How do you apply the ratio test to determine if #Sigma (n!)^3/((3n)!)# from #n=[1,oo)# is convergent to divergent?

1 Answer
Jun 20, 2017

The series:

#sum_(n=1)^oo (n!)^3/((3n)!)#

is convergent.

Explanation:

Evaluate the ratio:

#abs (a_(n+1)/a_n) = abs ( ( ((n+1)!)^3/((3(n+1))!) ) / ( (n!)^3/((3n)!) ) ) #

#abs (a_(n+1)/a_n) = ( ((n+1)!)^3/(n!)^3) ( ((3n)!)/ ( (3n+3)!) ) #

#abs (a_(n+1)/a_n) = (((n+1)!)/(n!))^3 ( ((3n)!)/ ( (3n+3)!) ) #

#abs (a_(n+1)/a_n) = (n+1)^3/((3n+3)(3n+2)(3n+1))#

So:

#lim_(n->oo) abs (a_(n+1)/a_n) = 1/27 < 1#

which means the series is convergent.