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Suppose $\Psi (x)=x+O(xE(x))$ and ${\sum }_{j\ge 0}{\sup }_{2^j\le u<2^{j+1}}|E(u)|<\infty$. The Möbius divisor-sum identity gives ${\sum }_{d\le x}\mu (d)/d=O(1)$. The exact Dirichlet convolution identity $\mu \log =-\mu \ast \Lambda$ then gives ${\sum }_{n\le x}\mu (n)\log n=O(x)$, because its error is a convergent dyadic harmonic sum. Since ${\sum }_{n\le x}\log (x/n)=O(x)$, 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.

Write $1(n)=1$ and let $\epsilon$ be the Identity for Dirichlet convolution. The Möbius function uses the empty-product convention $\mu (1)=1$; the Von Mangoldt function has $\Lambda (1)=0$. We first establish all the arithmetic function identities needed below. Every integer divisor of $n$ contributes one to $\tau (n)$, so $\tau =1\ast 1$. If $n>1$ has $r$ distinct prime factors, the squarefree divisors correspond to their subsets andAt $n=1$ the sum is one. Thus $\mu \ast 1=\epsilon$. Writing $n={\prod }_p{p}^{v_p(n)}$, the only divisors with nonzero Von Mangoldt function are $p,p^2,\dots ,p^{v_p(n)}$, and henceThe Dirichlet convolution is commutative by replacing $d$ with $n/d$, and associative because either bracketing sums $f(a)g(b)h(c)$ over $abc=n$. ConsequentlyThese are consequences of the definitions, rather than unproved invocations of Möbius inversion.

To identify the constant, integrate on each interval $[j,j+1)$:The tail is $O(1/J)$, since the integrand is at most $2/x^2$ for $x\ge 2$. Summation therefore gives $\gamma ={\lim }_{J\to \infty }(H_J-\log (J+1))$, where $H_J$ is the harmonic number, and $H_J=\log J+\gamma +O(1/J)$. Thus this is Euler's constant.

Put $r=\lfloor \sqrt{N}\rfloor$. Counting pairs of positive integers whose product is at most $N$, and subtracting the square counted twice, gives the Dirichlet hyperbola method identitySince $r=\sqrt{N}+O(1)$, the harmonic number estimate yieldsThe increasing natural logarithm also satisfies ${\sum }_{n\le N}\log n=N\log N-N+O(\log (2N))$ by comparison with its integral. Subtracting proves

For the second implication, denote the Mertens function by $M(x)$ and the Second Chebyshev function by $\psi (x)$. Apply the proved Dirichlet convolution identities to $b(n)=\tau (n)-\log n-2\gamma 1(n)$:Summing this identity over integers up to $N$ gives the exact relationFix an integer $K\ge 2$. The contribution to $E(N)$ from $d\le N/K$ has absolute value at mostFor $d>N/K$, group by $j=\lfloor N/d\rfloor$. This gives exactlyunder the assumed cancellation $M(x)=o(x)$; extending that assumption from integers to real $x$ uses $M(x)=M(\lfloor x\rfloor )$. Therefore ${\mathrm{limsup}}_{N\to \infty }|E(N)|/N\ll K^{-1/2}$. Letting $K\to \infty$ proves the prime number asymptotic from Mertens cancellation:

Let $\mu$ be the Möbius function. Its Möbius divisor-sum identity isIndeed, for $n>1$ the sum is the expansion of ${\prod }_{p\mid n}(1-1)$; nonsquarefree divisors contribute zero. Thus, for arithmetic functions $a,b$,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 $\mu \ast 1=\epsilon$, where $\epsilon$ is supported at one.

