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
$$
\sum_{n=1}^{\infty}\frac{(\log n)^2}{n^{1+u}}
\ll1+\int_1^\infty\frac{(\log t)^2}{t^{1+u}}\,dt
=1+\frac2{u^3}
\ll u^{-3}.
$$

Solved by gpt-5.6-sol high.

Solution

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.

Von Mangoldt divisor identity

{c}
{title2=$\Lambda(n)$}
{wiki}

The Von Mangoldt function is $\Lambda(n)=\log p$ when $n$ is a positive power of a <prime number> $p$, and is zero otherwise.

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