algèbre lycée recueilchapitre 1problème 1.1récurrence

Two equal sums

Recueil COMIMa — Techniques de résolution de problèmes · 2026 · Madagascar · ★★★★★

Statement

Montrez que pour tout $n \in \mathbb{N}$, nous avons $f(n) = g(n)$, où \[ f(n) = 1 - \frac{1}{2} + \frac{1}{3} - \cdots + \frac{1}{2n+1} - \frac{1}{2n}, \quad g(n) = \frac{1}{n+1} + \cdots + \frac{1}{2n}. \]

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Origin : Recueil COMIMa — Techniques de résolution de problèmes · 2026 · Madagascar

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