It is known that \(\displaystyle \lim_{k \to \infty} \left( \sum_i x_i^k \right)^{1/k} = \max_i x_i\) for \(x_i \in \mathbb{R}^+\). So, to extract the squarefree divisors of \(n\) only, we make use of the Möbius function and take the following sum:

\[\sum_{d|n} \mu^2(d)d^k = n^k \sum_{d|n} \mu^2(d)\left( \frac{n}{d} \right)^{-k} = n^k (\mu^2 * \text{Id}_{-k})(n)\]

Taking the \(k\)th root and letting \(k \to \infty\):

\[\max_{d|n} \mu^2(d)d = \lim_{k \to \infty} \left( n^k (\mu^2 * \text{Id}_{-k})(n) \right)^{1/k} = n \lim_{k \to \infty} (\mu^2 * \text{Id}_{-k})(n)^{1/k}\]

Note:

\(*\) is the Dirichlet convolution, a binary operator defined on functions \(\mathbb{N} \longrightarrow \mathbb{C}\) such that \(\displaystyle (f*g)(n) = \sum_{d|n} f(d)g(\frac{n}{d})\).

\(\text{Id}_{k}(n) = n^k\) for \(n \in \mathbb{N}\), \(k \in \mathbb{C}\).

## Comments

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TopNewestMind if you explain the first step? The

\(\displaystyle \lim_{k \to \infty} \left( \sum_i x_i^k \right)^{1/k} = \max_i x_i\)

? – Julian Poon · 1 year, 9 months ago

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– Jake Lai · 1 year, 9 months ago

It just means \(\displaystyle \lim_{k \to \infty} \left( \sum_{i=1}^n x_i^k \right)^{1/k} = \max_{i=1,2,\ldots ,n} x_i = \max \{ x_1,x_2, \ldots ,x_n \}\).Log in to reply

– Julian Poon · 1 year, 9 months ago

No, as to why it is so. Also mind if you also define what \(\text{Id}\) means? ThanksLog in to reply

This is a clean, nice proof. Also I've added clarifications to the note. – Jake Lai · 1 year, 9 months ago

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