Soit (Un)n ightarrow+infinity une suite numérique définie par Un = e^-n. Montrer que n ->+infinity U_n=0.

April 24, 2026Mathematics

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April 24, 2026

Welcome back — been a while! Let's pick up where you left off.

Exercice 1: 1. Convergence de $(U_n)_{n \in \mathbb{N}}$ vers 0: La suite $(U_n)_{n \in \mathbb{N}}$ est définie par $U_n = e^{-n}$ . Pour montrer que $\lim_{n \to \infty} U_n = 0$ , nous devons prouver que pour tout $\epsilon > 0$ , il existe un entier $N$ tel que pour tout $n \ge N$ , $|U_n - 0| < \epsilon$ . $|e^{-n} - 0| < \epsilon$ $e^{-n} < \epsilon$

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Welcome back — been a while! Let's pick up where you left off. Exercice 1: 1.

Asked on April 24, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

ScanSolve AI AnswerAnswered on April 24, 2026

Welcome back — been a while! Let's pick up where you left off. Exercice 1: 1. Convergence de (U_n)_n N vers 0: La suite (U_n)_n N est définie par U_n = e^-n. Pour montrer que _n U_n = 0, nous devons prouver que pour tout > 0, il existe un entier N tel que pour tout n N, |U_n - 0| < . |e^-n - 0| < e^-n <