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

Question

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

1078 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-09T18:15:31

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#

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.