Let $\mathbf1(n)=1$. The basic <Von Mangoldt divisor identity> is
$$
\log n=\sum_{d\mid n}\Lambda(d),
$$
because if $n=\prod_pp^{v_p(n)}$, the right-hand side is $\sum_pv_p(n)\log p=\log n$. In terms of <Dirichlet convolution>, this says $\log=\mathbf1*\Lambda$. Since the <Möbius function> is the convolution inverse of $\mathbf1$, convolving with $\mu$ gives $\Lambda=\mu*\log$. Consequently
$$
\Lambda(n)=\sum_{d\mid n}\mu(d)\log\frac nd.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 150 / 1 / a / Solution

The <Von Mangoldt function> is nonnegative and satisfies $\Lambda(m)\leq\log m$. Using the <Von Mangoldt divisor identity>,
$$
(\Lambda*\Lambda)(n)
=\sum_{d\mid n}\Lambda(d)\Lambda(n/d)
\leq\sum_{d\mid n}\Lambda(d)\log(n/d)
\leq\log n\sum_{d\mid n}\Lambda(d)
=(\log n)^2.
$$

For $0<u\leq1$, a comparison with an <improper integral> gives
$$
\sum_{n=1}^{\infty}\frac{(\log n)^2}{n^{1+u}}
\ll1+\int_1^\infty\frac{(\log t)^2}{t^{1+u}}\,dt
=1+\frac2{u^3}
\ll u^{-3}.
$$

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 150 / 3 / c / Solution

{c}

For every <positive integer> $n$,
$$
\sum_{d\mid n}\Lambda(d)=\log n.
$$
Equivalently, the constant-one function convolved with the <Von Mangoldt function> is the natural logarithm.

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 Incoming links: Von Mangoldt divisor identity