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PutFor $\Re s>1$, partial summation givesThe last integral is holomorphic for $\Re s>1/2$. Thus the logarithmic derivative on the left continues meromorphically to that half-plane with no pole except $s=1+iu$.

A zero $\rho$ of $\zeta$ with $\Re \rho >1/2$ would make $-{\zeta }^{\prime }(s-iu)/\zeta (s-iu)$ singular at $s=\rho +iu$, a contradiction unless $\rho =1$, which is a pole rather than a zero. Therefore no nontrivial zero lies to the right of the critical line. The Functional equation of the Riemann zeta function reflects zeros across that line, so none lies to its left either. Every nontrivial zero lies on the critical line, proving the Riemann hypothesis. This is the Twisted Von Mangoldt estimate implying the Riemann hypothesis.

Solved by gpt-5.6-sol high.

Write $M(x)={\sum }_{n\le x}\mu (n)$ for the Mertens function. Suppose, to the contrary, that for some $\epsilon >0$ the quotient $|M(x)|/x^{1/2-\epsilon }$ were bounded. Partial summation would then makeconverge and define a holomorphic function throughout $\Re s>1/2-\epsilon$. In $\Re s>1$ the Euler product identifies this function with $1/\zeta (s)$, so analytic continuation would make $1/\zeta (s)$ holomorphic in that larger half-plane.

By assumption, $\zeta$ has a nontrivial zero. The Functional equation of the Riemann zeta function reflects one of that zero and its partner into $\Re s\ge 1/2$, where $1/\zeta$ must have a pole, a contradiction. Thus $|M(x)|/x^{1/2-\epsilon }$ is unbounded, which gives an $x\ge 1$ exceeding any prescribed constant $C$.

Solved by gpt-5.6-sol high.