Calcule underset(n -> + oo)(l i m) (a_n + 1)/(a_n) e underset(n -> + oo)(l i m) root(n)(a_n)

Letra a)

a_n = (n !)/(n^n)

Passo 1

Quando a_n = (n !)/(n^n) , o limite underset(n -> + oo)(l i m) (a_n + 1)/(a_n) será.

underset(n -> + oo)(l i m) (((n + 1) !)/((n + 1)^(n + 1)))/((n !)/(n^n)) = underset(n -> + oo)(l i m) ((n + 1) !)/((n + 1)^(n + 1)) * (n^n)/(n !)

Abrindo o termo fatorial

underset(n -> + oo)(l i m) ((n + 1) * n !)/((n + 1)^(n + 1)) * (n^n)/(n !) -> underset(n -> + oo)(l i m) ((n + 1) * n^n)/((n + 1)^(n + 1))

Fornece-nos ainda

underset(n -> + oo)(l i m) ((n)/(n + 1))^n = underset(n -> + oo)(l i m) ((n + 1)/(n))^(- n) = underset(n -> + oo)(l i m) [(1 + (1)/(n))^(- n)]^(- 1)

A ultima igualdade é igual a

underset(n -> + oo)(l i m) [(1 + (1)/(n))^(- n)]^(- 1) = e^(- 1) = (1)/(e)

Passo 2

No exercício 6 vimos que

underset(n -> + oo)(l i m) root(n)(a_n) = (1)/(e)

Então o limite quando a_n = (n !)/(n^n) é

underset(n -> + oo)(l i m) root(n)((n !)/(n^n)) = (1)/(e)

Resposta

underset(n -> + oo)(l i m) (a_n + 1)/(a_n) = (1)/(e)

underset(n -> + oo)(l i m) root(n)(a_n) = (1)/(e)

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