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The logarithmic derivative of a nonzero differentiable function $f$ is $f^{\prime }/f$. For $\Re s>1$, the Euler product for the Riemann zeta function gives

Complete multiplicativity and absolute convergence give the Euler productSet $\sigma =1+1/\log x$. Taking logarithms of absolute values and expanding the local factors gives, uniformly in real $t$,The prime powers with exponent at least two contribute $O(1)$; changing $p^{-\sigma }$ to $p^{-1}$ below $x$ and estimating the tail above $x$ also cost $O(1)$. By Mertens theorem,Exponentiating yields

Solved by gpt-5.6-sol high.

If $f$ is a multiplicative function with $|f(n)|\le 1$, then for $\Re s>1$ its Dirichlet series has the Euler productIndeed, expanding the product over a finite set of primes and using unique prime factorization gives the sum over integers having no other prime factors. Moreover,so absolute convergence permits rearrangement and passage to the limit over all primes. This proves the formula.

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.

$\zeta (s)={\sum }_{n\ge 1}{n}^{-s}$ for $\Re s>1$; its Euler product is ${\prod }_p(1-p^{-s}{)}^{-1}$.