The Prime number theorem with classical zero-free-region error states that some absolute $c>0$ satisfiesHere $\Lambda$ is the Von Mangoldt function. An equivalent prime-counting form, after decreasing $c$ if needed, is $\pi (x)=\mathrm{Li}(x)+O(x{e}^{-c\sqrt{\log x}})$. The contributions of proper prime powers are $O(\sqrt{x}{\log }^{2}x)$ 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 givesSecond, the exact Dirichlet convolution identityfollows from $\log =1\ast \Lambda$: multiplication of a Dirichlet convolution by $\log n$ differentiates its two factors, so applying it to $\mu \ast 1=\epsilon$ and convolving again with $\mu$ gives this formula. Summing it yieldsThe 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 $\sum 1/d=O(1)$, and its exponential factor is at most $e^{-c\sqrt{j\log 2}}$. Their sum over $j\ge 0$ converges. The finitely many terms with $1\le x/d<2$ obey the same estimate after adjusting the constant in the prime-number-theorem error. ThereforeFor the Mertens function $M(x)={\sum }_{n\le x}\mu (n)$,The second sum has absolute value at most ${\sum }_{n\le x}\log (x/n)=O(x)$, by an integral comparison or the Stirling formula. ConsequentlyThe absolute value in this conclusion is present in the original PDF.

Let $M(x)={\sum }_{n\le x}\mu (n)$ be the Mertens function. Partial summation givesFor any $\Re s>0.9$, choose a sufficiently small $ϵ>0$ with $0.9+ϵ<\Re s$. The assumed bound makes the boundary term tend to zero and makes the integral absolutely convergent. Moreover, on each compact set of this half-plane, one can choose the same $ϵ$, so the integral has locally uniform convergence. Henceis a holomorphic function on $\Re s>0.9$, agreeing with the reciprocal Dirichlet series from part (b) on $\Re s>1$.

The identity theorem applied to the holomorphic functions $(s-1)\zeta (s)F(s)$ and $s-1$ extends their equality throughout $\Re s>0.9$; the assumed simple pole makes the first function holomorphic at $1$. Thus $\zeta (s)F(s)=1$ away from $1$, proving a zero-free half-plane:

To obtain the strongest symmetric conclusion, use the following basic facts explicitly: the functional equation of the Riemann zeta function gives the entire function $\xi (s)=\frac{1}{2}s(s-1){\pi }^{-s/2}\Gamma (s/2)\zeta (s)$ with $\xi (s)=\xi (1-s)$, and the trivial zeros of the Riemann zeta function are the negative even integers. Every Nontrivial zero of the Riemann zeta function is therefore accompanied by its reflection $1-s$. A Nontrivial zero of the Riemann zeta function with real part below $0.1$ would reflect to one above $0.9$, which is impossible. Consequently all nontrivial zeros lie in the closed stripThe $ϵ$ in the hypothesis prevents this argument from excluding either boundary line; the trivial zeros of the Riemann zeta function remain present.

The truncated Perron formula says that, for $c>1$ and $T\ge 2$,where an endpoint has half weight and the error is controlled by

Apply this with $a_n=\mu (n)$, $c=1+1/\log x$, andPart (b) makes the integrand $x^s/(s\zeta (s))$. Move the contour tousing a contour that stays inside the zero-free region of the Riemann zeta function near small $|t|$. The estimates supplied in the question give $1/\zeta (s)\ll \log (|t|+3)$ on the new contour. Its vertical segment is thereforeafter decreasing $c_3$. The horizontal segments and the Perron truncation error areNo residue is crossed because $1/\zeta (s)$ has a zero, rather than a pole, at $s=1$. Thus the Mertens function satisfiesfor some absolute $c>0$.

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 $C$.

Let $\Delta (N)={\sum }_{n\le N}(\tau (n)-\log n-2\gamma )$. The Möbius divisor-sum identity and the Von Mangoldt divisor identity give $\mu \ast \tau =1$ and $\mu \ast \log =\Lambda$, soHere $\ast$ is Dirichlet convolution. The divisor summatory asymptotic bounds the terms with $d\le N/K$ in absolute value by $O(N/\sqrt{K})$. For $d>N/K$, group by $j=\lfloor N/d\rfloor <K$: the remaining sum is ${\sum }_{j<K}\Delta (j)(M(N/j)-M(N/(j+1)))=o_K(N)$. First let $N$ tend to infinity, then $K$ tend to infinity. This proves the Second Chebyshev function asymptotic from cancellation of the Mertens function.