October 2018 1 16 Report

Witam, mam obiczyć monotoniczność ciągu a_n = \frac{2n+3}{n+1}

Pytanie to: czy dobrze "dochodzę" do rozwiązania?

a_{n-1} - a_n\\...\frac{2n^24n + 2 - 2n^2 - 7n -6} {n(n+3+\frac{2}{n})} = \\ \frac{-3n -4}{n^2+3n+2} = \frac{-3n}{n^2+3n}-2 = -\frac{3n}{n^2+3m} - 2 = \\ -1 n^2 -2 = -n^2 -2 == \\ |n|^2 - 2 < 0


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