better to contain a proof of it.
Best Answer
Abel's summation formula gives $$ G(x):=\sum\limits_{n \le x} {\frac{{\mu (n)}}{n}} = \frac{{M(x)}}{x} + \int_1^x {\frac{{M(t)}}{{t^2 }}{\rm d}t} . $$ Assume RH. Then $M(x)=\mathcal{O}(x^{1/2+\varepsilon})$. By the prime number theorem, $$ \sum\limits_{n = 0}^\infty {\frac{{\mu (n)}}{n}} = 0. $$ Thus $$ \sum\limits_{n \le x} {\frac{{\mu (n)}}{n}} = \frac{{M(x)}}{x} - \int_x^{ + \infty } {\frac{{M(t)}}{{t^2 }}{\rm d}t} , $$ where \begin{align*} \frac{{M(x)}}{x} - \int_x^{ + \infty } {\frac{{M(t)}}{{t^2 }}{\rm d}t} & = \mathcal{O}(x^{ - 1/2 + \varepsilon } ) + \mathcal{O}(1)\int_x^{ + \infty } {\frac{{\rm d}t}{{t^{3/2 - \varepsilon } }}} \\ & = \mathcal{O}(x^{ - 1/2 + \varepsilon } ) + \mathcal{O}(x^{ - 1/2 + \varepsilon } ) = \mathcal{O}(x^{ - 1/2 + \varepsilon } ). \end{align*} Now assume $G(x)=\mathcal{O}(x^{ - 1/2 + \varepsilon } )$. Then, by Abel's summation formula, $$ M(x):=\sum\limits_{n \le x} {\mu (n)} = \sum\limits_{n \le x} {\frac{{\mu (n)}}{n}n} = x\,G(x) + \int_1^x {G(t)\,{\rm d}t} . $$ Using our assumption, it follows that $M(x)=\mathcal{O}(x^{1/2+\varepsilon})$, which we know implies RH.