Question

Is the series converge or diverge?

`sum_(n=1)^oo ((n-1)!)/((n+2)!)`

Answer 1

It converges

Explanation:

Here, I'm going to use the definition of `n!` to simplify the question
`n! =1*2*3*4*...*n-2*n-1*n`

`(n-1)! =1*2*3*4*...*n-2*n-1`
So
`n! =n*(n-1)! =n*(n-1)*(n-2)!` and so on.

Now on to the series:

`\sum_{n=1}^\infty ((n-1)!)/((n+2)!)`
`=\sum_{n=1}^\infty ((n-1)!}/{n+2*n+1*n*(n-1)!`
`=\sum_{n=1}^\infty (1}/{n+2*n+1*n`
`=\sum_{n=1}^\infty (1}/{n^3+3n^2+2n`

From here, we can prove convergence a couple different ways, but here's one:

It has been proven that

`\sum_{n=1}^\infty (1}/{n^3` converges.

In fact, all `\sum_{n=1}^\infty (1}/{n^p` converge for `p>1` (see The p-series Test section of the link, about 3/4 of the way down the page).

So since `\sum_{n=1}^\infty (1}/{n^3+3n^2+2n] < \sum_{n=1}^\infty (1}/{n^3`,
(because the denominator will be larger for all n greater than or equal to 1)

Your series converges.