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.