= Solution
Let $\mu$ be the <Möbius function>. Its <Möbius divisor-sum identity> is
$$
\sum_{d\mid n}\mu(d)=\begin{cases}1,&n=1,\\0,&n>1.\end{cases}
$$
Indeed, for $n>1$ the sum is the expansion of $\prod_{p\mid n}(1-1)$; nonsquarefree <divisors> contribute zero. Thus, for <arithmetic functions> $a,b$,
$$
\boxed{b(n)=\sum_{d\mid n}a(d)\iff a(n)=\sum_{d\mid n}\mu(d)b(n/d).}
$$
To prove <Möbius inversion>, substitute the first formula into the second and collect the coefficient of $a(e)$. It is $\sum_{d\mid n/e}\mu(d)$, which is one for $e=n$ and zero otherwise. Conversely the same <divisor> interchange recovers $b$ from the displayed formula for $a$. In <Dirichlet convolution> notation this is simply $\mu*\mathbf1=\varepsilon$, where $\varepsilon$ is supported at one.
The <Prime number theorem with classical zero-free-region error> states that some absolute $c>0$ satisfies
$$
\boxed{\Psi(x):=\sum_{n\leq x}\Lambda(n)=x+O\bigl(xe^{-c\sqrt{\log x}}\bigr).}
$$
Here $\Lambda$ is the <Von Mangoldt function>. An equivalent prime-counting form, after decreasing $c$ if needed, is $\pi(x)=\operatorname{Li}(x)+O(xe^{-c\sqrt{\log x}})$. The contributions of proper <prime powers> are $O(\sqrt x\log^2x)$ and can be absorbed into this error. We use $\Psi$ to keep it distinct from the <Fourier transform> appearing in Question 5.
Here is a direct deduction of the requested <Mertens bound from a log-integrable Chebyshev error> from the stated <Prime number theorem>. First, the <divisor> identity gives
$$
1=\sum_{d\leq x}\mu(d)\left\lfloor\frac xd\right\rfloor,
\qquad \left|\sum_{d\leq x}\frac{\mu(d)}d\right|\leq\frac1x+\frac{\lfloor x\rfloor}{x}\leq2.
$$
Second, the exact <Dirichlet convolution> identity
$$
\mu(n)\log n=-(\mu*\Lambda)(n)
$$
follows from $\log=\mathbf1*\Lambda$: multiplication of a <Dirichlet convolution> by $\log n$ differentiates its two factors, so applying it to $\mu*\mathbf1=\varepsilon$ and convolving again with $\mu$ gives this formula. Summing it yields
$$
\sum_{n\leq x}\mu(n)\log n=-\sum_{d\leq x}\mu(d)\Psi(x/d)
=-x\sum_{d\leq x}\frac{\mu(d)}d+O\left(x\sum_{d\leq x}\frac{e^{-c\sqrt{\log(x/d)}}}{d}\right).
$$
The last harmonic sum is $O(1)$. To see this without an endpoint approximation, split $x/d$ into dyadic ranges $[2^j,2^{j+1})$. In each range $\sum1/d=O(1)$, and its exponential factor is at most $e^{-c\sqrt{j\log2}}$. Their sum over $j\geq0$ converges. The finitely many terms with $1\leq x/d<2$ obey the same estimate after adjusting the constant in the prime-number-theorem error. Therefore
$$
\sum_{n\leq x}\mu(n)\log n=O(x).
$$
For the <Mertens function> $M(x)=\sum_{n\leq x}\mu(n)$,
$$
M(x)\log x=\sum_{n\leq x}\mu(n)\log n+\sum_{n\leq x}\mu(n)\log(x/n).
$$
The second sum has absolute value at most $\sum_{n\leq x}\log(x/n)=O(x)$, by an integral comparison or the <Stirling formula>. Consequently
$$
\boxed{\frac{|M(x)|}{x}\ll\frac1{\log x}\qquad(x\geq2).}
$$
The absolute value in this conclusion is present in the original PDF.
Back to article page
