How do you test the series #Sigma (3n^2+1)/(2n^4-1)# from n is #[1,oo)# for convergence?

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Answer 1 · 2017-02-10T01:44:00

#sum_(n=1)^oo(3n^2+1)/(2n^4-1)#

is convergent based on the direct comparison test.

We can test the series by direct comparison. As:

#sum_(n=1)^oo a_n = sum_(n=1)^oo(3n^2+1)/(2n^4-1)#

is a series with positive terms, we need to find a convergent series #sum_(n=1)^oo b_n# such that:

#a_n < b_n# for #n>N#

Now we have:

#(3n^2+1)/(2n^4-1) < (3n^2)/(2n^4-1) < (3n^2)/(2n^4) < 1/n^2#

But #sum_(n=1)^oo 1/n^2 # is convergent based on the p-series test, so also:

#sum_(n=1)^oo(3n^2+1)/(2n^4-1)#

is convergent