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

sum_(n=0)^oo ((2n)!)/((3^n)(n!)^2

1 Answer
Apr 9, 2018

The series is divergent, because the limit of this ratio is >1

lim_(n->oo)a_(n+1)/a_n=lim_(n->oo)(4(n+1/2))/(3(n+1))=4/3>1

Explanation:

Let a_n be the n-th term of this series:
a_n=((2n)!)/(3^n(n!)^2)

Then
a_(n+1)=((2(n+1))!)/(3^(n+1)((n+1)!)^2)

=((2n+2)!)/(3*3^n((n+1)!)^2)

=((2n)!(2n+1)(2n+2))/(3*3^n(n!)^2(n+1)^2)

=((2n)!)/(3^n(n!)^2)*((2n+1)(2n+2))/(3(n+1)^2)

=a_n*((2n+1)2(n+1))/(3(n+1)^2)

a_(n+1)=a_n*(2(2n+1))/(3(n+1))

a_(n+1)/a_n=(4(n+1/2))/(3(n+1))

Taking limit of this ratio

lim_(n->oo)a_(n+1)/a_n=lim_(n->oo)(4(n+1/2))/(3(n+1))=4/3>1

So the series is divergent.