Use Ratio Test to find the convergence of the following series?

Question

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

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Answer 1 · 2018-04-01T14:30:38

The series:

#sum_(n=0)^oo (( n+3)!) / ( ( 3!)(n!)(3^n))#

is convergent.

Evaluate the ratio:

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

#a_(n+1)/a_n = ((n+4)!)/((n+3)!) (3!)/(3!) (n!)/((n+1)!) 3^n/3^(n+1)#

#a_(n+1)/a_n =1/3 (n+4)/(n+1)#

Hence:

#lim_(n->oo) a_(n+1)/a_n = 1/3#

and the series is convergent.