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

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Answer 1 · 2017-03-09T01:23:15

The series:

#sum_(n=1)^oo 1/sqrt(n!)#

is convergent.

Evaluate the ratio:

#abs(a_(n+1)/a_n) = abs ((1/sqrt((n+1)!))/(1/sqrt(n!))) = sqrt((n!)/((n+1)!)) = 1/sqrt(n+1)#

We have that:

#lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) 1/sqrt(n+1) = 0#

so the series is convergent.