How do you apply the ratio test to determine if #Sigma (4^n(n!)^2)/((2n)!)# from #n=[1,oo)# is convergent to divergent?

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Answer 1 · 2017-02-20T01:50:30

The ratio test is indeterminate for this series.

For the series:

#sum_(n=1)^oo a_n = sum_(n=1)^oo (4^n(n!)^2)/((2n)!)#

Evaluate the ratio:

#abs(a_(n+1)/a_n) = abs ( ( (4^(n+1)((n+1)!)^2)/((2(n+1))!) ) / ((4^n(n!)^2)/((2n)!)))#

#abs(a_(n+1)/a_n) = 4^(n+1)/4^n ((n+1)!)^2/ ((n!)^2) ((2n)!)/ ((2n+2)!) #

#abs(a_(n+1)/a_n) = (4 (n+1)^2) / ((2n+2)(2n+1)) #

#abs(a_(n+1)/a_n) = (4n^2+8n+4) / (4n^2+6n+2) #

So we have:

#lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo)(4n^2+8n+4) / (4n^2+6n+2) = 1 #

which means the ratio test is indecisive and we cannot determine whether the series is convergent or not.