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Cálculo IV (2022)
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[Resolução] 10.4.1
Disciplina de Cálculo IV do ICMC
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juliafrare
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[Resolução] 10.4.1
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juliafrare
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13 Oct 2022 07:57
\(f(n) = \frac{n}{n+1} - \frac{n+1}{n} = \frac{n^2-(n+1)^2}{n^2+n} = \frac{2n+1}{n^2+n}
\)
\(\lim_{n\to \infty}\frac{2n+1}{n^2+n} = \lim_{n\to \infty}\frac{2+1/n}{n+1} = \lim_{n\to \infty}\frac{2}{n} = 0
\)
Logo, a sequência converge.
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