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.

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