How do you use the ratio test to test the convergence of the series #∑ 3^n/(4n³+5)# from n=1 to infinity?

1 Answer
Oct 18, 2015

Answer:

The series is divergent, see the explanation.

Explanation:

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

#L=lim_(n->oo) |(3^(n+1)/(4(n+1)^3+5))/(3^n/(4n^3+5))| = lim_(n->oo) |((3^n*3)/(4(n+1)^3+5))/(3^n/(4n^3+5))|#

#L=lim_(n->oo) (3(4n^3+5))/(4(n+1)^3+5) = lim_(n->oo) (12n^3+15)/(4(n^3+3n^2+3n+1)+5)#

#L=lim_(n->oo) (12n^3+15)/(4n^3+12n^2+12n+9) = lim_(n->oo) (12+15/n^3)/(4+12/n+12/n^2+9/n^3)#

#L=12/4=3 > 1 =># the series is divergent