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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: