= Mertens bound from a log-integrable Chebyshev error
{c}
{title2=$M(x)\ll x/\log x$}
Suppose $\Psi(x)=x+O(xE(x))$ and $\sum_{j\geq0}\sup_{2^j\leq u<2^{j+1}}|E(u)|<\infty$. The <Möbius divisor-sum identity> gives $\sum_{d\leq x}\mu(d)/d=O(1)$. The exact <Dirichlet convolution> identity $\mu\log=-\mu*\Lambda$ then gives $\sum_{n\leq x}\mu(n)\log n=O(x)$, because its error is a convergent dyadic harmonic sum. Since $\sum_{n\leq x}\log(x/n)=O(x)$, the displayed <Mertens function> bound follows. The classical <Prime number theorem> error $E(x)=e^{-c\sqrt{\log x}}$ meets the summability condition. This is a deduction from the Chebyshev error itself, without a separate estimate for the reciprocal <zeta function>.
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