Is the series converge or diverge?

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

1 Answer
Apr 23, 2017

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.