Question

Considere , uma sequência (xn), n N de números reais e positivos tal que xn+1= xn/n+1. Neste caso lim n-+=xn converge para qual valor e por que:

Answer 1

Asumiré que:
$x_{n+1}=\dfrac{x_n}{n+1}\\ \\ x_2=\dfrac{x_1}{2}\\ \\ x_3=\dfrac{x_2}{3}=\dfrac{x_1}{2\cdot 3}\\ \\ x_4=\dfrac{x_3}{4}=\dfrac{x_1}{2\cdot 3\cdot 4}\\ \\ \boxed{x_n=\dfrac{x_1}{n!}}$

$\lim\limits_{n\to +\infty}x_n=\lim\limits_{n\to +\infty}\dfrac{x_1}{n!}\\ \\ \lim\limits_{n\to +\infty}x_n=x_1 \lim\limits_{n\to +\infty}\dfrac{1}{n!}\\ \\ \lim\limits_{n\to +\infty}x_n=x_1 \cdot 0\\ \\ \boxed{\lim\limits_{n\to +\infty}x_n=0}$

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$\lim\limits_{n\to +\infty}\dfrac{1}{n!}\\ \\ \text{Principio de Arqu\'imedes:}\\ \\ \forall \varepsilon \in \mathbb R,\exists n \in \mathbb N: n\ \textgreater \ \dfrac{1}{\varepsilon}\\ \\ \text{Por ende: }\\ \\ n! \ \textgreater \ \dfrac{1}{\varepsilon}\\ \\ \boxed{\dfrac{1}{n!}\ \textless \ \varepsilon }$

Supongamos que $\exists n_0 \in \mathbb N \,|\, n\ \textgreater \ n_0 \;\&\; \forall \varepsilon \ \textgreater \ 0 \text{ se cumpla}:\;\dfrac{1}{n!}\ \textless \ \varepsilon$

Esto precisamente significa que 1/n! converge a 0