L(u) = u - \fracu^33, \quad a_n = \sum_k=1^n L\left(\frackn\right) - AMAZONAWS

📅 March 6, 2026 👤 scraface

Mar 06, 2026

$L(u) = u - \frac{u^3}{3}, \quad a_n = \sum_{k=1}^{n} L\left(\frac{k}{n}\right)$