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

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Answer 1 · 2017-06-20T10:11:28

The series:

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

is convergent.

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.