How do you use the direct comparison test to determine if #Sigma 1/(3^n+1)# from #[0,oo)# is convergent or divergent?

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Answer 1 · 2017-12-12T14:02:46

The series is convergent

Let #a_n=1/(3^n+1)#

We know that

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

#1/(3^n+1)<1/3^n#

Let #b_n=1/3^n#

#sum_(n=1)^oob_n=3/2# Geometric series with the common ratio #r=1/3#

As the series #sum_(n=1)^oob_n# converges, we can conclude by the comparison test that the series #sum_(n=1)^ooa_n# converges