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

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Answer 1 · 2018-05-14T19:56:21

The series:

$sum_(n=0)^oo (n+3)/(n(n+2))$

is divergent.

Note that:

$(n+3)/(n+2) = (n+2+1)/(n+2) = 1+1/(n+2) > 1$

Then:

$(n+3)/(n(n+2)) > 1/n$

$sum_(n=0)^oo 1/n$

is divergent, by direct comparison also the series:

$sum_(n=0)^oo (n+3)/(n(n+2))$

is divergent.

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