# How do you determine whether the sequence a_n=(n!+2)/((n+1)!+1) converges, if so how do you find the limit?

Jun 18, 2017

The sequence ${a}_{n}$ converges to $0$. The series (i.e. sum) diverges.

#### Explanation:

Let:

a_n = (n!+2)/((n+1)!+1)

color(white)(a_n) = (n!+2)/((n+1)n!+1)

color(white)(a_n) = (1+2/(n!))/(n+1+1/(n!))

As $n \to \infty$:

2/(n!)->0

1/(n!)->0

So:

${a}_{n} \to \frac{1 + 0}{\infty + 1 + 0} = 0$

So the sequence ${a}_{1} , {a}_{2} , {a}_{3} , \ldots$ tends to $0$ as $n \to \infty$

a_n = (n!+2)/((n+1)!+1) > (n!)/((n+1)!+1) = 1/(n+1+1/(n!)) >= 1/(n+2)
${\sum}_{n = 1}^{\infty} {a}_{n} \ge {\sum}_{n = 1}^{\infty} \frac{1}{n + 2} = {\sum}_{n = 3}^{\infty} \frac{1}{n}$
which diverges, since the harmonic series ${\sum}_{n = 1}^{\infty} \frac{1}{n}$ diverges.