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

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Answer 1 · 2017-01-27T13:30:55

The series:

#sum_(n=1)^oo (2n^2-1)/(3n^5+2n+1)#

is convergent.

Given:

#a_n = (2n^2-1)/(3n^5+2n+1)#

we have to find a convergent series #sum_(n=1)^oo b_n# such that #a_n < b_n#

We can now consider that:

#2n^2-1 < 2n^2#

and

# (3n^5+2n+1) > 3n^5#

so that:

#(2n^2-1)/(3n^5+2n+1) < (2n^2)/(3n^5)#

#(2n^2-1)/(3n^5+2n+1) < 1/n^3#

The series #sum_(n=1)^oo 1/n^3# is known to be convergent based on the p-series test, therefore also the series:

#sum_(n=1)^oo (2n^2-1)/(3n^5+2n+1)#

is convergent.