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Question Number 204174 by patrice last updated on 08/Feb/24
Answered by Faetmaaa last updated on 27/Feb/24
3)−∀n∈N★Sn=1n∑nk=1ln(kn)=1−0n∑nk=1ln(0+k1−0n)→n→∞∫01ln(x)dx=[x(lnx−1)]01=−1
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