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Question Number 197349 by SANOGO last updated on 14/Sep/23
calcul∑+oon=1(−1)n2n+1n(n+1)
Answered by MM42 last updated on 14/Sep/23
2n+1n(n+1)=1n+1n+1sn=−(11+12)+(12+13)−(13+14)−...sn=−1±1n+1⇒s=limn→∞sn=−1
Commented by SANOGO last updated on 14/Sep/23
merci
Commented by MathematicalUser2357 last updated on 16/Sep/23
ronaldo
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