Is the series converge or diverge?

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

1 Answer
C. McCracken
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.

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