How do you use the limit comparison test to determine if #Sigma 2/(3^n-5)# from #[1,oo)# is convergent or divergent?

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Answer 1 · 2017-07-20T17:41:11

Use the geometric series of ratio #1/3#:

#sum_(n=0)^oo (1/3)^n = sum_(n=0)^oo 1/3^n = 1/(1-1/3) = 3/2#

as comparison:

#lim_(n->oo) (1/3^n)/(2/(3^n-5)) = 1/2 lim_(n->oo) (3^n-5)/3^n = 1/2lim_(n->oo) 1-5/3^n = 1/2#

As the limit is finite the two series have the same character so also:

#sum_(n=0)^oo 2/(3^n-5)#

is convergent.