Calculus of a Single Variable, Chapter 9, 9.3, Section 9.3, Problem 71

To evaluate the series sum_(n=1)^oo 1/(3n-2) , we may apply Direct Comparison test.
Direct Comparison test is applicable when sum a_n and sum b_n are both positive series for all n where a_n lt=b_n .
If sum b_n converges then sum a_n converges.
If sum a_n diverges so does the sum b_n diverges.
Let b_n=1/(3n-2) and a_n =1/(3n) since
It follows that a_n < b_n
Apply sum c* a_n = c sum a_n , we get:
sum_(n=1)^oo 1/(3n) =1/3sum_(n=1)^oo 1/(n)
Apply he p-series test: sum_(n=1)^oo 1/n^p is convergent if pgt1 and divergent if plt=1 .
For the sum_(n=1)^oo 1/n or sum_(n=1)^oo 1/n^1 , we have the corresponding value p=1 . It satisfies the condition plt=1 then the series sum_(n=1)^oo 1/n diverges. Therefore, the sum_(n=1)^oo 1/(3n) is a divergent series.
When the sum a_n = sum_(n=1)^oo 1/(3n) is a divergent series then sum b_n= sum_(n=1)^oo 1/(3n-2) is also a divergent series.