How do you apply the ratio test to determine if #Sigma (3^n(n!)^2)/((2n)!)# from #n=[1,oo)# is convergent to divergent?

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Answer 1 · 2017-01-21T01:01:35

Prove that: #lim_(n->oo) abs (a_(n+1)/a_n) <1#

The ratio test states that a necessary condition for the series:

#sum_(n=1)^oo a_n#

to converge is that

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

If #L<1# the condition is also sufficient, while for #L=1# the test is inconclusive.

Calculate the expression of the ratio for this series:

#abs (a_(n+1)/a_n) = ( (3^(n+1) ((n+1)!)^2) / (2(n+1)!)) / ( (3^n (n!)^2)/ ((2n)!)) = 3^(n+1)/3^n ((n+1)!)^2/((n!)^2) ((2n)!)/ ((2(n+1))!)#

#abs (a_(n+1)/a_n) = 3 (n+1)^2/((2n+2)(2n+1))=3/2 (n+1)/(2n+1)#

Passing to the limit:

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

so the series is convergent.